Find all subarrays with sum in the given range

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.

Python3

# Python3 program for the above approach

# Function to find subarrays with sums
# in the given range
def findSubarrays(arr, N, L, R):

    for i in range(N):
        sums = 0;
        for j in range(i, N):
            sums += arr[j];

            # If the sums is in the range then
            # insert it into the answer
            if (sums >= L and sums <= R):
                print(i, j);

N = 4;
arr = [ 2, 3, 5, 8 ];
L = 4
R = 13;

# Function call
findSubarrays(arr, N, L, R);

# This code is contributed by phasing17.

C#

// C# program for the above approach

using System;
using System.Collections.Generic;

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)
          Console.WriteLine(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 phasing17.

Javascript

// JS program for the above approach

// Function to find subarrays with sum
// in the given range
function findSubarrays(arr, N, L, R)
{

    for (let i = 0; i < N; i++) {
        let sum = 0;
        for (let 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)
                console.log(i + " " + j);
        }
    }
}

let N = 4;
let arr = [ 2, 3, 5, 8 ];
let L = 4, R = 13;

// Function call
findSubarrays(arr, N, L, R);

// This code is contributed by phasing17.

Output

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

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