Partition Array Into Three Parts With Equal Sum LT1013

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

Idea 1: S + S +... = 3*S = S + S + (-S) + S + S

Time complexity: O(n), 2 scan

Space complexity: O(1)

 class Solution {
     public boolean canThreePartsEqualSum(int[] A) {
         int totalSum = 0;
         for(int num: A) {
             totalSum += num;
         }

         if(totalSum%3 != 0) {
             return false;
         }

         int target = totalSum/3;

         int part = 0;
         int sum = 0;
         for(int i = 0; i < A.length; ++i) {
             sum += A[i];
             if(sum == target) {
                 ++part;
                 if(part >= 2) {
                     return true;
                 }
                 sum = 0;
             }

         }

         return false;
     }
 }