Split Array with Equal Sum in C++

Suppose we have an array with n integers, we have to find if there are triplets (i, j, k) which follows these conditions −

• 0 < i, i + 1 < j, j + 1 < k < n - 1

• Sum of subarrays (0, i - 1), (i + 1, j - 1), (j + 1, k - 1) and (k + 1, n - 1) will be same.

The subarray (L, R) is a slice of the original array starting from the element indexed L to the element indexed R.

So, if the input is like [1,2,1,2,1,2,1], then the output will be True, as i = 1, j = 3, k = 5.

sum(0, i - 1) = 1 sum(0, 0) = 1
sum(i + 1, j - 1) = 1 sum(2, 2) = 1
sum(j + 1, k - 1) = 1 sum(4, 4) = 1
sum(k + 1, n - 1) = 1 sum(6, 6) = 1

To solve this, we will follow these steps −

• n := size of nums

• Define an array sums of size n

• sums[0] := nums[0]

• for initialize i := 1, when i < n, update (increase i by 1), do −

  • sums[i] := sums[i] + (nums[i] + sums[i - 1])

• for initialize j := 3, when j − n, update (increase j by 1), do −

  • Define one set s

  • for initialize i := 1, when i < j - 1, update (increase i by 1), do: −

    • sum1 := sums[i - 1]

    • sum2 := sums[j - 1] - sums[i]

    • if sum1 is same as sum2, then −

      • insert sum1 into s

  • for initialize k := j + 2, when k < n - 1, update (increase k by 1), do −

    • sum1 := sums[k - 1] - sums[j]

    • sum2 := sums[n - 1] - sums[k]

    • if sum1 is same as sum2 and sum1 is in s, then −

      • return true

• return false

Example 

Let us see the following implementation to get better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
   bool splitArray(vector<int>& nums) {
      int n = nums.size();
      vector<int> sums(n);
      sums[0] = nums[0];
      for (int i = 1; i < n; i++) {
         sums[i] += (nums[i] + sums[i - 1]);
      }
      for (int j = 3; j < n; j++) {
         set<int> s;
         for (int i = 1; i < j - 1; i++) {
            int sum1 = sums[i - 1];
            int sum2 = sums[j - 1] - sums[i];
            if (sum1 == sum2)
               s.insert(sum1);
         }
         for (int k = j + 2; k < n - 1; k++) {
            int sum1 = sums[k - 1] - sums[j];
            int sum2 = sums[n - 1] - sums[k];
            if (sum1 == sum2 && s.count(sum1))
               return true;
            }
         }
         return false;
      }
};
main(){
   Solution ob;
   vector<int> v = {1,2,1,2,1,2,1};
   cout << (ob.splitArray(v));
}

Input

{1,2,1,2,1,2,1}

Output

1