(Easy) Partition Array Into Three Parts With Equal Sum - LeetCode

Posted 2022-10-03 codingyangmao

Description:

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

Accepted

18,270

Submissions

32,514

Solution:

class Solution 
    public boolean canThreePartsEqualSum(int[] A) 
        
        // int Sum of each part 
        int sum = 0;
        
        for(int i = 0; i< A.length; i++)
            
            sum = sum + A[i];    
        
        
        sum = sum / 3; 
        
        
        int find_i =-1;
        int find_j = -1;
        int cur_sum = 0;
        for(int i = 0; i< A.length; i++)
            
            cur_sum = cur_sum +A[i];
            
            if(cur_sum == sum)
                
                find_i = i;
             
                cur_sum = 0;
                break;
            
        
        
        for(int k = find_i+1; k<A.length; k++)
            cur_sum = cur_sum+A[k];
            
            if(cur_sum ==sum)
                
                find_j = k;
                break;
            
            
            
        
    
       if(find_i <find_j && find_i>=0&& find_j <A.length)
           return true;
       
      return false;