Find Pivot Index

Compute totalSum in one pass, then scan again with leftSum initially zero. At index i, the right side is totalSum minus leftSum minus nums[i]. If leftSum equals that quantity, i is a pivot because neither side includes the pivot element. Otherwise add nums[i] to leftSum and continue. The invariant before examining i is that leftSum equals the sum of all elements strictly before i. This works with negative values; do not use a two-pointer balancing trick that assumes sums change monotonically. Clarify whether an empty left or right side counts as sum zero, which is the usual definition. Test a pivot at index zero, a pivot at the last index, no pivot, all zeros, and large integers requiring a wider accumulator. The algorithm is O(n) time and O(1) extra space. A prefix array is valid but unnecessary unless other range queries also need it.