Partition Array Into Three Parts With Equal Sum

Given an array A of integers, return true if and only if we can partition the array into three non-empty parts with equal sums.

Formally, we can partition the array if we can find indexes i+1 < j with (A[0] + A[1] + ... + A[i] == A[i+1] + A[i+2] + ... + A[j-1] == A[j] + A[j-1] + ... + A[A.length - 1])

Example 1:

Input: A = [0,2,1,-6,6,-7,9,1,2,0,1]
Output: true
Explanation: 0 + 2 + 1 = -6 + 6 - 7 + 9 + 1 = 2 + 0 + 1

Example 2:

Input: A = [0,2,1,-6,6,7,9,-1,2,0,1]
Output: false

Example 3:

Input: A = [3,3,6,5,-2,2,5,1,-9,4]
Output: true
Explanation: 3 + 3 = 6 = 5 - 2 + 2 + 5 + 1 - 9 + 4

Constraints:

3 <= A.length <= 50000
-10^4 <= A[i] <= 10^4

Solution:

class Solution {
    public boolean canThreePartsEqualSum(int[] A) {
        int sum = 0;
        for (int i : A) sum += i;
        if (sum % 3 != 0) return false;
        int left = 0;
        int right = A.length - 1;
        int leftSum = A[left];
        int rightSum = A[right];
        int midSum = sum - leftSum - rightSum;
        while (left < right - 1) {
            // System.out.println(leftSum + ", " + midSum + ", " + rightSum);
            if (leftSum == midSum && midSum == rightSum) {
                return true;
            }
            if (leftSum != sum / 3) {
                leftSum += A[++left];
            } else if (rightSum != sum / 3) {
                rightSum += A[--right];
            }
            midSum = sum - leftSum - rightSum;
        }
        return false;
    }
}