# Print all subarrays with sum in a given range

• Last Updated : 02 Dec, 2021

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:

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

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 ` `using` `namespace` `std;`   `// Function to find subarrays in given range` `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) {` `                ``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 Code` `int` `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 approach` `import` `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 range` `def` `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 Code` `arr ``=` `[``15``, ``2``, ``4``, ``8``, ``9``, ``5``, ``10``, ``23``]` `N ``=` `len``(arr)` `L ``=` `10` `R ``=` `23` `subArraySum(arr, N, L, R)`   `# This code is contributed by Saurabh Jaiswal`

## C#

 `// C# code for the above approach` `using` `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

 ``

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)

My Personal Notes arrow_drop_up
Recommended Articles
Page :