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Pivot index

class Solution {
    public int pivotIndex(int[] nums) {
        int sum = 0, leftsum = 0;
        for (int x: nums) sum += x;
        for (int i = 0; i < nums.length; ++i) {
            if (leftsum == sum - leftsum - nums[i]) return i;
            leftsum += nums[i];
        return -1;
nums = [1, 7, 3, 6, 5, 6]
Output: 3
The sum of the numbers to the left of index 3 (nums[3] = 6) is equal to the sum of numbers to the right of index 3.
Also, 3 is the first index where this occurs.

Intuition and Algorithm

We need to quickly compute the sum of values to the left and the right of every index.

Let's say we knew S as the sum of the numbers, and we are at index i. If we knew the sum of numbers leftsum that are to the left of index i, then the other sum to the right of the index would just be S - nums[i] - leftsum.

As such, we only need to know about leftsum to check whether an index is a pivot index in constant time. Let's do that: as we iterate through candidate indexes i, we will maintain the correct value of leftsum.

Complexity Analysis

  • Time Complexity: O(N), where N is the length of nums.

  • Space Complexity: O(1), the space used by leftsum and S.

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