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Find Sum of all unique sub-array sum for a given array.
  • Difficulty Level : Medium
  • Last Updated : 14 Jun, 2021

Given an array of n-positive elements. The sub-array sum is defined as the sum of all elements of a particular sub-array, the task is to find the sum of all unique sub-array sum. 
Note: Unique Sub-array sum means no other sub-array will have the same sum value. 
Examples:

Input : arr[] = {3, 4, 5} 
Output : 40 
Explanation: All possible unique sub-array with their sum are as: 
(3), (4), (5), (3+4), (4+5), (3+4+5). Here all are unique so required sum = 40
Input : arr[] = {2, 4, 2} 
Output : 12 
Explanation: All possible unique sub-array with their sum are as: 
(2), (4), (2), (2+4), (4+2), (2+4+2). Here only (4) and (2+4+2) are unique.

Method 1 (Sorting Based) 
1- Calculate the cumulative sum of an array. 
2- Store all sub-array sum in vector. 
3- Sort the vector. 
4- Mark all duplicate sub-array sum to zero 
5- Calculate and return totalSum.

C++




// C++ for finding sum of all unique
// subarray sum
#include <bits/stdc++.h>
using namespace std;
 
// function for finding grandSum
long long int findSubarraySum(int arr[], int n)
{
    int i, j;
 
    // calculate cumulative sum of array
    // cArray[0] will store sum of zero elements
    long long int cArray[n + 1] = { 0 };
    for (i = 0; i < n; i++)
        cArray[i + 1] = cArray[i] + arr[i];
 
    vector<long long int> subArrSum;
 
    // store all subarray sum in vector
    for (i = 1; i <= n; i++)
        for (j = i; j <= n; j++)
            subArrSum.push_back(cArray[j] -
                                cArray[i - 1]);
 
    // sort the vector
    sort(subArrSum.begin(), subArrSum.end());
 
    // mark all duplicate sub-array
    // sum to zero
    long long totalSum = 0;
    for (i = 0; i < subArrSum.size() - 1; i++)
    {
        if (subArrSum[i] == subArrSum[i + 1])
        {
            j = i + 1;
            while (subArrSum[j] == subArrSum[i] &&
                                   j < subArrSum.size())
            {
                subArrSum[j] = 0;
                j++;
            }
            subArrSum[i] = 0;
        }
    }
 
    // calculate total sum
    for (i = 0; i < subArrSum.size(); i++)
        totalSum += subArrSum[i];
 
    // return totalSum
    return totalSum;
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 2, 3, 1, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << findSubarraySum(arr, n);
    return 0;
}

Java




// Java for finding sum of all unique
// subarray sum
import java.util.*;
 
class GFG{
 
// Function for finding grandSum
static int findSubarraySum(int arr[], int n)
{
    int i, j;
 
    // Calculate cumulative sum of array
    // cArray[0] will store sum of zero elements
    int cArray[] = new int[n + 1];
    for(i = 0; i < n; i++)
        cArray[i + 1] = cArray[i] + arr[i];
 
    Vector<Integer> subArrSum = new Vector<Integer>();
 
    // Store all subarray sum in vector
    for(i = 1; i <= n; i++)
        for(j = i; j <= n; j++)
            subArrSum.add(cArray[j] -
                          cArray[i - 1]);
 
    // Sort the vector
    Collections.sort(subArrSum);
 
    // Mark all duplicate sub-array
    // sum to zero
    int totalSum = 0;
    for(i = 0; i < subArrSum.size() - 1; i++)
    {
        if (subArrSum.get(i) ==
            subArrSum.get(i + 1))
        {
            j = i + 1;
            while (subArrSum.get(j) ==
                   subArrSum.get(i) &&
               j < subArrSum.size())
            {
                subArrSum.set(j, 0);
                j++;
            }
            subArrSum.set(i, 0);
        }
    }
 
    // Calculate total sum
    for(i = 0; i < subArrSum.size(); i++)
        totalSum += subArrSum.get(i);
 
    // Return totalSum
    return totalSum;
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 3, 2, 3, 1, 4 };
    int n = arr.length;
     
    System.out.print(findSubarraySum(arr, n));
}
}
 
// This code is contributed by Amit Katiyar

Python3




# Python3 for finding sum of all
# unique subarray sum
 
# function for finding grandSum
def findSubarraySum(arr, n):
     
    # calculate cumulative sum of array
    # cArray[0] will store sum of zero elements
    cArray = [0 for i in range(n + 1)]
    for i in range(0, n, 1):
        cArray[i + 1] = cArray[i] + arr[i]
 
    subArrSum = []
 
    # store all subarray sum in vector
    for i in range(1, n + 1, 1):
        for j in range(i, n + 1, 1):
            subArrSum.append(cArray[j] -
                             cArray[i - 1])
 
    # sort the vector
    subArrSum.sort(reverse = False)
 
    # mark all duplicate sub-array
    # sum to zero
    totalSum = 0
    for i in range(0, len(subArrSum) - 1, 1):
        if (subArrSum[i] == subArrSum[i + 1]):
            j = i + 1
            while (subArrSum[j] == subArrSum[i] and
                           j < len(subArrSum)):
                subArrSum[j] = 0
                j += 1
            subArrSum[i] = 0
 
    # calculate total sum
    for i in range(0, len(subArrSum), 1):
        totalSum += subArrSum[i]
 
    # return totalSum
    return totalSum
 
# Drivers code
if __name__ == '__main__':
    arr = [3, 2, 3, 1, 4]
    n = len(arr)
    print(findSubarraySum(arr, n))
     
# This code is contributed by
# Sahil_Shelangia

C#




// C# for finding sum of all unique
// subarray sum
using System;
using System.Collections.Generic;
class GFG{
 
// Function for finding grandSum
static int findSubarraySum(int []arr,
                           int n)
{
  int i, j;
 
  // Calculate cumulative sum
  // of array cArray[0] will
  // store sum of zero elements
  int []cArray = new int[n + 1];
   
  for(i = 0; i < n; i++)
    cArray[i + 1] = cArray[i] + arr[i];
 
  List<int> subArrSum = new List<int>();
 
  // Store all subarray sum in vector
  for(i = 1; i <= n; i++)
    for(j = i; j <= n; j++)
      subArrSum.Add(cArray[j] -
                    cArray[i - 1]);
 
  // Sort the vector
  subArrSum.Sort();
 
  // Mark all duplicate
  // sub-array sum to zero
  int totalSum = 0;
  for(i = 0; i < subArrSum.Count - 1; i++)
  {
    if (subArrSum[i] ==
        subArrSum[i + 1])
    {
      j = i + 1;
      while (subArrSum[j] ==
             subArrSum[i] &&
             j < subArrSum.Count)
      {
        subArrSum[j] = 0;
        j++;
      }
      subArrSum[i] = 0;
    }
  }
 
  // Calculate total sum
  for(i = 0; i < subArrSum.Count; i++)
    totalSum += subArrSum[i];
 
  // Return totalSum
  return totalSum;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = {3, 2, 3, 1, 4};
    int n = arr.Length;   
    Console.Write(findSubarraySum(arr, n));
}
}
 
// This code is contributed by Rajput-Ji
Output: 
41

 

Time Complexity: O(N^2 + N * logN)

Auxiliary Space: O(N)

Method 2 (Hashing Based) The idea is to make an empty hash table. We generate all subarrays. For every subarray, we compute its sum and increment count of the sum in the hash table. Finally, we add all those sums whose count is 1.



C++




// C++ for finding sum of all unique subarray sum
#include <bits/stdc++.h>
using namespace std;
 
// function for finding grandSum
long long int findSubarraySum(int arr[], int n)
{
    int res = 0;
 
    // Go through all subarrays, compute sums
    // and count occurrences of sums.
    unordered_map<int, int> m;
    for (int i = 0; i < n; i++) {
        int sum = 0;
        for (int j = i; j < n; j++) {
            sum += arr[j];
            m[sum]++;
        }
    }
 
    // Print all those sums that appear
    // once.
    for (auto x : m)
        if (x.second == 1)
            res += x.first;
 
    return res;
}
 
// Driver code
int main()
{
    int arr[] = { 3, 2, 3, 1, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << findSubarraySum(arr, n);
    return 0;
}

Java




// Java for finding sum of all
// unique subarray sum
import java.util.*;
 
class GFG
{
 
// function for finding grandSum
static int findSubarraySum(int []arr, int n)
{
    int res = 0;
 
    // Go through all subarrays, compute sums
    // and count occurrences of sums.
    HashMap<Integer,
            Integer> m = new HashMap<Integer,
                                     Integer>();
    for (int i = 0; i < n; i++)
    {
        int sum = 0;
        for (int j = i; j < n; j++)
        {
            sum += arr[j];
            if (m.containsKey(sum))
            {
                m.put(sum, m.get(sum) + 1);
            }
            else
            {
                m.put(sum, 1);
            }
        }
    }
 
    // Print all those sums that appear
    // once.
    for (Map.Entry<Integer,
                   Integer> x : m.entrySet())
        if (x.getValue() == 1)
            res += x.getKey();
 
    return res;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 3, 2, 3, 1, 4 };
    int n = arr.length;
    System.out.println(findSubarraySum(arr, n));
}
}
 
// This code is contributed by PrinciRaj1992

Python3




# Python3 for finding sum of all
# unique subarray sum
 
# function for finding grandSum
def findSubarraySum(arr, n):
 
    res = 0
 
    # Go through all subarrays, compute sums
    # and count occurrences of sums.
    m = dict()
    for i in range(n):
        Sum = 0
        for j in range(i, n):
            Sum += arr[j]
            m[Sum] = m.get(Sum, 0) + 1
         
    # Print all those Sums that appear
    # once.
    for x in m:
        if m[x] == 1:
            res += x
 
    return res
 
# Driver code
arr = [3, 2, 3, 1, 4]
n = len(arr)
print(findSubarraySum(arr, n))
 
# This code is contributed by mohit kumar

C#




// C# for finding sum of all
// unique subarray sum
using System;
using System.Collections.Generic;
     
class GFG
{
 
// function for finding grandSum
static int findSubarraySum(int []arr, int n)
{
    int res = 0;
 
    // Go through all subarrays, compute sums
    // and count occurrences of sums.
    Dictionary<int,
               int> m = new Dictionary<int,
                                       int>();
    for (int i = 0; i < n; i++)
    {
        int sum = 0;
        for (int j = i; j < n; j++)
        {
            sum += arr[j];
            if (m.ContainsKey(sum))
            {
                m[sum] = m[sum] + 1;
            }
            else
            {
                m.Add(sum, 1);
            }
        }
    }
 
    // Print all those sums that appear
    // once.
    foreach(KeyValuePair<int, int> x in m)
        if (x.Value == 1)
            res += x.Key;
 
    return res;
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 3, 2, 3, 1, 4 };
    int n = arr.Length;
    Console.WriteLine(findSubarraySum(arr, n));
}
}
 
// This code is contributed by Rajput-Ji
Output: 
41

 

Time Complexity: O(N^2)

Auxiliary Space: O(N)

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