Sum of unique sub-array sums

Last Updated : 26 Feb, 2025

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.

[Naive Approach] – Sorting – O(n² Log n) Time and O(n²) Space

Calculate the cumulative sum of an array.
Store all sub-array sum in vector.
Sort the vector.
Mark all duplicate sub-array sum to zero
Calculate and return totalSum.

C++

#include <bits/stdc++.h>
using namespace std;

int findSubarraySum(vector<int> &arr) {
    int n = arr.size();
    vector<int> subArrSum;

    // Generate all subarray sums and store them in a vector
    for (int i = 0; i < n; i++) {
        int sum = 0;
        for (int j = i; j < n; j++) {
            sum += arr[j];
            subArrSum.push_back(sum);
        }
    }

    // Sort the vector to group duplicate sums together
    sort(subArrSum.begin(), subArrSum.end());

    // Sum only unique subarray sums
    int m = subArrSum.size();
    int totalSum = 0;
    
    if (m == 1) return subArrSum[0];
    
    // Explicitely handle the first element
    if (subArrSum[0] != subArrSum[1])
        totalSum += subArrSum[0];

    // Process Middle Elements
    for (int i = 1; i < m - 1; i++) {
        if (subArrSum[i] != subArrSum[i+1] && 
            subArrSum[i] != subArrSum[i-1]) {
            totalSum += subArrSum[i];
        }
    }
    
    // Explicitely handle the last element
    if (subArrSum[m-1] != subArrSum[m-2])
        totalSum += subArrSum[m-1];

    return totalSum;
}

int main() {
    vector<int> arr = {3, 2, 3, 1, 4};
    cout << findSubarraySum(arr); 
    return 0;
}

Java

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

public class Main {
    public static int findSubarraySum(int[] arr) {
        int n = arr.length;
        List<Integer> subArrSum = new ArrayList<>();

        // Generate all subarray sums and store them in a list
        for (int i = 0; i < n; i++) {
            int sum = 0;
            for (int j = i; j < n; j++) {
                sum += arr[j];
                subArrSum.add(sum);
            }
        }

        // Sort the list to group duplicate sums together
        Collections.sort(subArrSum);

        // Sum only unique subarray sums
        int m = subArrSum.size();
        int totalSum = 0;
        
        if (m == 1) return subArrSum.get(0);
        
        // Explicitely handle the first element
        if (!subArrSum.get(0).equals(subArrSum.get(1)))
            totalSum += subArrSum.get(0);

        // Process Middle Elements
        for (int i = 1; i < m - 1; i++) {
            if (!subArrSum.get(i).equals(subArrSum.get(i + 1)) && 
                !subArrSum.get(i).equals(subArrSum.get(i - 1))) {
                totalSum += subArrSum.get(i);
            }
        }
        
        // Explicitely handle the last element
        if (!subArrSum.get(m - 1).equals(subArrSum.get(m - 2)))
            totalSum += subArrSum.get(m - 1);

        return totalSum;
    }

    public static void main(String[] args) {
        int[] arr = {3, 2, 3, 1, 4};
        System.out.println(findSubarraySum(arr));
    }
}

Python

def find_subarray_sum(arr):
    n = len(arr)
    sub_arr_sum = []

    # Generate all subarray sums and store them in a list
    for i in range(n):
        sum_ = 0
        for j in range(i, n):
            sum_ += arr[j]
            sub_arr_sum.append(sum_)

    # Sort the list to group duplicate sums together
    sub_arr_sum.sort()

    # Sum only unique subarray sums
    m = len(sub_arr_sum)
    total_sum = 0
    
    if m == 1:
        return sub_arr_sum[0]
    
    # Explicitly handle the first element
    if m > 1 and sub_arr_sum[0] != sub_arr_sum[1]:
        total_sum += sub_arr_sum[0]

    # Process Middle Elements
    for i in range(1, m - 1):
        if sub_arr_sum[i] != sub_arr_sum[i + 1] and sub_arr_sum[i] != sub_arr_sum[i - 1]:
            total_sum += sub_arr_sum[i]
    
    # Explicitly handle the last element
    if m > 1 and sub_arr_sum[m - 1] != sub_arr_sum[m - 2]:
        total_sum += sub_arr_sum[m - 1]

    return total_sum

arr = [3, 2, 3, 1, 4]
print(find_subarray_sum(arr))

// C# implementation to find the sum of unique subarray sums
using System;
using System.Collections.Generic;
using System.Linq;

class Program
{
    static int FindSubarraySum(int[] arr)
    {
        int n = arr.Length;
        List<int> subArrSum = new List<int>();

        // Generate all subarray sums and store them in a list
        for (int i = 0; i < n; i++)
        {
            int sum = 0;
            for (int j = i; j < n; j++)
            {
                sum += arr[j];
                subArrSum.Add(sum);
            }
        }

        // Sort the list to group duplicate sums together
        subArrSum.Sort();

        // Sum only unique subarray sums
        int m = subArrSum.Count;
        int totalSum = 0;
        
        if (m == 1) return subArrSum[0];
        
        // Explicitly handle the first element
        if (subArrSum[0] != subArrSum[1])
            totalSum += subArrSum[0];

        // Process Middle Elements
        for (int i = 1; i < m - 1; i++)
        {
            if (subArrSum[i] != subArrSum[i + 1] && 
                subArrSum[i] != subArrSum[i - 1])
            {
                totalSum += subArrSum[i];
            }
        }
        
        // Explicitly handle the last element
        if (subArrSum[m - 1] != subArrSum[m - 2])
            totalSum += subArrSum[m - 1];

        return totalSum;
    }

    static void Main()
    {
        int[] arr = { 3, 2, 3, 1, 4 };
        Console.WriteLine(FindSubarraySum(arr));
    }
}

JavaScript

function findSubarraySum(arr) {
    let n = arr.length;
    let subArrSum = [];

    // Generate all subarray sums and store them in an array
    for (let i = 0; i < n; i++) {
        let sum = 0;
        for (let j = i; j < n; j++) {
            sum += arr[j];
            subArrSum.push(sum);
        }
    }

    // Sort the array to group duplicate sums together
    subArrSum.sort((a, b) => a - b);

    // Sum only unique subarray sums
    let m = subArrSum.length;
    let totalSum = 0;
    
    if (m === 1) return subArrSum[0];
    
    // Explicitely handle the first element
    if (subArrSum[0] !== subArrSum[1])
        totalSum += subArrSum[0];

    // Process Middle Elements
    for (let i = 1; i < m - 1; i++) {
        if (subArrSum[i] !== subArrSum[i + 1] && 
            subArrSum[i] !== subArrSum[i - 1]) {
            totalSum += subArrSum[i];
        }
    }
    
    // Explicitely handle the last element
    if (subArrSum[m - 1] !== subArrSum[m - 2])
        totalSum += subArrSum[m - 1];

    return totalSum;
}

const arr = [3, 2, 3, 1, 4];
console.log(findSubarraySum(arr));

Output

[Efficient Approach] – Sorting – O(n²) Time and O(n²) Space

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++

#include <bits/stdc++.h>
using namespace std;

// function for finding grandSum
long long int findSubarraySum(vector<int>& arr)
{
	long long int res = 0;

	// Go through all subarrays, compute sums
	// and count occurrences of sums.
	unordered_map<int, int> m;
	int n = arr.size();
	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()
{
	vector<int> arr = { 3, 2, 3, 1, 4 };
	cout << findSubarraySum(arr);
	return 0;
}

Java

// function for finding grandSum
import java.util.HashMap;
import java.util.Map;

public class Main {
    public static int findSubarraySum(int[] arr) {
        int res = 0;

        // Go through all subarrays, compute sums
        // and count occurrences of sums.
        HashMap<Integer, Integer> m = new HashMap<>();
        int n = arr.length;
        for (int i = 0; i < n; i++) {
            int sum = 0;
            for (int j = i; j < n; j++) {
                sum += arr[j];
                m.put(sum, m.getOrDefault(sum, 0) + 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;
    }

    public static void main(String[] args) {
        int[] arr = { 3, 2, 3, 1, 4 };
        System.out.println(findSubarraySum(arr));
    }
}

Python

# function for finding grandSum
def find_subarray_sum(arr):
    res = 0

    # Go through all subarrays, compute sums
    # and count occurrences of sums.
    m = {}
    n = len(arr)
    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]
print(find_subarray_sum(arr))

using System;
using System.Collections.Generic;

public class MainClass {
    public static int FindSubarraySum(int[] arr) {
        int res = 0;

        // Go through all subarrays, compute sums
        // and count occurrences of sums.
        Dictionary<int, int> m = new Dictionary<int, int>();
        int n = arr.Length;
        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]++;
                } else {
                    m[sum] = 1;
                }
            }
        }

        // Print all those sums that appear
        // once.
        foreach (var x in m) {
            if (x.Value == 1)
                res += x.Key;
        }

        return res;
    }

    public static void Main(string[] args) {
        int[] arr = { 3, 2, 3, 1, 4 };
        Console.WriteLine(FindSubarraySum(arr));
    }
}

JavaScript

// function for finding grandSum
function findSubarraySum(arr) {
    let res = 0;

    // Go through all subarrays, compute sums
    // and count occurrences of sums.
    const m = {};
    const n = arr.length;
    for (let i = 0; i < n; i++) {
        let sum = 0;
        for (let j = i; j < n; j++) {
            sum += arr[j];
            m[sum] = (m[sum] || 0) + 1;
        }
    }

    // Print all those sums that appear
    // once.
    for (const x in m) {
        if (m[x] === 1) {
            res += Number(x);
        }
    }

    return res;
}

// Driver code
const arr = [3, 2, 3, 1, 4];
console.log(findSubarraySum(arr));

Output

Length of longest strict bitonic subsequence

Shivam.Pradhan

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