Meta Python Developer Interview Questions & Answers

I use two pointers, one for each string, and a list to build the result. While either string still has characters, I append the next character from word1 when available, then the next character from word2 when available. This keeps the required alternating order. When one string ends, the loop continues with the remaining characters of the other string. The time complexity is O(m + n), and the auxiliary space complexity is O(m + n).

The problem gives us two strings and asks us to create one new string. We take one character from word1, then one from word2, and repeat. If one string ends first, we append the remaining characters from the longer string. Two pointers work well because each pointer tracks the next unused character in one input string.

1. What input sizes, value ranges, and edge cases should the solution handle?
2. What output should be returned for empty, invalid, or duplicate input?
3. Should I prioritize execution time or memory use, and may I use the standard library?

The inputs are two strings named word1 and word2. The output is one new merged string.

In the diagram, word1 is "abcde" and word2 is "pqr". The expected output is "apbqcrde".

The characters must keep their original order inside each input string. We only alternate the string from which we take the next character.

I use two integer pointers named i and j. Pointer i tracks the next unused character in word1. Pointer j tracks the next unused character in word2.

I also use a list named result. Appending characters to a Python list is efficient. After processing both strings, I join the list into one final string.

The main invariant is that result contains every character before index i in word1 and every character before index j in word2 in the required alternating order.

Set i = 0 and j = 0. Both pointers start at the first character of their strings.

Create result = []. It is empty because no characters have been processed yet.

The loop continues while i is inside word1 or j is inside word2. The or condition is important because processing must continue after one string ends.

Start with word1 = "abcde", word2 = "pqr", i = 0, j = 0, and result = [].

Step 1: i is valid. Append word1[0], which is "a". The result becomes ["a"]. Increase i to 1.

Step 2: j is valid. Append word2[0], which is "p". The result becomes ["a", "p"]. Increase j to 1.

Step 3: append word1[1], which is "b". The result becomes ["a", "p", "b"]. Increase i to 2.

Step 4: append word2[1], which is "q". The result becomes ["a", "p", "b", "q"]. Increase j to 2.

Step 5: append word1[2], which is "c". The result becomes ["a", "p", "b", "q", "c"]. Increase i to 3.

Step 6: append word2[2], which is "r". The result becomes ["a", "p", "b", "q", "c", "r"]. Increase j to 3.

Now word2 has ended, but word1 still has two characters.

Step 7: append word1[3], which is "d". The result becomes ["a", "p", "b", "q", "c", "r", "d"]. Increase i to 4.

Step 8: append word1[4], which is "e". The result becomes ["a", "p", "b", "q", "c", "r", "d", "e"]. Increase i to 5.

Both strings are now finished. Joining the list returns "apbqcrde".

During each loop iteration, the algorithm appends the next unused character from word1 when one exists. It then appends the next unused character from word2 when one exists.

This preserves the order of characters inside both strings. It also creates the required alternating order while both strings still have characters.

When one string ends, its condition becomes false. The condition for the other string can still be true, so its remaining characters are appended in order.

Therefore, every input character is added exactly once, and the returned string is correct.

The function initializes i, j, and result. The while condition uses or so the loop continues until both strings are fully processed.

The first if statement checks whether word1 still has an unused character. If it does, the code appends word1[i] and increments i.

The second if statement checks whether word2 still has an unused character. If it does, the code appends word2[j] and increments j.

Finally, "".join(result) combines the collected characters and returns the merged string.

Let m be the length of word1 and n be the length of word2. Each input character is processed once, so the time complexity is O(m + n).

The result list can hold m + n characters, so the auxiliary space complexity is O(m + n).

The same code handles an empty string, two empty strings, strings with very different lengths, and strings that each contain one character.

Use one pointer for each string and one list for the merged output. During each loop iteration, append word1[i] if i is valid, then append word2[j] if j is valid. Increment only the pointer whose character was appended. The central invariant is that result contains all characters before i in word1 and all characters before j in word2 in the correct alternating order. The loop uses or, so it continues until both strings have been completely processed.

Let m be the length of word1 and n be the length of word2. Each character from both strings is appended exactly once, so the time complexity is O(m + n). The result list grows to contain all m + n characters before they are joined, so the auxiliary space complexity is O(m + n).

A common mistake is using and instead of or in the while condition. That stops the loop when the shorter string ends and loses the remainder of the longer string. Another mistake is appending word2[i] inside the word1 branch or incrementing the wrong pointer. Candidates may also forget the boundary checks and cause an IndexError. Another mistake is changing the original order of characters. It is also incorrect to claim O(1) auxiliary space because the result list grows with the total input size.

Before writing the loop, explain that i and j always point to the next unused characters and that result already contains all earlier characters in the correct order.

I would use linear search because the list is unsorted. I start at index 0 and compare each value with the target. If they match, I return the current index immediately. This returns the first occurrence when duplicates exist. If the loop finishes without a match, I return -1. I process each element at most once and stop when the answer is found. The worst-case time is O(n), and the auxiliary space is O(1).

The problem asks us to find the first position of a target value in an unsorted list. If the target is missing, we return -1. Because the values are not ordered, binary search cannot safely remove half of the search area. Linear search fits the problem because it checks the values from left to right and can stop as soon as it finds a match.

1. What input sizes, value ranges, and edge cases should the solution handle?
2. What output should be returned for empty, invalid, or duplicate input?
3. Should I prioritize execution time or memory use, and may I use the standard library?

The input contains a list of integers called nums and an integer called target. The output is the index of the first target occurrence. An index is the position of a value in the list. If the target does not occur, the function returns -1.

In the diagram, nums = [4, 1, 7, 3, 9] and target = 3. The expected output is 3 because nums[3] = 3.

The list is unsorted. This means there is no ordering that supports binary search. The direct method is linear search. We read the values from left to right and compare each one with the target.

The central invariant is: before checking index i, every earlier index from 0 through i - 1 has already been checked, and none of those earlier values equals the target.

Traversal begins at index 0, which is the leftmost position. The target is 3. At the start, no indices have been checked.

The Python code uses enumerate(nums). This provides the current index and the current value during each loop step.

At index 0, the current value is 4. We check 4 == 3. The condition is false, so we continue. The checked indices become [0]. The target has not been found yet.

At index 1, the current value is 1. We check 1 == 3. The condition is false, so we continue. The checked indices become [0, 1]. The target has not been found yet.

At index 2, the current value is 7. We check 7 == 3. The condition is false, so we continue. The checked indices become [0, 1, 2]. The target has not been found yet.

At index 3, the current value is 3. We check 3 == 3. The condition is true, so the function returns 3 and stops. The checked indices are [0, 1, 2, 3].

Index 4 is not processed because the algorithm stops immediately after finding the target. The algorithm processes 4 of the 5 elements.

All positions before index 3 were checked and did not contain the target. At index 3, the value equals the target. Therefore, index 3 is the correct returned position. Because the search moves from left to right and stops at the first match, it also returns the first occurrence when duplicates exist.

If the loop finishes without returning, every element has been checked and none equals the target. Returning -1 is then correct.

The function accepts nums and target. The for loop uses enumerate to get each index and value. The condition value == target checks whether the current value is the answer. When the condition is true, the function returns the index immediately. If the loop ends without a match, the function returns -1.

The worst-case time complexity is O(n) because the target may be absent or may appear at the last position. The best-case time is O(1) when the first element matches. The auxiliary space complexity is O(1) because the algorithm uses only a few variables and does not create another collection that grows with the input.

Relevant edge cases include an empty list, a one-element list, the target at the first position, the target being absent, duplicate target values, negative values, and zero.

The key insight is that an unsorted list gives us no safe way to skip values. We therefore scan from left to right with linear search. At each index, we compare the current value with the target. We return immediately when they match. The invariant is that before checking index i, all earlier indices have already been checked and none contains the target. This proves that the first match is the first occurrence. If no match is found after the loop, the target is absent.

Let n be the number of elements in nums. In the worst case, the algorithm checks all n elements, so the time complexity is O(n). In the best case, the first element matches, so the time is O(1). We process the input at most once and may stop early. The auxiliary space is O(1) because the algorithm stores only the current index and value. It does not build a list, set, or dictionary.

Common mistakes include returning the matching value instead of its index, continuing after a match instead of returning immediately, returning the last duplicate instead of the first occurrence, forgetting to return -1 when the target is absent, trying to use binary search on an unsorted list, and claiming that the algorithm always processes every element even though it may stop early.

Say the invariant before writing the loop: every earlier index has already been checked and does not contain the target. Then place the return directly inside the matching condition to show the early stop.

I first handle the empty-list case by returning 0.0. For a non-empty list, I keep a running total and add each book price to it. After processing every price, I divide the total by the number of prices. For [10.0, 20.5, 15.5, 30.0], the total is 76.0 and the average is 19.0. This is correct because an arithmetic mean is the sum divided by the count. The time complexity is O(n), and the auxiliary space complexity is O(1).

The problem gives a list of book prices and asks for their arithmetic mean. Arithmetic mean means the total of all prices divided by the number of prices. A running total is enough, so we can solve the problem with one loop and constant extra memory. We must check for an empty list first because dividing by zero is invalid.

1. What input sizes, value ranges, and edge cases should the solution handle?
2. What output should be returned for empty, invalid, or duplicate input?
3. Should I prioritize execution time or memory use, and may I use the standard library?

The input is a list of book prices. The function returns one floating-point value.

For a non-empty list, the returned value is:

sum of all prices / number of prices

For an empty list, the required result is 0.0.

The diagram uses this example:

prices = [10.0, 20.5, 15.5, 30.0]

The expected result is 19.0.

First, check whether prices is empty. If it is empty, return 0.0 immediately.

For a non-empty list, initialize total to 0.0. The variable total stores the sum of all prices processed so far.

The central invariant is simple: after each loop iteration, total equals the sum of every price already processed.

Start with:

total = 0.0

Process the values from left to right.

Add 10.0. The total changes from 0.0 to 10.0.

Add 20.5. The total changes from 10.0 to 30.5.

Add 15.5. The total changes from 30.5 to 46.0.

Add 30.0. The total changes from 46.0 to 76.0.

After the loop, all four prices have been processed.

The final total is 76.0. The number of prices is 4.

average = total / count

average = 76.0 / 4

average = 19.0

The function returns 19.0.

At the start, total is 0.0, which is the sum of zero processed prices.

Each loop iteration adds the current price to total. Therefore, the invariant remains true after every iteration.

When the loop finishes, total is the sum of the full list. Dividing that sum by the list length gives the arithmetic mean by definition.

The condition if not prices handles the empty-list case and prevents division by zero.

The variable total begins at 0.0. The for loop adds each price to it. After the loop, len(prices) gives the count. The function returns total / len(prices).

Let n be the number of prices. The loop processes all n prices once, so the time complexity is O(n).

The function uses only the running total and normal local variables. Its auxiliary space complexity is O(1).

For an empty list, it returns 0.0. For a list containing one value, the average is that value.

Use a running sum. Begin with total = 0.0, then add each price to it. The invariant is that total always equals the sum of all prices processed so far. When the loop ends, total is the sum of the complete list. Dividing it by len(prices) gives the arithmetic mean. This approach directly follows the definition of average and does not need an additional data structure.

Let n be the number of book prices. The loop visits each of the n prices once, so the time complexity is O(n). The function keeps only a running total and a few local values. This extra memory does not grow when the input grows, so the auxiliary space complexity is O(1).

A common mistake is forgetting the empty-list check, which can cause division by zero. Another mistake is dividing during each loop iteration instead of waiting until the complete total is known. Candidates may also divide by the wrong count, return the sum instead of the average, or claim O(1) time even though every input price must be processed.

Explain the invariant before coding: after each iteration, total is the sum of all prices processed so far. This makes the loop and the correctness argument easy to follow.

I use a stack to build the canonical path. I split the absolute path by "/" and process each token from left to right. I ignore empty tokens and ".". For "..", I pop one directory only when the stack is not empty. Every normal directory name is pushed onto the stack. Finally, I join the stack with "/" and add one leading slash. This takes O(n) time and uses O(n) auxiliary space.

The problem asks us to simplify an absolute Unix-style path. We must remove repeated separators, ignore ".", and process ".." as moving to the parent directory. A stack fits this problem because it stores the valid path from the root to the current location.

1. What input sizes, value ranges, and edge cases should the solution handle?
2. What output should be returned for empty, invalid, or duplicate input?
3. Should I prioritize execution time or memory use, and may I use the standard library?

The input is an absolute Unix-style path. It starts at the root directory.

We must return its canonical form. The result has one leading slash. It has no repeated separators, no trailing slash unless the result is the root, and no special "." or ".." components.

For the example input "/home//foo/../bar/./baz/", the expected output is "/home/bar/baz".

The stack stores the directory names in the current canonical path.

Each stack entry is one real directory name. The bottom entry is closest to the root. The top entry is the current directory.

The central invariant is: after each token is processed, the stack contains the canonical directory names for the processed part of the path.

Start with an empty stack.

Split the path by "/". The example produces these tokens: "", "home", "", "foo", "..", "bar", ".", "baz", and "".

Process the tokens from left to right.

Ignore an empty token because it comes from a repeated, leading, or trailing separator.

Ignore "." because it means stay in the current directory.

For "..", pop one directory only when the stack is not empty. If the stack is already empty, remain at the root.

Push every other token because it is a normal directory name.

The stack starts as [].

Step 0 processes an empty token. Ignore it. The stack remains [].

Step 1 processes "home". Push it. The stack changes from [] to ["home"].

Step 2 processes another empty token. Ignore it. The stack remains ["home"].

Step 3 processes "foo". Push it. The stack changes from ["home"] to ["home", "foo"].

Step 4 processes "..". Pop the most recent directory, "foo". The stack changes from ["home", "foo"] to ["home"].

Step 5 processes "bar". Push it. The stack changes from ["home"] to ["home", "bar"].

Step 6 processes ".". Ignore it because the current directory does not change. The stack remains ["home", "bar"].

Step 7 processes "baz". Push it. The stack changes from ["home", "bar"] to ["home", "bar", "baz"].

Step 8 processes the final empty token. Ignore it. The final stack is ["home", "bar", "baz"].

The stack always represents the canonical path for the part already processed.

Empty tokens and "." do not change the current location, so ignoring them is correct.

A ".." removes exactly one previous directory when possible. It never moves above the root because the code pops only when the stack is not empty.

A normal directory name adds one valid path component.

After every token has been processed, joining the stack with "/" and adding one leading slash gives the canonical path "/home/bar/baz".

The code calls path.split("/") to create the tokens.

It skips empty tokens and ".".

For "..", it pops only when the stack contains a directory.

For every other token, it appends the directory name.

At the end, it returns "/" plus the stack joined by "/". When the stack is empty, this expression correctly returns "/".

The code processes each token once. Splitting the input and joining the final stack are also linear in the path length. The total time is O(n), where n is the number of characters in the input path.

The stack may hold directory names whose total size grows with the input, so the auxiliary space is O(n).

Important edge cases include the root path "/", repeated separators, attempts to move above the root such as "/../../", paths ending with a slash, and names such as "...". Only "." and ".." are special. A name like "..." is a normal directory name.

The key idea is to keep only the directory names that are still part of the current canonical path. A stack is a natural fit because ".." removes the most recently added directory. Empty tokens and "." are ignored. A normal directory is pushed. A ".." token pops only when the stack is not empty. The invariant is that after each token, the stack contains the canonical directory sequence for the processed prefix of the path. Joining the stack with "/" and adding one leading slash produces the final answer.

Let n be the number of characters in the input path. Splitting the path, processing its tokens, and joining the final directory names take O(n) time in total. The stack may store directory names whose total length grows with the input, so the auxiliary space is O(n).

Common mistakes include treating "..." as a special token even though only "." and ".." are special. Another mistake is popping from an empty stack when the path tries to move above the root. Candidates may forget to ignore empty tokens created by repeated separators. They may also return a trailing slash or forget the required leading slash. A final mistake is claiming constant auxiliary space even though the stack can grow with the input.

State the stack invariant before coding: after each token, the stack contains the canonical directory names for the processed part of the path. Then each ignore, push, and pop operation is easy to explain.

I would use iterative quickselect. First, I convert the kth largest position into the ascending index len(nums) - k. Then I partition the current inclusive range using its last value as the pivot. The pivot moves to its final sorted index. I compare that index with the target and keep only the side that can still contain the answer. The average time is O(n), the worst case is O(n²), and the auxiliary space is O(1).

The problem asks us to return the kth largest value in an integer array. We do not need to fully sort the array. Quickselect searches for one final sorted position. For an array of length n, the kth largest value belongs at ascending index n - k. Each partition places one pivot at its final sorted index and lets us discard the side that cannot contain the answer.

1. What input sizes, value ranges, and edge cases should the solution handle?
2. What output should be returned for empty, invalid, or duplicate input?
3. Should I prioritize execution time or memory use, and may I use the standard library?

The input is an integer array nums and an integer k. The output is the kth largest value, not an index.

For nums = [3, 2, 1, 5, 6, 4] and k = 2, the length is 6. The target index in ascending order is:

target_index = len(nums) - k = 6 - 2 = 4

If the array were fully sorted as [1, 2, 3, 4, 5, 6], index 4 would contain 5. Therefore, 5 is the second largest value.

I set left = 0 and right = 5. These are inclusive boundaries, so the current search interval is [0, 5].

The main invariant is that the target index always stays inside the current interval [left, right].

The partition function uses nums[right] as the pivot. It moves values less than or equal to the pivot to the left side. Values greater than the pivot stay on the right side. It then puts the pivot into its final sorted position and returns that position as pivot_index.

The first search interval is [0, 5]. The pivot is nums[5], which is 4.

Before partitioning, nums is [3, 2, 1, 5, 6, 4]. After partitioning, nums becomes [3, 2, 1, 4, 6, 5]. The pivot 4 is now at index 3.

The target index is 4. Since pivot_index 3 is smaller than target_index 4, the answer must be on the right side. I update left to pivot_index + 1, which is 4. The new interval is [4, 5].

The current interval contains [6, 5]. The pivot is nums[5], which is 5.

After partitioning this interval, nums becomes [3, 2, 1, 4, 5, 6]. The pivot 5 is now at index 4.

Now pivot_index equals target_index. The algorithm stops and returns nums[4], which is 5. Only two partition rounds were executed.

Every partition places one pivot at the same index it would have in a fully sorted array.

If pivot_index is smaller than target_index, the target cannot be at the pivot or to its left. It must be on the right. If pivot_index is larger than target_index, the target must be on the left.

Therefore, each update removes only positions that cannot contain the answer. When pivot_index reaches target_index, nums[pivot_index] is the kth largest value.

The main function computes target_index and starts with the complete array interval. It repeatedly calls partition.

The partition function keeps a variable named store_index. This marks where the next value less than or equal to the pivot should be placed. The loop reads every value in the current interval except the pivot. When a value is less than or equal to the pivot, the code swaps it into store_index and moves store_index one position right.

After the loop, the pivot is swapped into store_index. The function returns store_index as pivot_index.

The main loop returns immediately when pivot_index equals target_index. Otherwise, it changes either left or right and repeats.

The average time is O(n). Quickselect normally removes a large part of the remaining search interval after each partition.

The worst-case time is O(n²). This can happen when the last value repeatedly creates very uneven partitions and removes only one position at a time.

The algorithm modifies nums in place and uses O(1) auxiliary space.

When k = 1, it returns the maximum value. When k = len(nums), it returns the minimum value. Duplicate values, negative values, and zero are handled correctly.

The key idea is to find one ranked position instead of sorting the complete array. The kth largest value belongs at ascending index len(nums) - k. Quickselect partitions the current inclusive interval around a pivot. After partitioning, the pivot is at its final sorted index. The central invariant is that the target index always remains inside [left, right]. If the pivot index is too small, only the right side can contain the target. If it is too large, only the left side can contain the target. The search stops when pivot_index equals target_index.

Let n be the number of values in nums. The average time is O(n) because quickselect usually reduces the remaining search interval after each partition. The worst-case time is O(n²). This happens when the chosen pivot repeatedly creates a very uneven split, so only one position is removed in each round. The implementation changes the array in place and uses only a few variables. Therefore, its auxiliary space is O(1).

A common mistake is using k directly as an ascending index instead of computing len(nums) - k. Another mistake is moving the wrong boundary after comparing pivot_index with target_index. Candidates may forget that both left and right are inclusive. In the partition function, using the wrong comparison can break duplicate handling. It is also incorrect to claim guaranteed O(n) time because the worst case is O(n²). Finally, this implementation changes the input array, which should be stated.

State the invariant before writing code: the target index always remains inside [left, right], and every completed partition places its pivot at its final sorted index.

I would store each vector as a dictionary from index to non-zero value. For the dot product, I iterate through the smaller dictionary and look up the same index in the other one. When both vectors contain that index, I multiply the values and add the product to the total. This avoids work on zero entries. Building the sparse maps takes O(n) time. The dot product takes O(min(k1, k2)) expected time, and storage is O(k1 + k2).

The problem asks us to store two vectors efficiently when most values are zero and then calculate their dot product. A dense scan checks every position, including many positions that contribute nothing. Instead, we store only non-zero values in dictionaries. Each dictionary maps an index to the value at that index. We then iterate through the smaller dictionary and check matching indices in the other one.

1. What input sizes, value ranges, and edge cases should the solution handle?
2. What output should be returned for empty, invalid, or duplicate input?
3. Should I prioritize execution time or memory use, and may I use the standard library?

The input is two vectors of the same length. The output is one number: their dot product.

For the example:

v1 = [1, 0, 0, 2, 3]

v2 = [0, 3, 0, 4, 5]

The expected output is 23.

A dot product multiplies values at equal indices and adds the products:

1 × 0 + 0 × 3 + 0 × 0 + 2 × 4 + 3 × 5 = 23.

For each vector, I build a dictionary that maps an index to its non-zero value.

The first vector becomes:

{0: 1, 3: 2, 4: 3}

The second vector becomes:

{1: 3, 3: 4, 4: 5}

If an index is missing from a dictionary, the original value at that position is zero.

I compare the number of stored entries in both dictionaries. I iterate through the smaller dictionary. This reduces the number of lookups when one vector contains fewer non-zero values.

In this example, both dictionaries contain three entries. The walkthrough iterates through the first vector's dictionary.

The invariant is: after each executed step, total equals the sum of the products for all processed stored indices that also appear in the other vector.

Start with total = 0.

At index 0, the first vector stores the value 1. The second dictionary has no entry at index 0. Its original value there is zero, so this position contributes 0. We skip it, and total stays 0.

At index 3, the first vector stores 2. The second vector stores 4 at the same index. We calculate 2 × 4 = 8 and update total from 0 to 8.

At index 4, the first vector stores 3. The second vector stores 5 at the same index. We calculate 3 × 5 = 15 and update total from 8 to 23.

No stored entries remain in the chosen dictionary, so we return 23.

Only values at equal indices contribute to a dot product. Removing zero entries does not change the sum because every product containing zero contributes nothing.

For each stored index in the smaller dictionary, the algorithm checks the same index in the other dictionary. If the index exists, it adds the exact product. If the index is missing, the other value is zero and nothing is added.

Therefore, the final total is exactly the dot product.

The constructor uses enumerate to read each index and value. It stores an entry only when the value is not zero.

The dotProduct method compares the two dictionary sizes and assigns them to smaller and larger. It initializes total to zero. It then loops through each stored index and value in smaller.

When the same index exists in larger, it multiplies the two values and adds the result to total. After all stored entries in smaller have been processed, the method returns total.

Building both sparse representations takes O(n) time overall when each equal-length vector has length n. Each original position is examined once per vector.

The dot product takes O(min(k1, k2)) expected time, where k1 and k2 are the numbers of non-zero entries. Python dictionary lookup is O(1) on average.

The two dictionaries store k1 + k2 entries, so their space use is O(k1 + k2).

Relevant edge cases include all-zero vectors, vectors with no overlapping non-zero indices, negative values, and vectors with very different non-zero counts.

The key insight is that zero values never change a dot-product sum, so they do not need to be stored. Each sparse vector uses a dictionary with index as the key and the non-zero value as the value. The dot product iterates through the smaller dictionary and looks up the same index in the other dictionary. The invariant is that after every processed entry, total equals the sum of the matching products seen so far. This avoids scanning every dense position when most entries are zero.

Let n be the length of each vector. Let k1 and k2 be the numbers of non-zero entries in the two vectors. Building the sparse dictionaries takes O(n) time overall because each vector position is checked once. The dot product takes O(min(k1, k2)) expected time because it loops through the smaller dictionary. A Python dictionary lookup is O(1) on average. The two dictionaries store k1 + k2 entries, so the space use is O(k1 + k2).

A common mistake is scanning every position in the dense vectors, which loses the sparsity benefit. Another mistake is storing zero values in the dictionary and wasting memory. Some candidates iterate through the larger sparse dictionary, which causes unnecessary lookups. Another error is matching equal values instead of matching equal indices. Candidates may also claim guaranteed O(1) dictionary operations, but Python dictionary lookup and insertion are O(1) on average.

First state exactly what the dictionary stores: index maps to non-zero value. Then explain that you iterate through the smaller dictionary and look up the same index in the other one. This makes both the correctness argument and the complexity easy to explain.

I would store every node on p’s path to the root in a set. Then I would move upward from q and check each node against that set. The first match is the lowest common ancestor because q is checked from its nearest ancestor to its farthest ancestor. For p = F and q = D, the first match is B. The expected time is O(h_p + h_q), and the auxiliary space is O(h_p).

The problem gives two node references, p and q. Each node contains only a parent reference. We must return the lowest node that is an ancestor of both nodes. The diagram uses an ancestor set. First, we record p and every node above p. Then we climb from q and return the first node that appears in that set.

1. What input sizes, value ranges, and edge cases should the solution handle?
2. What output should be returned for empty, invalid, or duplicate input?
3. Should I prioritize execution time or memory use, and may I use the standard library?

The inputs are two node references named p and q. Each node has a value and a parent reference.

The function returns a node reference, not only the node’s value.

If p or q is None, the function returns None. If p and q are the same node object, the function returns that node immediately. If the nodes are in different trees, the second traversal reaches None and the function returns None.

I use a Python set named ancestors.

The set stores node references from p’s path to the root. It does not store only node values because two different nodes could have the same value.

The central invariant is this: after the first traversal, ancestors contains exactly p and every ancestor of p.

I then climb from q. Because I check q’s path from the nearest node to the farthest node, the first node found in ancestors is the lowest common ancestor.

In the diagram, p = F and q = D.

The set starts empty:

ancestors = set()

The first traversal starts at F. At each step, I add the current node to the set and then move to its parent.

The parent relationships are:

B.parent = A C.parent = A D.parent = B E.parent = B F.parent = E G.parent = E

Step 1: The current p-side node is F. The set is empty. Add F. The set becomes {F}. Continue.

Step 2: Move to E. The set is {F}. Add E. The set becomes {F, E}. Continue.

Step 3: Move to B. The set is {F, E}. Add B. The set becomes {F, E, B}. Continue.

Step 4: Move to A. The set is {F, E, B}. Add A. The set becomes {F, E, B, A}. Continue until A.parent is None.

Now begin the second traversal at q = D.

Step 5: Check whether D is in {F, E, B, A}. It is not. Move to D.parent, which is B. Continue.

Step 6: Check whether B is in {F, E, B, A}. It is. Return B and stop immediately.

No node after B is processed because the answer has already been found.

The set contains the complete ancestor chain of F:

F -> E -> B -> A

The upward path from D is:

D -> B -> A

We inspect D’s path from nearest to farthest. D is not shared. B is the first shared node. Therefore, B is the lowest common ancestor.

A higher shared node such as A cannot be the answer because B is below A and is already an ancestor of both F and D.

The function first checks the invalid-input case. It then checks whether p and q are the same object.

Next, it creates an empty set. The first while loop starts at p and adds each node reference before moving to its parent.

The second while loop starts at q. It checks whether the current node is in the set. If it is, the function returns that node immediately. Otherwise, it moves to the parent.

If the second traversal reaches None, the nodes do not share a common ancestor in the same tree, so the function returns None.

Let h_p be the number of nodes from p to the root, including p. Let h_q be the number of nodes from q to the root, including q.

The expected time is O(h_p + h_q). Python set insertion and lookup are O(1) on average.

The auxiliary space is O(h_p) because the set stores p and its ancestors.

Important edge cases are a None input, identical node references, one node already being an ancestor of the other, and nodes that belong to different trees.

The key idea is to save p’s full ancestor chain in a set. Each set entry is a node reference. After the first traversal, the invariant is that the set contains exactly p and every node above p. We then move upward from q in nearest-to-farthest order. The first node found in the set must be the lowest common ancestor. The set is useful because each membership check takes O(1) time on average in Python.

Let h_p be the number of nodes on p’s path to the root, including p. Let h_q be the number of nodes on q’s path to the root, including q. We insert at most h_p node references into the set and check at most h_q node references from q. Python set insertion and lookup are O(1) on average, so the expected time is O(h_p + h_q). The set can hold h_p nodes, so the auxiliary space is O(h_p).

One mistake is storing node values instead of node references. Different nodes can have the same value. Another mistake is forgetting to add p itself to the set. A candidate may also continue climbing after finding B and incorrectly return A. Other common errors are assuming the tree is a binary search tree, forgetting the identical-input case, or describing Python set operations as guaranteed O(1) instead of O(1) on average.

State the invariant before writing code: after the first loop, the set contains exactly p and every ancestor of p. Then explain that checking q from nearest to farthest makes the first set match the lowest common ancestor.

I would first combine all workshop records into a dictionary that maps each year to its total number of classes. Then I examine only years whose previous year is missing, because each of those years starts one consecutive run. From each start, I move forward year by year, add the class counts, and keep the largest total. Each active year is included in one run expansion. The expected time is O(n), and the auxiliary space is O(u) for u unique years.

The problem gives workshop records as pairs of year and class count. We need the largest sum across a consecutive run of active years. A dictionary is a good fit because it combines records for the same year and lets us quickly check whether the previous or next year exists.

1. What input sizes, value ranges, and edge cases should the solution handle?
2. What output should be returned for empty, invalid, or duplicate input?
3. Should I prioritize execution time or memory use, and may I use the standard library?

The input is a list of pairs. Each pair contains a year and a number of classes.

For the example:

records = [(2018, 3), (2019, 5), (2021, 4), (2022, 6), (2023, 2), (2025, 7)]

The output is one number. It is the largest total classes across any consecutive run of active years.

The active runs are:

2018 to 2019: 3 + 5 = 8

2021 to 2023: 4 + 6 + 2 = 12

2025 to 2025: 7

The required result is 12.

I create a dictionary named classes_by_year.

Each key is a year.

Each value is the total number of classes for that year.

If the same year appears more than once, I add those counts together.

For the example, the dictionary becomes:

{2018: 3, 2019: 5, 2021: 4, 2022: 6, 2023: 2, 2025: 7}

I also initialize best_total to 0.

For each year, I check whether year - 1 exists in the dictionary.

If the previous year exists, the current year is already inside a run. I skip it.

For example, 2019 is skipped because 2018 exists.

The same rule skips 2022 because 2021 exists and skips 2023 because 2022 exists.

A year starts a new run only when its previous year is missing.

When I find a run start, I set current_year to that year and current_total to 0.

Then I move forward while current_year exists in the dictionary.

For the run starting at 2021:

At 2021, I add 4. The running total becomes 4.

At 2022, I add 6. The running total becomes 10.

At 2023, I add 2. The running total becomes 12.

The year 2024 is missing, so the run stops.

I compare 12 with best_total and keep the larger value.

The year 2018 has no active predecessor because 2017 is missing. It starts a run containing 2018 and 2019. The total is 8, so best_total becomes 8.

The year 2019 is skipped because 2018 exists.

The year 2021 has no active predecessor because 2020 is missing. It starts a run containing 2021, 2022, and 2023. The total is 12, so best_total becomes 12.

The years 2022 and 2023 are skipped because each has an active predecessor.

The year 2025 starts a run because 2024 is missing. Its run total is 7, so best_total stays 12.

The function returns 12.

Every consecutive active run has exactly one first year. That first year is the only year in the run whose previous year is missing.

The algorithm starts only from those first years. It then adds every year in that run exactly once.

Because it calculates every run total and keeps the largest one, the final answer is correct.

The first loop builds the year-to-total dictionary.

The second loop checks each unique year. It skips years that are not run starts.

The while loop expands one full consecutive run.

Python dictionary lookup and insertion are O(1) on average.

The expected time is O(n), where n is the number of input records. The auxiliary space is O(u), where u is the number of unique years.

The key insight is that every consecutive active run has exactly one start year. A start year is an active year whose previous year is not present. The dictionary stores year -> total classes for that year. This gives average O(1) checks for whether the previous or next year exists. The central invariant is that only a year with no active predecessor starts a forward scan. Therefore, every active year is counted exactly once as part of one run, and every run total is considered.

The expected time complexity is O(n), where n is the number of input records. Building the dictionary takes O(n). Python dictionary lookup and insertion are O(1) on average. The forward scans take O(u) total because each of the u unique active years belongs to one run expansion. Since u cannot be larger than n, the total expected time is O(n). The auxiliary space is O(u) for the dictionary.

A common mistake is starting a forward scan from every year. That repeats work for years inside the same run. Another mistake is forgetting to combine duplicate records for the same year. Some candidates skip the predecessor check and count the same run more than once. Another error is resetting best_total instead of keeping the maximum. Candidates may also claim guaranteed O(n) time even though Python dictionary operations are O(1) only on average.

State the invariant before writing code: only a year whose previous year is missing can start a run. This makes the skip condition, correctness proof, and expected O(n) analysis easy to explain.

At a high level, I would separate normal comment operations from realtime delivery. Clients sign in through the Auth Service, receive a JWT, and send HTTPS requests through the API Gateway. The gateway validates the JWT, enforces rate limits, and routes requests to the Live Comments API Service. That service creates, lists, edits, or soft-deletes comments in the Comment Store. After a successful write, it publishes an event to the Realtime Subscription Hub. Clients receive ordered events through SSE or WebSocket. This adds operational complexity, but it gives clear security, stable ordering, pagination, and scalable live delivery.

The API must support durable comment operations and immediate live updates. The difficult parts are authorization, stable ordering, pagination, rate control, and consistent error handling. I would explain the design by following the numbered request, response, storage, and subscription flows in the diagram.

1. Which clients and core use cases must the API support?
2. What authentication, authorization, and data-validation rules should I assume?
3. What scale, error handling, idempotency, and versioning requirements matter?

I would start by separating login from normal comment requests. A viewer, broadcaster, or moderator signs in or refreshes a session through the Auth Service. The Auth Service issues a JWT, which is a signed token that identifies the caller.

The client then includes that JWT as a Bearer token on every comment operation. The request uses HTTPS, so the traffic is encrypted while moving across the network.

The main request enters the API Gateway using HTTPS plus the Bearer JWT. The gateway validates the JWT and routes the request. It can obtain validation keys or use token introspection from the Auth Service when needed.

The gateway also enforces rate limits. The Rate Limiter tracks callers per user, live stream, or IP address. It limits writes, reads, and subscriptions. When a caller exceeds the allowed rate, the API returns 429 Too Many Requests.

The visible endpoints are:

• POST /live/{liveId}/comments to create a comment.
• GET /live/{liveId}/comments?cursor=...&limit=...&order=asc|desc to list comments.
• PATCH /comments/{commentId} to edit a comment.
• DELETE /comments/{commentId} to soft-delete a comment.
• GET /live/{liveId}/comments/stream for an SSE subscription.
• WebSocket /ws/live/{liveId}/comments for a live connection.

The gateway forwards the validated request to the Live Comments API Service. This service owns the business rules, authorization, ordering, pagination, and error handling.

Authentication proves who the caller is. Authorization decides what that caller may do. A comment owner may edit or delete their own comment. A moderator may edit or delete any comment allowed by the moderation rules.

The service handles common failures consistently. 401 means authentication is missing or invalid. 403 means the user is authenticated but not allowed. 404 means the requested resource was not found. 409 means the request conflicts with the current state. A limit breach returns 429, while unexpected server failures return a 5xx response.

Errors use a JSON body with error_code, message, and details.

The service reads and writes comment rows in the Comment Store. Each row contains comment_id, live_id, user_id, text, created_at, updated_at, deleted, and server_sequence.

A delete is a soft delete. The row remains stored, but the deleted field marks it as removed.

The server_sequence value is unique within one live stream. It provides stable ordering. The design can use created_at plus comment_id as a fallback stable sort.

The list endpoint uses cursor-based pagination. A cursor identifies the next position in the ordered result. The service returns comment rows plus next_cursor. This works better than page numbers when new comments arrive continuously.

After the service completes the operation, it returns rows, next_cursor, an acknowledgment, or an error to the API Gateway. The gateway then sends the JSON response back to the client.

This is the synchronous path. The request travels from the client to the gateway and service. The response returns from the service through the gateway to the client.

After a successful create, update, or delete, the Live Comments API Service publishes an event to the Realtime Subscription Hub. The supported event names are comment.created, comment.updated, and comment.deleted.

A client opens an SSE or WebSocket subscription for one live stream. The subscription hub manages those connections and fans out events to subscribed clients. The events are delivered in the order defined by the comment sequence.

The normal JSON response and the realtime event are separate flows. The client receives the operation result through the gateway. Other subscribed clients receive the change through the hub.

The benefit is clear responsibility. The gateway handles token checks and rate control. The comments service owns business rules. The store owns durable data. The hub owns realtime fan-out.

The downside is more services and more operational work. We accept this because a busy live stream needs stable ordering, controlled writes, efficient pagination, and many long-lived subscriptions.

The design separates work so each part has one clear job. The gateway checks JWTs, applies rate limits, and routes requests. The comments service handles validation, authorization, ordering, and pagination. The store keeps durable rows. The subscription hub sends live events. Cursor pagination stays stable when new comments arrive, while page numbers may shift. A per-stream sequence gives reliable order, but the service must assign that value carefully. SSE is simple for server-to-client updates. WebSocket supports a longer two-way connection, but it needs more connection management. Soft delete keeps history, but stored data continues growing. The benefit is safer scaling and clearer ownership. The downside is more components, monitoring, and failure points. We accept that cost because live comments create both heavy write traffic and many realtime connections.

At a high level, I would place a versioned REST gateway before four resource APIs: posts, follows, profiles, and feeds. The client first gets a JWT from the Auth Service. It then sends HTTPS requests with that token to the gateway. The gateway validates the token, enforces authorization, applies rate limits, adds a request ID, and routes the call. Each API uses the required data store. Feed requests read follow relationships and recent posts. Cursor pagination keeps changing feeds stable. The trade-off is extra service coordination for clearer ownership and safer request handling.

The goal is to support posting, following, profiles, and feeds through clear APIs. The main challenge is protecting each request while keeping the resource boundaries simple. I would explain the design by following the exact request and response paths in the diagram.

1. Which clients and core use cases must the API support?
2. What authentication, authorization, and data-validation rules should I assume?
3. What scale, error handling, idempotency, and versioning requirements matter?

I would begin with the mobile or web client. The client sends OAuth login credentials to the Auth Service. The Auth Service checks the login and returns an access token as a JWT.

A JWT is a signed token that identifies the caller. The client places it in the Authorization: Bearer JWT header. It then sends the API request over HTTPS.

The request enters the Instagram API Platform through the REST API Gateway and Router. The gateway is the main boundary for the resource APIs.

The gateway accepts HTTPS requests containing JSON. All routes use the /v1 prefix for API versioning.

The gateway validates the Bearer JWT before routing the request. It also enforces authorization before routing. Authentication checks who the caller is. Authorization checks whether that caller may perform the requested action.

The gateway applies rate limiting to control excessive traffic. It also adds a request_id for tracing failures. After these checks, it routes the request to the correct resource API.

To create a post, the gateway sends POST /v1/posts to the Posts API. The request body contains caption and media_url.

The Posts API writes the post to the Posts Store. The store returns the stored post record to the service. The Posts API then returns 201 Created with {post_id, created_at}.

That response moves back to the gateway. The gateway returns the final JSON response to the client.

To follow a user, the gateway sends POST /v1/users/{id}/follow to the Follow API. To unfollow that user, it sends DELETE /v1/users/{id}/follow.

The Follow API creates or removes a follow edge in the Follow Graph Store. A follow edge represents one user following another user. The store returns the resulting follow state.

The Follow API returns either 200 OK or 204 No Content. The response returns through the gateway to the client.

For a profile request, the gateway sends GET /v1/profiles/{username} to the Profile API. The Profile API reads the profile from the Profiles Store. The store returns the profile data. The API then returns 200 OK with profile JSON.

For a feed request, the gateway sends GET /v1/feed?limit=20&cursor=abc123 to the Feed API. The Feed API reads followed accounts from the Follow Graph Store. That store returns the following list.

The Feed API then reads recent posts from the Posts Store. The store returns the post items. The Feed API combines the results and returns {items[], next_cursor} with 200 OK.

The cursor marks where the next page begins. Cursor-based pagination is preferred over offset pagination because feeds change often. It produces more stable pages when new posts arrive.

Validation, authentication, authorization, rate-limit, and service failures are mapped through the error-handling path. The response uses {code, message, details, request_id}.

The diagram shows these common codes. 400 means invalid input. 401 means authentication is missing or invalid. 403 means the caller is authenticated but not allowed. 404 means a resource was not found. 409 means a state conflict. 429 means the rate limit was exceeded. 500 means an unexpected server error.

The benefit of this design is clear ownership and consistent protection. The downside is extra routing and coordination between services. We accept that cost because the boundaries make the system easier to secure, maintain, and extend.

The benefit of separate Posts, Follow, Profile, and Feed APIs is clear ownership. Each API handles one main resource and uses the stores shown in the design. The gateway gives one place for JWT validation, authorization, rate limiting, request IDs, routing, and /v1 versioning. This reduces repeated work inside every API. The downside is that the Feed API needs data from two stores. It reads followed accounts first, then recent posts. That can increase response time. Cursor pagination works well for changing feeds because new posts do not shift every page. Its downside is that cursors are less simple than page numbers. Standard JSON errors help clients handle failures consistently. We accept the extra service coordination because it improves security, ownership, and long-term maintenance.