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Find all subarrays with sum in the given range

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  • Difficulty Level : Hard
  • Last Updated : 22 Sep, 2022
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Given an unsorted array of size, N. Find subarrays that add to a sum in the given range L-R.

Examples:

Input: arr[] = {2, 3, 5, 8}, L = 4, R = 13
Output: The indexes of subarrays are {0, 1}, {0, 2}, {1, 2}, {2, 2}, {2, 3}, {3, 3}

Input: arr[] = {1, 4, 6}, L = 3, R = 8
Output: The indexes of subarrays are {0, 1}, {1, 1}, {2, 2}

Naive approach: Follow the given steps to solve the problem using this approach:

  • Generate every possible subarray using two loops
  • If the sum of that subarray lies in the range [L, R], then push the starting and ending index into the answer array
  • Print the subarrays
     

Below is the implementation of the above approach:  

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find subarrays with sum
// in the given range
void findSubarrays(vector<int> &arr, vector<pair<int,int>> &ans, int L, int R)
{
    int N = arr.size();
 
      for(int i=0; i<N; i++)
    {
        int sum = 0;
        for(int j=i; j<N; j++)
        {
            sum += arr[j];
           
              // If the sum is in the range then
              // insert it into the answer
            if(sum >= L && sum <= R)
                ans.push_back({i, j});
        }
    }
}
 
void printSubArrays(vector<pair<int,int>> &ans)
{
    int size = ans.size();
      for(int i=0; i<size; i++)
        cout<<ans[i].first<<" "<<ans[i].second<<endl;
}
 
// Driver Code
int main()
{
    vector<int> arr = {2, 3, 5, 8};
    int L = 4, R = 13;
      vector<pair<int,int>> ans;
     
    // Function call
    findSubarrays(arr, ans, L, R);
      printSubArrays(ans);
   
    return 0;
}


Java




// Java program for the above approach
import java.io.*;
 
class GFG {
 
  // Function to find subarrays with sum
  // in the given range
  static void findSubarrays(int arr[], int N, int L,
                            int R)
  {
 
    for (int i = 0; i < N; i++) {
      int sum = 0;
      for (int j = i; j < N; j++) {
        sum += arr[j];
 
        // If the sum is in the range then
        // insert it into the answer
        if (sum >= L && sum <= R)
          System.out.println(i + " " + j);
      }
    }
  }
 
  public static void main(String[] args)
  {
    int N = 4;
    int arr[] = { 2, 3, 5, 8 };
    int L = 4, R = 13;
 
    // Function call
    findSubarrays(arr, N, L, R);
  }
}
 
// This code is contributed by mudit148.


Output

0 1
0 2
1 2
2 2
2 3
3 3

Time complexity: O(N2)
Auxiliary Space: O(N2)


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