Find subarray with given sum | Set 1 (Nonnegative Numbers)

Difficulty Level : Medium
Last Updated : 17 Jun, 2022

Given an unsorted array arr of nonnegative integers and an integer sum, find a continuous subarray which adds to a given sum. There may be more than one subarrays with sum as the given sum, print first such subarray.
Examples :

Input: arr[] = {1, 4, 20, 3, 10, 5}, sum = 33
Output: Sum found between indexes 2 and 4
Sum of elements between indices 2 and 4 is 20 + 3 + 10 = 33

Input: arr[] = {1, 4, 0, 0, 3, 10, 5}, sum = 7
Output: Sum found between indexes 1 and 4
Sum of elements between indices 1 and 4 is 4 + 0 + 0 + 3 = 7

Input: arr[] = {1, 4}, sum = 0
Output: No subarray found
There is no subarray with 0 sum

Simple Approach: A simple solution is to consider all subarrays one by one and check the sum of every subarray. Following program implements the simple solution. Run two loops: the outer loop picks a starting point I and the inner loop tries all subarrays starting from i.

Algorithm:

Traverse the array from start to end.
From every index start another loop from i to the end of array to get all subarray starting from i, keep a variable sum to calculate the sum.
For every index in inner loop update sum = sum + array[j]
If the sum is equal to the given sum then print the subarray.

C++

/* A simple program to print subarray 
with sum as given sum */
#include <bits/stdc++.h>
using namespace std;
 
/* Returns true if the there is a subarray 
of arr[] with sum equal to 'sum' otherwise 
returns false. Also, prints the result */
int subArraySum(int arr[], int n, int sum)
{
    int curr_sum, i, j;
 
    // 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 == sum) {
                cout << "Sum found between indexes "
                     << i << " and " << j - 1;
                return 1;
            }
            if (curr_sum > sum || j == n)
                break;
            curr_sum = curr_sum + arr[j];
        }
    }
 
    cout << "No subarray found";
    return 0;
}
 
// Driver Code
int main()
{
    int arr[] = { 15, 2, 4, 8, 9, 5, 10, 23 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int sum = 23;
    subArraySum(arr, n, sum);
    return 0;
}
 
// This code is contributed
// by rathbhupendra

C

/* A simple program to print 
subarray with sum as given sum */
#include <stdio.h>
 
/* Returns true if the there is a subarray 
of arr[] with a sum equal to 'sum'
   otherwise returns false.  Also, prints 
the result */
int subArraySum(int arr[], int n, int sum)
{
    int curr_sum, i, j;
 
    // 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 == sum) {
                printf(
                    "Sum found between indexes %d and %d",
                    i, j - 1);
                return 1;
            }
            if (curr_sum > sum || j == n)
                break;
            curr_sum = curr_sum + arr[j];
        }
    }
 
    printf("No subarray found");
    return 0;
}
 
// Driver program to test above function
int main()
{
    int arr[] = { 15, 2, 4, 8, 9, 5, 10, 23 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int sum = 23;
    subArraySum(arr, n, sum);
    return 0;
}

Java

class SubarraySum {
    /* Returns true if the there is a 
subarray of arr[] with a sum equal to
       'sum' otherwise returns false.  
Also, prints the result */
    int subArraySum(int arr[], int n, int sum)
    {
        int curr_sum, i, j;
 
        // 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 == sum) {
                    int p = j - 1;
                    System.out.println(
                        "Sum found between indexes " + i
                        + " and " + p);
                    return 1;
                }
                if (curr_sum > sum || j == n)
                    break;
                curr_sum = curr_sum + arr[j];
            }
        }
 
        System.out.println("No subarray found");
        return 0;
    }
 
    public static void main(String[] args)
    {
        SubarraySum arraysum = new SubarraySum();
        int arr[] = { 15, 2, 4, 8, 9, 5, 10, 23 };
        int n = arr.length;
        int sum = 23;
        arraysum.subArraySum(arr, n, sum);
    }
}
 
// This code has been contributed by Mayank Jaiswal(mayank_24)

Python3

# Returns true if the
# there is a subarray
# of arr[] with sum
# equal to 'sum' 
# otherwise returns
# false. Also, prints
# the result 
def subArraySum(arr, n, sum_):
     
    # Pick a starting 
    # point
    for i in range(n):
        curr_sum = arr[i]
     
        # try all subarrays
        # starting with 'i'
        j = i + 1
        while j <= n:
         
            if curr_sum == sum_:
                print ("Sum found between")
                print("indexes % d and % d"%( i, j-1))
                 
                return 1
                 
            if curr_sum > sum_ or j == n:
                break
             
            curr_sum = curr_sum + arr[j]
            j += 1
 
    print ("No subarray found")
    return 0
 
# Driver program 
arr = [15, 2, 4, 8, 9, 5, 10, 23]
n = len(arr)
sum_ = 23
 
subArraySum(arr, n, sum_)
 
# This code is Contributed by shreyanshi_arun.

C#

// C# code to Find subarray
// with given sum
using System;
 
class GFG {
    // Returns true if the there is a
    // subarray of arr[] with sum
    // equal to 'sum' otherwise returns
    // false. Also, prints the result
    int subArraySum(int[] arr, int n,
                    int sum)
    {
        int curr_sum, i, j;
 
        // 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 == sum) {
                    int p = j - 1;
                    Console.Write("Sum found between "
                                  + "indexes " + i + " and " + p);
                    return 1;
                }
                if (curr_sum > sum || j == n)
                    break;
                curr_sum = curr_sum + arr[j];
            }
        }
 
        Console.Write("No subarray found");
        return 0;
    }
 
    // Driver Code
    public static void Main()
    {
        GFG arraysum = new GFG();
        int[] arr = { 15, 2, 4, 8, 9, 5, 10, 23 };
        int n = arr.Length;
        int sum = 23;
        arraysum.subArraySum(arr, n, sum);
    }
}
 
// This code has been contributed
// by nitin mittal

PHP

<?php
// A simple program to print subarray
// with sum as given sum 
 
/* Returns true if the there is
   a subarray of arr[] with 
   sum equal to 'sum'
   otherwise returns false. 
   Also, prints the result */
function subArraySum($arr, $n, $sum)
{
    $curr_sum; $i; $j;
 
    // 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 == $sum)
            {
                echo "Sum found between indexes ", $i, " and ", $j-1 ;
                return 1;
            }
            if ($curr_sum > $sum || $j == $n)
                break;
        $curr_sum = $curr_sum + $arr[$j];
        }
    }
 
    echo "No subarray found";
    return 0;
}
 
    // Driver Code
    $arr= array(15, 2, 4, 8, 9, 5, 10, 23);
    $n = sizeof($arr);
    $sum = 23;
    subArraySum($arr, $n, $sum);
    return 0;
     
// This code is contributed by AJit
?>

Javascript

<script>
 
/* A simple program to print subarray 
with sum as given sum */
 
 
/* Returns true if the there is a subarray 
of arr[] with sum equal to 'sum' otherwise 
returns false. Also, prints the result */
function subArraySum(arr, n, sum)
{
    let curr_sum=0;
 
    // Pick a starting point
    for (let i = 0; i < n; i++) {
        curr_sum = arr[i];
 
        // try all subarrays starting with 'i'
        for (let j = i + 1; j <= n; j++) {
            if (curr_sum == sum) {
                document.write("Sum found between indexes "+i+" and "+(j - 1));
                return;
            }
            if (curr_sum > sum || j == n)
                break;
            curr_sum = curr_sum + arr[j];
        }
    }
 
    document.write("No subarray found");
    return;
}
 
// Driver Code
 
let arr= [15, 2, 4, 8, 9, 5, 10, 23];
let n = arr.length;
let sum = 23;
subArraySum(arr, n, sum);
 
</script>

Output

Sum found between indexes 1 and 4

Complexity Analysis:

Time Complexity: O(n^2) in worst case.
Nested loop is used to traverse the array so the time complexity is O(n^2)
Space Complexity: O(1).
As constant extra space is required.

Efficient Approach: There is an idea if all the elements of the array are positive. If a subarray has sum greater than the given sum then there is no possibility that adding elements to the current subarray the sum will be x (given sum). Idea is to use a similar approach to a sliding window. Start with an empty subarray, add elements to the subarray until the sum is less than x. If the sum is greater than x, remove elements from the start of the current subarray.
Algorithm:

Create two variables, l=0, sum = 0
Traverse the array from start to end.
Update the variable sum by adding current element, sum = sum + array[i]
If the sum is greater than the given sum, update the variable sum as sum = sum – array[l], and update l as, l++.
If the sum is equal to given sum, print the subarray and break the loop.

C++

/* An efficient program to print 
subarray with sum as given sum */
#include <iostream>
using namespace std;
 
/* Returns true if the there is a subarray of 
arr[] with a sum equal to 'sum' otherwise 
returns false. Also, prints the result */
int subArraySum(int arr[], int n, int sum)
{
    /* Initialize curr_sum as value of 
    first element and starting point as 0 */
    int curr_sum = arr[0], start = 0, i;
 
    /* Add elements one by one to curr_sum and 
    if the curr_sum exceeds the sum,
    then remove starting element */
    for (i = 1; i <= n; i++) {
        // If curr_sum exceeds the sum,
        // then remove the starting elements
        while (curr_sum > sum && start < i - 1) {
            curr_sum = curr_sum - arr[start];
            start++;
        }
 
        // If curr_sum becomes equal to sum,
        // then return true
        if (curr_sum == sum) {
            cout << "Sum found between indexes "
                 << start << " and " << i - 1;
            return 1;
        }
 
        // Add this element to curr_sum
        if (i < n)
            curr_sum = curr_sum + arr[i];
    }
 
    // If we reach here, then no subarray
    cout << "No subarray found";
    return 0;
}
 
// Driver Code
int main()
{
    int arr[] = { 15, 2, 4, 8, 9, 5, 10, 23 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int sum = 23;
    subArraySum(arr, n, sum);
    return 0;
}
 
// This code is contributed by SHUBHAMSINGH10

C

/* An efficient program to print 
subarray with sum as given sum */
#include <stdio.h>
 
/* Returns true if the there is a 
subarray of arr[] with a sum 
equal to 'sum' otherwise returns 
false.  Also, prints the result */
int subArraySum(int arr[], int n, int sum)
{
    /* Initialize curr_sum as 
       value of first element and 
starting point as 0 */
    int curr_sum = arr[0], start = 0, i;
 
    /* Add elements one by one to 
curr_sum and if the curr_sum 
       exceeds the sum, then remove 
starting element */
    for (i = 1; i <= n; i++) {
        // If curr_sum exceeds the sum,
        // then remove the starting elements
        while (curr_sum > sum && start < i - 1) {
            curr_sum = curr_sum - arr[start];
            start++;
        }
 
        // If curr_sum becomes equal to sum,
        // then return true
        if (curr_sum == sum) {
            printf(
                "Sum found between indexes %d and %d",
                start, i - 1);
            return 1;
        }
 
        // Add this element to curr_sum
        if (i < n)
            curr_sum = curr_sum + arr[i];
    }
 
    // If we reach here, then no subarray
    printf("No subarray found");
    return 0;
}
 
// Driver program to test above function
int main()
{
    int arr[] = { 15, 2, 4, 8, 9, 5, 10, 23 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int sum = 23;
    subArraySum(arr, n, sum);
    return 0;
}

Java

class SubarraySum {
    /* Returns true if the there is 
a subarray of arr[] with sum equal to
       'sum' otherwise returns false.  
Also, prints the result */
    int subArraySum(int arr[], int n, int sum)
    {
        int curr_sum = arr[0], start = 0, i;
 
        // Pick a starting point
        for (i = 1; i <= n; i++) {
            // If curr_sum exceeds the sum,
            // then remove the starting elements
            while (curr_sum > sum && start < i - 1) {
                curr_sum = curr_sum - arr[start];
                start++;
            }
 
            // If curr_sum becomes equal to sum,
            // then return true
            if (curr_sum == sum) {
                int p = i - 1;
                System.out.println(
                    "Sum found between indexes " + start
                    + " and " + p);
                return 1;
            }
 
            // Add this element to curr_sum
            if (i < n)
                curr_sum = curr_sum + arr[i];
        }
 
        System.out.println("No subarray found");
        return 0;
    }
 
    public static void main(String[] args)
    {
        SubarraySum arraysum = new SubarraySum();
        int arr[] = { 15, 2, 4, 8, 9, 5, 10, 23 };
        int n = arr.length;
        int sum = 23;
        arraysum.subArraySum(arr, n, sum);
    }
}
 
// This code has been contributed by Mayank Jaiswal(mayank_24)

Python3

# An efficient program 
# to print subarray
# with sum as given sum 
 
# Returns true if the 
# there is a subarray 
# of arr[] with sum
# equal to 'sum' 
# otherwise returns 
# false. Also, prints 
# the result.
def subArraySum(arr, n, sum_):
     
    # Initialize curr_sum as
    # value of first element
    # and starting point as 0 
    curr_sum = arr[0]
    start = 0
 
    # Add elements one by 
    # one to curr_sum and 
    # if the curr_sum exceeds 
    # the sum, then remove 
    # starting element 
    i = 1
    while i <= n:
         
        # If curr_sum exceeds
        # the sum, then remove
        # the starting elements
        while curr_sum > sum_ and start < i-1:
         
            curr_sum = curr_sum - arr[start]
            start += 1
             
        # If curr_sum becomes
        # equal to sum, then
        # return true
        if curr_sum == sum_:
            print ("Sum found between indexes")
            print ("% d and % d"%(start, i-1))
            return 1
 
        # Add this element 
        # to curr_sum
        if i < n:
            curr_sum = curr_sum + arr[i]
        i += 1
 
    # If we reach here, 
    # then no subarray
    print ("No subarray found")
    return 0
 
# Driver program
arr = [15, 2, 4, 8, 9, 5, 10, 23]
n = len(arr)
sum_ = 23
 
subArraySum(arr, n, sum_)
 
# This code is Contributed by shreyanshi_arun.

C#

// An efficient C# program to print
// subarray with sum as given sum
using System;
 
class GFG {
 
    // Returns true if the
    // there is a subarray of
    // arr[] with sum equal to
    // 'sum' otherwise returns false.
    // Also, prints the result
    int subArraySum(int[] arr, int n,
                    int sum)
    {
        int curr_sum = arr[0],
            start = 0, i;
 
        // Pick a starting point
        for (i = 1; i <= n; i++) {
            // If curr_sum exceeds
            // the sum, then remove
            // the starting elements
            while (curr_sum > sum && start < i - 1) {
                curr_sum = curr_sum - arr[start];
                start++;
            }
 
            // If curr_sum becomes equal to
            // sum, then return true
            if (curr_sum == sum) {
                int p = i - 1;
                Console.WriteLine("Sum found between "
                                  + "indexes " + start + " and " + p);
                return 1;
            }
 
            // Add this element to curr_sum
            if (i < n)
                curr_sum = curr_sum + arr[i];
        }
        Console.WriteLine("No subarray found");
        return 0;
    }
 
    // Driver code
    public static void Main()
    {
        GFG arraysum = new GFG();
        int[] arr = new int[] { 15, 2, 4, 8,
                                9, 5, 10, 23 };
        int n = arr.Length;
        int sum = 23;
        arraysum.subArraySum(arr, n, sum);
    }
}
 
// This code has been contributed by KRV.

PHP

<?php
/* An efficient program to print 
subarray with sum as given sum */
 
/* Returns true if the there is a 
subarray of arr[] with sum equal 
to 'sum' otherwise returns false. 
Also, prints the result */
function subArraySum($arr, $n, $sum)
{
    /* Initialize curr_sum as 
    value of first element
    and starting point as 0 */
    $curr_sum = $arr[0]; 
    $start = 0; $i;
 
    /* Add elements one by one to 
    curr_sum and if the curr_sum 
    exceeds the sum, then remove
    starting element */
    for ($i = 1; $i <= $n; $i++)
    {
        // If curr_sum exceeds the sum, 
        // then remove the starting elements
        while ($curr_sum > $sum and
               $start < $i - 1)
        {
            $curr_sum = $curr_sum - 
                        $arr[$start];
            $start++;
        }
 
        // If curr_sum becomes equal 
        // to sum, then return true
        if ($curr_sum == $sum)
        {
            echo "Sum found between indexes",
                             " ", $start, " ", 
                           "and ", " ", $i - 1;
            return 1;
        }
 
        // Add this element
        // to curr_sum
        if ($i < $n)
        $curr_sum = $curr_sum + $arr[$i];
    }
 
    // If we reach here,
    // then no subarray
    echo "No subarray found";
    return 0;
}
 
// Driver Code
$arr = array(15, 2, 4, 8, 
              9, 5, 10, 23);
$n = count($arr);
$sum = 23;
subArraySum($arr, $n, $sum);
 
// This code has been
// contributed by anuj_67.
?>

Javascript

<script>
/* Returns true if the there is
a subarray of arr[] with sum equal to
       'sum' otherwise returns false. 
Also, prints the result */
     
    function subArraySum(arr,n,sum)
    {
        let curr_sum = arr[0], start = 0, i;
  
        // Pick a starting point
        for (i = 1; i <= n; i++) {
            // If curr_sum exceeds the sum,
            // then remove the starting elements
            while (curr_sum > sum && start < i - 1) {
                curr_sum = curr_sum - arr[start];
                start++;
            }
  
            // If curr_sum becomes equal to sum,
            // then return true
            if (curr_sum == sum) {
                let p = i - 1;
                document.write(
                    "Sum found between indexes " + start
                    + " and " + p+"<br>");
                return 1;
            }
  
            // Add this element to curr_sum
            if (i < n)
                curr_sum = curr_sum + arr[i];
        }
  
        document.write("No subarray found");
        return 0;
    }
     
    let arr=[15, 2, 4, 8, 9, 5, 10, 23 ];
    let n = arr.length;
    let sum = 23;
    subArraySum(arr, n, sum);
     
    // This code is contributed by unknown2108
</script>

Output

Sum found between indexes 1 and 4

Complexity Analysis:

Time Complexity: O(n).
- The Array is traversed only once to insert elements into the window. It will take O(N) time
- The Array is traversed again once to remove elements from the window. It will also take O(N) time.
- So the total time will be O(N) + O(N) = O(2*N), which is similar to O(N)
Space Complexity: O(1).
As constant extra space is required.

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The above solution doesn’t handle negative numbers. We can use hashing to handle negative numbers. See below set 2.

My Personal Notes

Find a peak element

Alternative Sorting

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