188 Data Scientist Interview Questions & Answers

I would use a dictionary with a 26-number frequency tuple as the key. For each string, I count how many times each lowercase letter appears, convert the count list to a tuple, and append the original string to that key's group. Anagrams produce the same frequency tuple, so they go into the same list. Finally, I return all dictionary groups. The expected time is O(n * k), and the space complexity shown by this solution is O(n * k).

We receive a list of lowercase English strings and need to put strings with exactly the same letters and letter counts into the same group. Every input occurrence must appear in the result, so duplicate strings must stay. We return new lists and do not change the input list. The order of the groups and the order inside each group do not matter. I use the count of each letter from a through z as a signature because anagrams always have the same counts.

1. Can I rely on every string containing only lowercase English letters? Yes, that is guaranteed by the problem.
2. Does the order of the groups or the strings inside each group matter? No, either order is accepted.
3. Should duplicate input strings be preserved? Yes, every occurrence must remain in the output.

The function receives strs: list[str]. There can be 1 to 10,000 strings. Each string has length 0 to 100 and contains only lowercase English letters. We return a new list[list[str]]. Strings with identical character counts belong in the same group. We must preserve duplicate strings, and we must not mutate strs.

For each string, I create a list of 26 zeros. Position 0 represents a, position 1 represents b, and so on through position 25 for z. I count every character in the string. After counting, I convert the list to a tuple. A tuple is hashable, so it can be used as a dictionary key.

The dictionary maps frequency tuple -> list of original strings. If two strings have the same 26 counts, they produce the same key and are appended to the same list. The central invariant is that all strings stored under one key have exactly the same character counts.

The input is ["eat", "tea", "tan", "ate", "nat", "bat"].

For "eat", the counts for a, e, and t are each 1. Its signature becomes key K1, so the dictionary starts K1 with ["eat"].

For "tea", the same three letters occur once, so it produces K1 and joins the same group. K1 becomes ["eat", "tea"].

For "tan", the letters a, n, and t each occur once. That produces a different signature K2, so K2 starts as ["tan"].

For "ate", the counts match K1, so K1 becomes ["eat", "tea", "ate"].

For "nat", the counts match K2, so K2 becomes ["tan", "nat"].

For "bat", the letters a, b, and t each occur once. That creates a third signature K3 with ["bat"].

The dictionary therefore contains K1 -> ["eat", "tea", "ate"], K2 -> ["tan", "nat"], and K3 -> ["bat"]. One valid returned result is [["eat", "tea", "ate"], ["tan", "nat"], ["bat"]].

Two lowercase strings are anagrams exactly when every letter appears the same number of times in both strings. The 26-number tuple records those counts. Therefore, anagrams create identical keys and enter the same group. Strings with different character counts create different keys. Every input string is appended exactly once, so no string is dropped or added and duplicates are preserved.

I use defaultdict(list) so a new signature automatically starts with an empty list. For every character, ord(ch) - ord('a') converts the letter to an index from 0 through 25. After the whole string is counted, tuple(count) creates the dictionary key and groups[key].append(s) stores the unchanged original string. Finally, list(groups.values()) returns the groups as a new outer list.

Let n be the number of strings and k be the average string length. Counting all characters takes O(n * k) work. Python dictionary lookup and insertion are O(1) on average, so the expected total time is O(n * k). The diagram gives O(n * k) space for storing the frequency keys and grouped strings. Empty strings, repeated characters, and duplicate strings all work with the same logic.

The key insight is that anagrams have identical character frequencies. For each string, build a canonical signature containing 26 counts, one for each lowercase English letter. Convert that count list to a tuple and use it as a dictionary key. The dictionary maps each signature to the original strings with that signature. The invariant is that every string stored under the same dictionary key has exactly the same 26 character counts. Because equal character counts mean the strings are anagrams, each dictionary value becomes one correct anagram group.

Let n be the number of strings and k be the average length of a string. We examine each character once while building its 26-number frequency signature, so the expected total time is O(n * k). Dictionary lookup and insertion are O(1) on average. The diagram shows O(n * k) space for the frequency keys and grouped strings. The returned groups also preserve every input occurrence, including duplicates.

One mistake is reusing the same 26-element count list for multiple strings, which mixes their frequencies. Another is trying to use the mutable count list directly as a dictionary key instead of converting it to a tuple. Using a set for each group is also wrong because it would remove duplicate input strings. Another mistake is changing the original strings or input list even though the function must not mutate strs. Finally, do not claim that the displayed output order is the only valid order.

Explain the invariant before coding: two strings go to the same dictionary group exactly when their 26 lowercase-letter counts are identical. Then make the code follow that statement directly: count letters, convert the counts to a tuple, and append the original string.

I would count how often each integer appears, then group the distinct values into buckets where the bucket index is the frequency. I scan those buckets from the highest frequency down and collect values until I have k of them. This avoids sorting the values by frequency. The bucket invariant makes the result correct because higher-frequency values are visited first. The solution uses O(n) auxiliary space and runs in O(n) expected time because Python dictionary operations are O(1) on average.

We are given a list of integers called nums and a number k. We need to return the k integer values that appear most often. We return the values themselves, not their positions. The order of the returned values does not matter. We must not change nums. The problem guarantees that the set of top-k values is unique, so no tie-breaking rule is needed at the boundary. Instead of sorting by frequency, we can count occurrences and place values into frequency buckets.

1. Can I return the k values in any order? Yes. The problem allows any order.
2. Is a tie-breaking rule needed when deciding the kth value? No. The set of k most frequent values is guaranteed to be unique.
3. Should nums remain unchanged? Yes. The function must not mutate nums.

The function is top_k_frequent(nums: list[int], k: int) -> list[int]. nums has from 1 to 100,000 integers. Each value is between -10^4 and 10^4. We must return k integer values, not indices. The returned order can be different. For nums = [1,1,1,2,2,3] and k = 2, one valid result is [1,2]. [2,1] is also valid.

First, use Counter to build a frequency map. The map stores value -> frequency. For the example, the final map is {1:3, 2:2, 3:1}. Next, create n + 1 empty buckets, where n = len(nums). Bucket index f represents frequency f. Each distinct value goes into the bucket that matches its frequency.

Before counting, the logical frequency state is empty. The result starts as []. After counting, the example has 1 -> 3, 2 -> 2, and 3 -> 1. The bucket invariant is: every value stored in buckets[f] occurs exactly f times in nums. For this example, buckets[3] = [1], buckets[2] = [2], and buckets[1] = [3].

Process nums = [1,1,1,2,2,3]. The running frequency states are {1:1}, {1:2}, {1:3}, {1:3, 2:1}, {1:3, 2:2}, and {1:3, 2:2, 3:1}. After building the buckets, scan frequencies from high to low. At f = 3, take 1, so result = [1]. At f = 2, take 2, so result = [1,2]. Now len(result) == k, so the function stops and returns [1,2].

Counter stores the exact occurrence count of every distinct integer. Therefore, buckets[f] contains exactly the values that occur f times. Scanning bucket indices from high to low visits values in non-increasing frequency order. The first k collected values are therefore the k most frequent values. The problem guarantees that no boundary tie-breaking rule is needed.

The code creates Counter(nums), allocates len(nums) + 1 bucket lists, and places each distinct value into the bucket matching its count. Then it scans bucket indices from len(nums) down to 1. Every encountered value is appended to result. The function returns immediately when result contains k values. The final return statement is only a defensive fallback because the stated problem guarantees that the earlier return will be reached.

The expected running time is O(n). Counter uses Python dictionary operations, which are O(1) on average. Creating the buckets, placing distinct values into them, and scanning the bucket structure are all linear in n. Auxiliary space is O(n) for the frequency map and buckets. Negative values and zero work normally as dictionary keys. Repeated values are counted correctly. k may equal the number of distinct values. nums is never changed.

The key idea is to avoid sorting all values by frequency. First, build a frequency map with Counter, where each key is an integer and each stored count is its frequency. Then create frequency buckets so buckets[f] contains every integer that occurs exactly f times. The central invariant is that a value in buckets[f] occurs exactly f times. Because larger bucket indices mean larger frequencies, scanning from n down to 1 exposes the most frequent values first. We stop when k values have been collected.

Let n be len(nums). Building Counter(nums) takes O(n) expected time because Python dictionary lookup and insertion are O(1) on average. Creating n + 1 buckets takes O(n). Placing every distinct value into one bucket takes at most O(n). Scanning the buckets also takes O(n). The total expected time is therefore O(n), which is better than O(n log n). The frequency map and bucket lists together use O(n) auxiliary space.

One common mistake is sorting the values by frequency and then claiming the solution is O(n). A full frequency sort can make the running time O(n log n) in the worst case. Another mistake is returning indices instead of integer values. Candidates may also create too few buckets and forget that a value can occur len(nums) times. Another error is continuing to scan after k values have already been collected. Finally, Python Counter relies on average O(1) hash operations, so the running time should be described as O(n) expected time rather than guaranteed worst-case O(n).

Before writing the descending loop, say what a bucket means: buckets[f] contains values that occur exactly f times. That invariant makes both the correctness argument and the O(n) expected-time explanation much easier.

I use two passes and store the answer directly in the result list. First, I move from left to right and store the product of all elements before each index. Then I move from right to left with a running suffix product and multiply it into the stored prefix value. This gives every index the product of all other elements without division. The algorithm takes O(n) time and O(1) auxiliary space, excluding the required output list.

We need to return a new list where each position contains the product of every input number except the number at that same position. We cannot use division, and we cannot change the input list. The example is nums = [1, 2, 3, 4], which must return [24, 12, 8, 6]. The main idea is to combine the product of everything on the left of each position with the product of everything on its right. Two passes let us do this in linear time without extra arrays.

1. Can I use the required output list as temporary storage for prefix products?
2. Should I assume every input satisfies the stated length, value, and prefix or suffix product limits?

The function receives nums: list[int] and returns a new list[int]. At index i, the returned value must equal the product of all input elements except nums[i]. Division is forbidden. The input list must not be changed. The list length is from 2 to 100,000. Each value is from -30 to 30. Every prefix or suffix product fits in a signed 32-bit integer.

For the example nums = [1, 2, 3, 4], the result is [24, 12, 8, 6]. For example, index 1 contains 12 because 1 * 3 * 4 = 12.

For each index i, I split the answer into two parts. The prefix product is the product of every value strictly to the left of i. The suffix product is the product of every value strictly to the right of i. Their product is exactly the product of every element except nums[i].

I do not create separate prefix and suffix arrays. The required result list stores the prefix products after the first pass. A single variable stores the running suffix product during the second pass.

I initialize left = 1. The value 1 is the multiplicative identity, so it correctly represents an empty product before index 0.

At index 0, left is 1, so result[0] = 1. Then left becomes 1 * 1 = 1. At index 1, left is 1, so result[1] = 1. Then left becomes 1 * 2 = 2. At index 2, left is 2, so result[2] = 2. Then left becomes 2 * 3 = 6. At index 3, left is 6, so result[3] = 6. Then left becomes 6 * 4 = 24.

After the first pass, result = [1, 1, 2, 6]. Each value is the product of the elements before that index.

I initialize right = 1. This represents the empty product to the right of the last index.

At index 3, right = 1. I calculate result[3] = 6 * 1 = 6. Then right = 1 * 4 = 4. At index 2, right = 4. I calculate result[2] = 2 * 4 = 8. Then right = 4 * 3 = 12. At index 1, right = 12. I calculate result[1] = 1 * 12 = 12. Then right = 12 * 2 = 24. At index 0, right = 24. I calculate result[0] = 1 * 24 = 24. Then right = 24 * 1 = 24.

The final result is [24, 12, 8, 6].

Before left is multiplied by nums[i], it contains exactly the product of all elements before index i. Therefore, the first pass stores the correct prefix product at every index.

Before right is multiplied by nums[i], it contains exactly the product of all elements after index i. During the second pass, multiplying the stored prefix by right combines every element before and after i. The current element is never included. Therefore, every final result value is correct.

The result list is initialized with ones so that every index exists before the prefix pass writes into it. The first loop moves left to right and stores prefix products. The second loop moves right to left and multiplies each prefix by the current suffix product. The input list is never modified, and division is never used.

There are two linear passes, so the time complexity is O(n). Apart from the required result list, the code uses only a constant number of variables. Therefore, auxiliary space is O(1), excluding the output. Negative numbers and zeros work naturally because the algorithm only uses multiplication.

The key insight is that the answer for index i equals the product of everything before i multiplied by the product of everything after i. The first pass stores each prefix product directly in the required result list. The second pass keeps one running suffix product and multiplies it into each stored prefix. The central invariant is that, before updating left with nums[i], left equals the product of nums[0:i]. Similarly, before updating right with nums[i], right equals the product of nums[i+1:n]. This gives the required answer without division or extra prefix and suffix arrays.

Let n be the number of elements. The algorithm makes one left-to-right pass and one right-to-left pass. Each pass does constant work for every element, so the total time is O(n). The required result list contains n values, but the problem tells us to exclude that output from auxiliary-space counting. Apart from the result list, the algorithm uses only n, left, right, and the loop index. Therefore, auxiliary space is O(1).

A common mistake is using division even though the problem forbids it. Another mistake is creating separate prefix and suffix arrays. That is correct logically, but it uses O(n) auxiliary space instead of the required O(1) auxiliary space excluding the output. A third mistake is updating left before assigning result[i], which incorrectly includes nums[i] in its own prefix product. The same error can happen in reverse by updating right before multiplying it into result[i]. Candidates may also accidentally modify nums instead of returning a new list, or claim O(1) total space without explaining that the required output list is excluded.

Explain the update order before writing the loops. Say that left must be stored before including nums[i], and right must be used before including nums[i]. That one invariant explains why the current element is excluded and makes the implementation much easier to verify.

I would binary search only the smaller array instead of merging both arrays. I choose a cut in the smaller array and derive the matching cut in the other array so the combined left side contains half of the elements. I compare the four values around the two cuts. If the partition is invalid, I move the binary-search range left or right. When both boundary conditions hold, I compute the median. This takes O(log(min(m, n))) time and O(1) auxiliary space.

We have two lists whose values are already in sorted order. We need the middle value we would get if we could view all values together in one sorted sequence. We must return that value as a float without changing either input list. Each list can be empty, but together they contain at least one value. Instead of building a combined list, we find where to split both lists so the values on the left and right meet in the correct place. Binary search makes this fast.

1. Can I assume both lists are already sorted in nondecreasing order?
2. Is the combined input guaranteed to contain at least one value?
3. Should the function always return a float, including when the total length is odd?
4. May I use positive and negative infinity as virtual boundary values when a partition touches an array end?

The function receives two sorted lists of integers. Each list has length from 0 to 1,000. Their combined length is at least 1. Values are between -10^6 and 10^6. We must return the mathematical median as a float. We must not mutate either list, and we may use only the Python standard library.

I first make nums1 the smaller working array by swapping only the local references when needed. This does not modify the caller's lists. I binary search a cut position i from 0 through m, where m is the length of the smaller array. The matching cut in nums2 is j = half - i, where half = (m + n + 1) // 2.

The two cuts always place exactly half of the combined values, rounded up, on the left side. A partition is valid when left_max1 <= right_min2 and left_max2 <= right_min1.

After nums1 is the smaller working array, set m = len(nums1), n = len(nums2), low = 0, and high = m. Set half = (m + n + 1) // 2. In each iteration, choose i = (low + high) // 2 and calculate j = half - i.

The four values around the cuts are left_max1, right_min1, left_max2, and right_min2. If a cut is at the start or end of an array, the code uses -infinity or +infinity as a virtual boundary value. These values are not elements of the inputs.

The supplied input is nums1 = [1, 3] and nums2 = [2]. Because nums1 is longer, the code swaps the working references. The binary search therefore uses nums1 = [2] and nums2 = [1, 3]. Now m = 1, n = 2, half = 2, low = 0, and high = 1.

First iteration: i = 0 and j = 2. The boundaries are left_max1 = -infinity, right_min1 = 2, left_max2 = 3, and right_min2 = +infinity. The condition left_max2 <= right_min1 fails because 3 > 2. The cut in nums1 is too far left, so low becomes i + 1 = 1.

Second iteration: low = 1 and high = 1, so i = 1 and j = 1. Now left_max1 = 2, right_min1 = +infinity, left_max2 = 1, and right_min2 = 3. Both partition conditions hold because 2 <= 3 and 1 <= +infinity. The total length is 3, so the median is max(2, 1) = 2.0. The function returns immediately.

The cuts are chosen so the combined left side always contains half = (m + n + 1) // 2 values. Because both input arrays are sorted, it is enough to compare the values directly around the cuts. When left_max1 <= right_min2 and left_max2 <= right_min1, no boundary value on the left is larger than a boundary value that belongs on the right. The partition is therefore correct. For an odd total length, the median is the largest value on the left. For an even total length, the median is the average of the largest left value and the smallest right value.

The code swaps the local references when nums1 is longer. It then binary searches the inclusive cut range from 0 to m. Each iteration calculates i and derives j. The code reads four boundary values and uses infinities when a cut touches an array end. If both cross-boundary conditions hold, it returns the median. If left_max1 > right_min2, the nums1 cut is too far right, so high becomes i - 1. Otherwise, the nums1 cut is too far left, so low becomes i + 1. Every update reduces the search interval.

The binary search runs only on the smaller list, so its tight running time is O(log(min(m, n))). This is also O(log(m+n)), which satisfies the required complexity. The algorithm uses only a fixed number of variables, so auxiliary space is O(1). It handles one empty array, non-overlapping value ranges, duplicate values, negative values, and both odd and even combined lengths without changing either input.

The key insight is that we do not need to build the complete merged order. We only need the boundary between the lower half and upper half of the combined sorted values. We binary search the cut i in the smaller array and calculate the other cut as j = (m + n + 1) // 2 - i. The central invariant is that the two cuts together always place the required number of elements on the combined left side. A partition is correct when left_max1 <= right_min2 and left_max2 <= right_min1. If left_max1 > right_min2, the nums1 cut is too far right, so the search moves left. Otherwise the nums1 cut is too far left, so the search moves right. Merging both arrays would take O(m+n) time, while this partition search takes O(log(min(m, n))) time and O(1) auxiliary space.

Let m and n be the working list lengths after nums1 is made the smaller array. Binary search looks only at possible cut positions in nums1. Each iteration removes about half of the remaining cut positions, so the tight running time is O(log(min(m, n))). This also satisfies the required O(log(m+n)) bound. The algorithm does not create a merged list or any growing data structure. It keeps only a fixed number of indices and boundary values. Auxiliary space is therefore O(1).

A common mistake is merging both lists first. That takes O(m+n) time and misses the required logarithmic target. Another mistake is not making nums1 the smaller working array before binary search. Candidates also sometimes calculate j incorrectly instead of using half - i. A valid partition requires both cross-boundary checks, not just one. The search direction is also easy to reverse: left_max1 > right_min2 means move high left, while the other invalid case means move low right. Finally, -infinity and +infinity are only virtual boundary sentinels. They are not actual values added to either list.

Before writing the loop, draw the two cuts and name left_max1, right_min1, left_max2, and right_min2. Then write the two valid-partition conditions. From those comparisons, derive the binary-search direction instead of trying to memorize it.

I would use a sliding window with two pointers and a Counter. I move the right pointer across s and track how many required character occurrences are still missing. When missing becomes zero, the current window is valid, so I move the left pointer right while it stays valid and record the shortest window. For the example, the answer is "BANC". The expected time is O(len(s) + len(t)), and auxiliary space is O(|Σ|), at most 52 character keys here.

We need the shortest continuous part of s that contains every character required by t. If t repeats a character, the returned part must contain enough copies of that character. Uppercase and lowercase letters are different. Both strings have length from 1 to 100,000. The answer is unique when it exists. If no valid substring exists, we return "". We must not change the input strings. A sliding window fits because it lets us grow a candidate substring and then remove unnecessary characters from its left side.

1. Should repeated characters in t be satisfied with the same multiplicity? The stated requirement says yes.
2. Should uppercase and lowercase letters be treated as different characters? The stated requirement says yes.
3. Should I return the substring itself rather than its indices? The stated contract says to return the substring.

The function is min_window(s: str, t: str) -> str. We return the shortest contiguous substring of s containing every character of t with the required multiplicity. The strings contain case-sensitive English letters. If no valid window exists, we return "". The inputs are not mutated. For s = "ADOBECODEBANC" and t = "ABC", the required counts are A:1, B:1, and C:1, and the final answer is "BANC".

I use two pointers, left and right, to describe the current window s[left:right + 1]. I use Counter(t) as need. A positive need[ch] means the current window is still missing that many copies of ch. Zero means the requirement for that character is exactly satisfied. A negative value means the window has extra copies. The variable missing counts the total required character occurrences that are still unsatisfied. The key invariant is: when missing == 0, the current window contains every required character with the required multiplicity.

We start with need = Counter(t), missing = len(t), left = 0, best_len = infinity, and best_start = 0. For t = "ABC", need begins as {A: 1, B: 1, C: 1} and missing begins as 3. No valid window has been recorded yet.

At right = 0, the character is A. Since need['A'] > 0, this A satisfies one requirement, so missing changes from 3 to 2. Then need['A'] is decremented.

At right = 3, the character is B. It satisfies another requirement, so missing becomes 1.

At right = 5, the character is C. It satisfies the last missing requirement, so missing becomes 0. The window [0, 5] is "ADOBEC". Its length is 6, so it becomes the first best window.

The window is now valid, so we shrink it. We remove the A at index

1. need['A'] becomes positive again, so missing becomes
2. The left pointer becomes 1, and the window is invalid.

The right pointer continues. At right = 9, another B enters the window, giving an extra B. At right = 10, A enters and missing becomes 0 again. The current valid window is [1, 10], or "DOBECODEBA".

We now shrink repeatedly while the window remains valid. We remove D at index 1, O at index 2, the extra B at index 3, and E at index 4. The window is still valid. When we remove C at index 5, need['C'] becomes positive, so missing becomes 1 and shrinking stops. The left pointer is now 6. The best length is still 6.

At right = 12, the character is C. It satisfies the missing C, so missing becomes 0. The valid window is [6, 12], or "ODEBANC". We remove O at index 6 and D at index 7 while the window stays valid. The window [8, 12] is "EBANC", so the best length improves to 5. We then remove E at index 8. The window [9, 12] is "BANC", so the best length improves to 4 and best_start becomes 9. Removing the B at index 9 makes need['B'] positive, so missing becomes 1 and shrinking stops. The scan is finished, so we return s[9:13], which is "BANC".

Whenever missing == 0, the current window contains every character from t with the required multiplicity. While that condition remains true, moving left right removes characters that are unnecessary or available in extra quantity. Before each removal, the code records the current window if it is shorter than the best one seen so far. Shrinking stops as soon as a required occurrence is lost. Therefore, the algorithm considers the smallest valid windows produced by each right boundary and keeps the shortest one overall.

The for loop expands the window with right. If need[ch] > 0, the new character satisfies one missing occurrence, so missing decreases. The code then decrements need[ch] because the character is inside the window. When missing == 0, the inner while loop measures the current window, updates the best answer when it is shorter, and removes s[left]. Removing a character increments its need value. If that value becomes positive, the window has lost a required occurrence, so missing increases. The function finally returns the best slice or "" if no valid window was found.

Building Counter(t) takes O(len(t)) expected time. The right pointer moves across s once. The left pointer also moves only forward, so each position leaves the window at most once. Counter operations use Python dictionary behavior and are O(1) on average. Therefore, the total expected time is O(len(s) + len(t)). Auxiliary space is O(|Σ|). Because the stated alphabet contains only uppercase and lowercase English letters, there are at most 52 distinct character keys. Relevant edge cases are t being longer than s, no valid window existing, repeated characters in t, and case-sensitive matching.

The key idea is to avoid testing every substring. The right pointer expands one sliding window. need stores the remaining requirement for each character, and missing stores the total number of required character occurrences that are still unsatisfied. The central invariant is that missing == 0 exactly when the current window satisfies every requirement from t. Once the window is valid, the left pointer moves right repeatedly to discard unnecessary or extra characters. The best answer is updated before each removal. This finds the smallest valid window for each useful right boundary and keeps the shortest one overall.

The expected time is O(len(s) + len(t)). Creating the Counter reads t once. The right pointer reads each character of s once. The left pointer also only moves forward, so each position is removed from the window at most once. Python Counter uses dictionary operations, whose lookups and updates are O(1) on average. Auxiliary space is O(|Σ|), meaning extra memory proportional to the number of distinct character keys. Under the stated case-sensitive English-letter alphabet, there are at most 52 possible keys.

A common mistake is treating the problem as a subsequence problem instead of requiring a contiguous substring. Another is tracking only whether a character appears and forgetting that repeated characters in t require repeated occurrences in the window. Candidates may also shrink the window only once instead of repeatedly while it remains valid. Updating the best answer after the window becomes invalid is another error. It is also easy to change missing for extra copies incorrectly. Finally, the complexity should be described as expected linear time because Counter operations use average O(1) dictionary behavior.

Explain missing before writing the loop. Say that missing == 0 means the current window is valid. Then describe the two actions in order: expand with right, and repeatedly shrink with left while the window remains valid. This makes the Counter updates and the best-window update much easier to reason about.

I would use two pointers, one at each end of the height array. I keep the highest bar seen from the left and from the right. At each step, I process the side with the lower current height because the opposite side already provides a sufficient boundary. I update that side's running maximum, add the water above the current bar, and move the pointer inward. Each bar is processed once, so the time is O(n) and the auxiliary space is O(1).

The input is a list of nonnegative bar heights. Every bar has width 1. After rain falls, lower bars between taller boundaries can hold water. We need to return the total number of water units that remain. We must not change the input list. The supplied example holds 6 units. The goal is O(n) time with O(1) extra space. A two-pointer method fits because it works inward from both ends and keeps only a few running values.

1. Can I assume the input is always valid and contains at least one bar?
2. Should I return only the total trapped-water volume, not the amount stored at each index?
3. Should I treat the requirement not to mutate height as strict?

The function receives height: list[int]. Each value is a nonnegative bar height, and each bar has width 1. The input length is between 1 and 20,000. Each height is at most 100,000. We return one integer: the total trapped-water volume. We do not return indices or a new height array. We also do not modify height.

For height = [0,1,0,2,1,0,1,3,2,1,2,1], the trapped-water amounts by index are [0,0,1,0,1,2,1,0,0,1,0,0]. Their sum is 6.

I put left at the first index and right at the last index. I keep left_max, the highest bar seen from the left, and right_max, the highest bar seen from the right.

On each loop, I compare height[left] with height[right]. If the left value is less than or equal to the right value, I process the left side. Otherwise, I process the right side.

The central invariant is that when I process the lower current side, the opposite current bar is at least as high as that side. Therefore the running maximum on the processed side is enough to determine the water at that position.

For the supplied example, left = 0 and right = 11. We set left_max = 0, right_max = 0, and water = 0.

One pointer moves inward on every loop. The loop stops after the pointers cross.

Step 1: left = 0, right = 11. The current heights are 0 and 1. We process the left side. left_max stays 0. We add 0 water. Total water is 0.

Step 2: left = 1, right = 11. Both current heights are 1. The condition uses <=, so we process the left side. left_max becomes 1. We add 0. Total water remains 0.

Step 3: left = 2, right = 11. The current left height is 0. left_max is 1, so we add 1 - 0 = 1. Total water becomes 1.

Step 4: left = 3, right = 11. The current heights are 2 and 1. The right side is lower, so we process index 11. right_max becomes 1. We add 0. Total water remains 1.

Step 5: left = 3, right = 10. Both current heights are 2. We process the left side. left_max becomes 2. We add 0.

Step 6: left = 4, right = 10. The height at index 4 is 1. left_max is 2, so we add 2 - 1 = 1. Total water becomes 2.

Step 7: left = 5, right = 10. The height at index 5 is 0. We add 2 - 0 = 2. Total water becomes 4.

Step 8: left = 6, right = 10. The height at index 6 is 1. We add 2 - 1 = 1. Total water becomes 5.

Step 9: left = 7, right = 10. The current heights are 3 and 2. We process the right side. right_max becomes 2. We add 0. Total water remains 5.

Step 10: left = 7, right = 9. The height at index 9 is 1. right_max is 2, so we add 2 - 1 = 1. Total water becomes 6.

Step 11: left = 7, right = 8. The height at index 8 is 2. right_max is 2, so we add 0. Total water remains 6.

Step 12: left = 7, right = 7. Both pointers are at the bar of height 3. The <= condition processes the left side. left_max becomes 3 and we add 0. Then left moves to 8. Now left > right, so the loop stops.

The function returns 6.

Suppose height[left] <= height[right]. The right current bar already gives a boundary at least as high as the current left bar. We can therefore finalize the left position using left_max. After updating left_max, the trapped water there is left_max - height[left]. The same reasoning works from the right when the right current bar is lower. This lets us safely finalize exactly one position per loop.

The code stores two indices, two running maximum heights, and the accumulated water. It updates the selected side's maximum before calculating its contribution. Because the running maximum includes the current bar, the subtraction cannot be negative. The code then moves that pointer inward. It only reads values from height, so the input is never changed.

Each loop moves exactly one pointer inward. Every position is therefore processed once, giving O(n) time. The algorithm stores only a fixed number of integer variables, giving O(1) auxiliary space.

A single bar returns 0. Monotone increasing or decreasing heights trap 0 water. An all-zero input also returns 0. Repeated equal heights work normally. The problem guarantees a valid input with at least one element.

Use two pointers and two running maximum heights. left starts at the beginning and right starts at the end. If height[left] <= height[right], process the left side. Otherwise, process the right side. Before adding water, update that side's running maximum. The invariant is that the lower current side can be finalized because the opposite current bar is at least as high as the side being processed. This avoids separate prefix and suffix arrays while still calculating each position correctly.

Let n be the number of bars. The time complexity is O(n). One pointer moves inward on every loop, so each position is processed once. The auxiliary space is O(1). The algorithm stores only left, right, left_max, right_max, and water. The amount of extra memory does not grow with n. The input list is read but never copied or modified.

One common mistake is moving the wrong pointer. This solution processes the left side when height[left] <= height[right]; otherwise it processes the right side. Another mistake is calculating the contribution before updating the corresponding running maximum. Candidates may also build full left-max and right-max arrays, which would violate the O(1) auxiliary-space target. Other mistakes are modifying the input unnecessarily, using a loop condition that skips the final position, or claiming the wrong complexity.

State the invariant before coding: process the lower current side because the opposite current bar already provides a sufficient boundary. Then explain that the running maximum on the processed side determines the water at that position. This makes both the pointer movement and the O(n) time with O(1) auxiliary space easy to justify.