Find all subarrays with sum in the given range (original) (raw)

Last Updated : 20 Dec, 2022

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 &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 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