LeetCode 1013 - Partition Array Into Three Parts With Equal Sum

https://leetcode.com/problems/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: [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: [0,2,1,-6,6,7,9,-1,2,0,1]
Output: false

Example 3:

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

Note:

1. 3 <= A.length <= 50000
2. -10000 <= A[i] <= 10000

https://leetcode.com/problems/partition-array-into-three-parts-with-equal-sum/discuss/261012/Java-O(n)-time-O(1)-space-explained

    public boolean canThreePartsEqualSum(int[] A) {
        int sum = 0;
        //calculate overall sum
        for (int n : A)
            sum += n;
        //check if it's divisible by 3
        if (sum % 3 != 0)
            return false;
        //exact sum of each segment
        sum /= 3;

        int curSum = 0;
        int numOfSegments = 0;
        for (int i = 0; i < A.length; i++) {
            curSum += A[i];
            //check if we can form a segment
            if (curSum == sum && numOfSegments <= 1) {
                numOfSegments++;
                curSum = 0;
            }
        }
        //if we have 2 segments formed greedily and sum of leftover is also 1/3 of overall sum 
        return (numOfSegments == 2 && curSum == sum);
    }

https://leetcode.com/problems/partition-array-into-three-parts-with-equal-sum/discuss/261732/Simple-Java-solution

    public boolean canThreePartsEqualSum(int[] A) {
        int total = Arrays.stream(A).reduce(0, Integer::sum);
        if (total % 3 != 0) return false;
        int count = 0, sum = 0;
        for (int i = 0; i < A.length; i++) {
            sum += A[i];
            if (sum == (total / 3)) {
                sum = 0;
                count++;
            }
        }
        return count >= 3;
    }