Find Sum of all unique subarray sum for a given array in C++

In this problem, we are given an array arr[] consisting of n integer values. Our task is to find the sum of all unique subarray sum for a given array. Subarray sum is the sum of elements of the given subarray.

Let's take an example to understand the problem,

Input : arr[] = {1, 2, 4}
Output : 23

Explanation −

All subarrays of the given array are :
(1), (2), (4), (1, 2), (2, 4), (1, 2, 4)
Sum of subarrays = 1 + 2 + 4 + (1+2) + (2+4) + (1+2+4) = 23

Solution Approach

A solution to the problem is by storing the subarray sum and then sorting them to find unique once. Then we will consider all unique subarrays for the sum.

Algorithm

Step 1 − Find the sum of all sub-arrays and store it in a vector.

Step 2 − Sort the vector.

Step 3 − Consider all the vectors which are unique and mark the sum of rest to 0.

Step 4 − Calculate and print the sum.

Example

Program to illustrate the working of our solution

#include <bits/stdc++.h>
using namespace std;
long int findSumOfSubArraySum(int arr[], int n){
   int i, j;
   long int sumArrayTill[n + 1] = { 0 };
   for (i = 0; i < n; i++)
      sumArrayTill[i + 1] = sumArrayTill[i] + arr[i];
   vector<long int> subArraySum;
   for (i = 1; i <= n; i++)
      for (j = i; j <= n; j++)
        subArraySum.push_back(sumArrayTill[j] - sumArrayTill[i - 1]);
   sort(subArraySum.begin(), subArraySum.end());
   for (i = 0; i < subArraySum.size() - 1; i++){
      if (subArraySum[i] == subArraySum[i + 1]) {
         j = i + 1;
         while (subArraySum[j] == subArraySum[i] && j < subArraySum.size()){
            subArraySum[j] = 0; j++;
         }
         subArraySum[i] = 0;
      }
   }
   long sum = 0;
   for (i = 0; i < subArraySum.size(); i++)
      sum += subArraySum[i];
   return sum;
}
int main(){
   int arr[] = { 1, 2, 4, 7, 9 };
   int n = sizeof(arr) / sizeof(arr[0]);
   cout<<"The sum of all unique subarray sum is "<<findSumOfSubArraySum(arr, n);
   return 0;
}

Output

The sum of all unique subarray sum is 144

Another approach using Iteration

Another approach to solve the problem is using a hash table. We will find the subarray sums and store them in a hash table and increment hash count. Then find the sum of all unique subarrays (subarray with hash count 1).

Example

Program to illustrate the working of our solution

#include <bits/stdc++.h>
using namespace std;
long int findSumOfSubArraySum(int arr[], int n){
   int sumSubArraySum = 0;
   unordered_map<int, int> sumSubArray;
   for (int i = 0; i < n; i++) {
      int sum = 0;
      for (int j = i; j < n; j++) {
         sum += arr[j];
         sumSubArray[sum]++;
      }
   }
   for (auto itr : sumSubArray)
      if (itr.second == 1)
         sumSubArraySum += itr.first;
      return sumSubArraySum;
}
int main(){
   int arr[] = { 1, 2, 4, 7, 5 };
   int n = sizeof(arr) / sizeof(arr[0]);
   cout<<"The sum of all unique subarray sum is "<<findSumOfSubArraySum(arr, n);
   return 0;
}

Output

The sum of all unique subarray sum is 124