First, compute the total sum and check if removing one element can make the remaining sum even. Then, for each possible removal, check if the remaining array can be split into two contiguous segments of equal sum using prefix sums or a sliding window. For the follow-up, collect all valid (index, cut) pairs by iterating over each removal and each possible cut position.
Pro tip: Clarify whether the two segments must be non-empty and whether the removed link can be at either end; these edge cases often trip up candidates. Also, mention that a naive O(n^2) solution is acceptable initially, but you can optimize to O(n) with prefix sums and a hash map.
Restate the problem to ensure clarity: remove exactly one link, then check if the remaining chain can be split into two contiguous segments of equal total weight. Discuss edge cases: empty array, single element, multiple valid removals, and whether segments must be non-empty.
Calculate the total sum of the array. For a removal to possibly work, the remaining sum must be even, so total sum minus removed element must be even. This gives a quick filter for candidate removals.
For each index i, compute the remaining array (or simulate removal) and check if there exists a cut position j such that the sum of the left segment equals the sum of the right segment. Use prefix sums to compute segment sums in O(1) after O(n) preprocessing.
For the follow-up, collect all valid (i, j) pairs. Optimize by precomputing prefix sums and using a hash map to find cut positions quickly, reducing time complexity from O(n^2) to O(n).
State the time and space complexity: O(n) time and O(n) space with prefix sums and hash map. Walk through a small example to verify correctness, including edge cases.
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