1013

1013. Partition Array Into Three Parts With Equal Sum (Easy)

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

Related Topics:
Array

Solution 1.

// OJ: https://leetcode.com/problems/partition-array-into-three-parts-with-equal-sum/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    bool canThreePartsEqualSum(vector<int>& A) {
        int total = accumulate(begin(A), end(A), 0), cnt = 0, sum = 0;
        if (total % 3) return false;
        total /= 3;
        for (int i = 0; i < A.size() - 1; ++i) {
            sum += A[i];
            if (sum == total) {
                sum = 0;
                ++cnt;
            }
            if (cnt == 2) return true;
        }
        return false;
    }
};