Maximize Subarray Sum After Removing All Occurrences of One Element

Maximize Subarray Sum After Removing All Occurrences of One Element

What is this problem about?

In this problem, you are given an array of integers. You are allowed to choose exactly one integer value X that exists in the array, and remove all occurrences of X from the array. Your goal is to find the maximum possible contiguous subarray sum in the resulting array. You cannot remove all elements (the resulting array must be non-empty).

Why is this asked in interviews?

This is a sophisticated variation of Kadane's Algorithm. Interviewers use it to see if a candidate can decouple the concept of "removing elements" from the physical mutation of an array. It tests your ability to use Prefix Sums or Segment Trees to calculate the impact of a global removal without actually doing an $O(N^2)$ simulation (removing each unique element and recalculating Kadane's).

Algorithmic pattern used

This problem leverages Dynamic Programming (Kadane's) combined with Prefix/Suffix Sums. To avoid $O(N^2)$ time, we can precalculate the maximum subarray sum ending at every index (max_end[i]) and starting at every index (max_start[i]). Then, for every unique element X in the array, we find all its indices. For a specific occurrence of X at index i, if we imagine it's removed, the best subarray bridging across this gap would simply be max_end[i-1] + max_start[i+1] (assuming both are positive). We track the maximum achievable sum if all Xs are removed simultaneously. Alternatively, advanced solutions use a Segment Tree to query the max subarray sum after point updates.

Example explanation

Array: [1, -2, 3, -2, 5] Let's trace standard Kadane's: Max is $3 + (-2) + 5 = 6$. What if we remove all occurrences of -2? The array becomes [1, 3, 5]. The new max subarray sum is $1 + 3 + 5 = 9$.

If we track it via prefix/suffix maxes (ignoring edge cases for simplicity): max_end (max sum ending at i): [1, -1, 3, 1, 6] max_start (max sum starting at i): [5, 6, 6, 3, 5] For the -2 at index 3: If we remove it, the bridge is max_end[2] (3) + max_start[4] (5) = 8. If we remove all -2s, we need to evaluate the segments between the -2s and sum the positive components.

Common mistakes candidates make

A common error is just running Kadane's algorithm, finding the minimum negative number in the optimal subarray, and subtracting it. This is flawed because removing a specific number removes all its occurrences globally, which might connect two entirely different, highly profitable subarrays that were previously separated by multiple instances of that negative number.

Interview preparation tip

When an interview question asks to "remove all occurrences of X to maximize Y" in an array, it is almost always a hint to use a Hash Map to group the array by values. Group the original indices of every unique value. Then, evaluate the "gaps" between these indices using precomputed Prefix/Suffix logic. This transforms global mutations into localized $O(1)$ math operations.