Partition Array Into Three Parts With Equal Sum

May 1, 2019

Introduction

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:

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

Solution

Simple solution:

func canThreePartsEqualSum(A []int) bool {
    
    n := len(A)
    
    sum := 0
    for _, v := range A {
        sum += v
    }
    
    firstSum := 0
    for i := 0; i < n; i++ {
        
        firstSum += A[i]
        
        secondSum := 0
        for j := i+1; j < n; j++ {
            
            secondSum += A[j]
            
            if firstSum == secondSum && firstSum == sum - firstSum - secondSum {
                return true
            }
        }
    }
    return false
}

Another way to solve this problem is to split the sum in two 3 parts.

Faster solution

func canThreePartsEqualSum(A []int) bool {
    
    n := len(A)
    
    sum := 0
    for _, v := range A {
        sum += v
    }
    
    if sum % 3 != 0 {
        return false
    }
    
    partSum := sum / 3
    j := 0
    
    s := 0
    for i := 0; i < n; i++ {
        
        s += A[i]
        if s == partSum {
            j++
            s = 0
        }
        
    }
    
    return s == 0 && j == 3

}

Explanation

If we precalculate the total sum of array, we can solve this problem by going through all combinations between first sub-array [0,i], second sub-array [i+1,j] and the third subarray [j+1,n-1]. If sum of all 3 subarrays ar equal, then we have a solution of partitioning the array.