Given an array arr[] of positive integers and two integers L and R defining the range [L, R]. The task is to print the subarrays having sum in the range L to R.
Examples:
Input: arr[] = {1, 4, 6}, L = 3, R = 8
Output: {1, 4}, {4}, {6}.
Explanation: All the possible subarrays are the following
{1] with sum 1.
{1, 4} with sum 5.
{1, 4, 6} with sum 11.
{4} with sum 4.
{4, 6} with sum 10.
{6} with sum 6.
Therefore, subarrays {1, 4}, {4}, {6} are having sum in range [3, 8].Input: arr[] = {2, 3, 5, 8}, L = 4, R = 13
Output: {2, 3}, {2, 3, 5}, {3, 5}, {5}, {5, 8}, {8}.
Approach: This problem can be solved by doing brute force and checking for each and every possible subarray using two loops. Below is the implementation of the above approach.
C++
// C++ program for above approach#include <bits/stdc++.h>using namespace std;// Function to find subarrays in given rangevoid subArraySum(int arr[], int n, int leftsum, int rightsum){ int curr_sum, i, j, res = 0; // Pick a starting point for (i = 0; i < n; i++) { curr_sum = arr[i]; // Try all subarrays starting with 'i' for (j = i + 1; j <= n; j++) { if (curr_sum > leftsum && curr_sum < rightsum) { cout << "{ "; for (int k = i; k < j; k++) cout << arr[k] << " "; cout << "}\n"; } if (curr_sum > rightsum || j == n) break; curr_sum = curr_sum + arr[j]; } }}// Driver Codeint main(){ int arr[] = { 15, 2, 4, 8, 9, 5, 10, 23 }; int N = sizeof(arr) / sizeof(arr[0]); int L = 10, R = 23; subArraySum(arr, N, L, R); return 0;} |
Java
// Java code for the above approachimport java.io.*;class GFG { // Function to find subarrays in given range static void subArraySum(int arr[], int n, int leftsum, int rightsum) { int curr_sum, i, j, res = 0; // Pick a starting point for (i = 0; i < n; i++) { curr_sum = arr[i]; // Try all subarrays starting with 'i' for (j = i + 1; j <= n; j++) { if (curr_sum > leftsum && curr_sum < rightsum) { System.out.print("{ "); for (int k = i; k < j; k++) System.out.print(arr[k] + " "); System.out.println("}"); } if (curr_sum > rightsum || j == n) break; curr_sum = curr_sum + arr[j]; } } } // Driver Code public static void main(String[] args) { int arr[] = { 15, 2, 4, 8, 9, 5, 10, 23 }; int N = arr.length; int L = 10, R = 23; subArraySum(arr, N, L, R); }}// This code is contributed by Potta Lokesh |
Python3
# Python program for above approach# Function to find subarrays in given rangedef subArraySum (arr, n, leftsum, rightsum): res = 0 # Pick a starting point for i in range(n): curr_sum = arr[i] # Try all subarrays starting with 'i' for j in range(i + 1, n + 1): if (curr_sum > leftsum and curr_sum < rightsum): print("{ ", end="") for k in range(i, j): print(arr[k], end=" ") print("}") if (curr_sum > rightsum or j == n): break curr_sum = curr_sum + arr[j] # Driver Codearr = [15, 2, 4, 8, 9, 5, 10, 23]N = len(arr)L = 10R = 23subArraySum(arr, N, L, R)# This code is contributed by Saurabh Jaiswal |
C#
// C# code for the above approachusing System;class GFG { // Function to find subarrays in given range static void subArraySum(int []arr, int n, int leftsum, int rightsum) { int curr_sum, i, j, res = 0; // Pick a starting point for (i = 0; i < n; i++) { curr_sum = arr[i]; // Try all subarrays starting with 'i' for (j = i + 1; j <= n; j++) { if (curr_sum > leftsum && curr_sum < rightsum) { Console.Write("{ "); for (int k = i; k < j; k++) Console.Write(arr[k] + " "); Console.WriteLine("}"); } if (curr_sum > rightsum || j == n) break; curr_sum = curr_sum + arr[j]; } } } // Driver Code public static void Main() { int []arr = { 15, 2, 4, 8, 9, 5, 10, 23 }; int N = arr.Length; int L = 10, R = 23; subArraySum(arr, N, L, R); }}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script> // JavaScript program for above approach // Function to find subarrays in given range const subArraySum = (arr, n, leftsum, rightsum) => { let curr_sum, i, j, res = 0; // Pick a starting point for (i = 0; i < n; i++) { curr_sum = arr[i]; // Try all subarrays starting with 'i' for (j = i + 1; j <= n; j++) { if (curr_sum > leftsum && curr_sum < rightsum) { document.write("{ "); for (let k = i; k < j; k++) document.write(`${arr[k]} `); document.write("}<br/>"); } if (curr_sum > rightsum || j == n) break; curr_sum = curr_sum + arr[j]; } } } // Driver Code let arr = [15, 2, 4, 8, 9, 5, 10, 23]; let N = arr.length; let L = 10, R = 23; subArraySum(arr, N, L, R); // This code is contributed by rakeshsahni</script> |
Output
{ 15 }
{ 15 2 }
{ 15 2 4 }
{ 2 4 8 }
{ 4 8 }
{ 4 8 9 }
{ 8 9 }
{ 8 9 5 }
{ 9 5 }
{ 5 10 }
Time Complexity: O(N^3)
Auxiliary Space: O(1)
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