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)

 1 class Solution {
 2     public boolean canThreePartsEqualSum(int[] A) {
 3         int totalSum = 0;
 4         for(int num: A) {
 5             totalSum += num;
 6         }
 7         
 8         if(totalSum%3 != 0) {
 9             return false;
10         }
11         
12         int target = totalSum/3;
13         
14         int part = 0;
15         int sum = 0;
16         for(int i = 0; i < A.length; ++i) {
17             sum += A[i];
18             if(sum == target) {
19                 ++part;
20                 if(part >= 2) {
21                     return true;
22                 }
23                 sum = 0;
24             }
25             
26         }
27         
28         return false;
29     }
30 }