Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
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 : 23Explanation −
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
457 Views
