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++ program for above approach #include <bits/stdc++.h> 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 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 |
# 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# 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. |
<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> |
{ 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)