Skip to content
Related Articles

Related Articles

Table of Contents

Improve Article
Save Article
Like Article

Smallest subarray of size greater than K with sum greater than a given value

  • Difficulty Level : Hard
  • Last Updated : 03 May, 2021

Given an array, arr[] of size N, two positive integers K and S, the task is to find the length of the smallest subarray of size greater than K, whose sum is greater than S.

Examples: 

Input: arr[] = {1, 2, 3, 4, 5}, K = 1, S = 8
Output: 2
Explanation: 
Smallest subarray with sum greater than S(=8) is {4, 5}
Therefore, the required output is 2.

Input: arr[] = {1, 3, 5, 1, 8, 2, 4}, K= 2, S= 13
Output: 3

 

 Approach: The problem can be solved using Sliding Window Technique. Follow the steps below to solve the problem:



  1. Initialize two variables say, i = 0 and j = 0 both pointing to the start of array i.e index 0.
  2. Initialize a variable sum to store the sum of the subArray currently being processed.
  3. Traverse the array, arr[] and by incrementing j and adding arr[j]
  4. Take our the window length or the length of the current subArray which is given by j-i+1 (+1 because the indexes start from zero) .
  5. Firstly, check if the size of the current subArray i.e winLen  here is greater than K. if this is not the case increment the j value and continue the loop.
  6. Else , we get that the size of the current subArray is greater than K, now we have to check if we meet the second condition i.e sum of the current Subarray is greater than S.
  7. If this is the case, we update minLength variable which stores the minimum length of the subArray satisfying the above conditions.
  8. At this time , we check if the size of the subArray can be reduced (by incrementing i )  such that it still is greater than K and sum is also greater than S. We constantly remove the ith element of the array from the sum to reduce the subArray size in the While loop and then increment j such that we move to the next element in the array .the 
  9. Finally, print the minimum length of required subarray obtained that satisfies the conditions.

Below is the implementation of the above approach:
 

C++




// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the length of the
// smallest subarray of size > K with
// sum greater than S
int smallestSubarray(int K, int S,
                     int arr[], int N)
{
    // Store the first index of
    // the current subarray
    int start = 0;
 
    // Store the last index of
    // the the current subarray
    int end = 0;
 
    // Store the sum of the
    // current subarray
    int currSum = arr[0];
 
    // Store the length of
    // the smallest subarray
    int res = INT_MAX;
 
    while (end < N - 1) {
 
        // If sum of the current subarray <= S
        // or length of current subarray <= K
        if (currSum <= S
            || (end - start + 1) <= K) {
            // Increase the subarray
            // sum and size
            currSum += arr[++end];
        }
 
        // Otherwise
        else {
 
            // Update to store the minimum
            // size of subarray obtained
            res = min(res, end - start + 1);
 
            // Decrement current subarray
            // size by removing first element
            currSum -= arr[start++];
        }
    }
 
    // Check if it is possible to reduce
    // the length of the current window
    while (start < N) {
        if (currSum > S
            && (end - start + 1) > K)
            res = min(res, (end - start + 1));
 
        currSum -= arr[start++];
    }
    return res;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int K = 1, S = 8;
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << smallestSubarray(K, S, arr, N);
}
Java



// Java program to implement
// the above approach
import java.io.*;
 
class GFG{
 
// Function to find the length of the
// smallest subarray of size > K with
// sum greater than S
public static int smallestSubarray(int k, int s,
                                   int[] array, int N)
{
     
        int i=0;
        int j=0;
        int minLen = Integer.MAX_VALUE;
        int sum = 0;
 
        while(j < N)
        {
            sum += array[j];
            int winLen = j-i+1;
            if(winLen <= k)
                j++;
            else{
                if(sum > s)
                {
                    minLen = Math.min(minLen,winLen);
                    while(sum > s)
                    {
                        sum -= array[i];
                        i++;
                    }
                    j++;
                }
            }
        }
        return minLen;
}
 
// Driver Code
public static void main(String[] args)
{
    int[] arr = { 1, 2, 3, 4, 5 };
    int K = 1, S = 8;
    int N = arr.length;
     
    System.out.print(smallestSubarray(K, S, arr, N));
}
}
 
// This code is contributed by akhilsaini
Python3



# Python3 program to implement
# the above approach
import sys
 
# Function to find the length of the
# smallest subarray of size > K with
# sum greater than S
def smallestSubarray(K, S, arr, N):
   
  # Store the first index of
  # the current subarray
  start = 0
 
  # Store the last index of
  # the the current subarray
  end = 0
 
  # Store the sum of the
  # current subarray
  currSum = arr[0]
 
  # Store the length of
  # the smallest subarray
  res = sys.maxsize
 
  while end < N - 1:
 
      # If sum of the current subarray <= S
      # or length of current subarray <= K
      if ((currSum <= S) or
         ((end - start + 1) <= K)):
           
          # Increase the subarray
          # sum and size
          end = end + 1;
          currSum += arr[end]
 
      # Otherwise
      else:
 
          # Update to store the minimum
          # size of subarray obtained
          res = min(res, end - start + 1)
 
          # Decrement current subarray
          # size by removing first element
          currSum -= arr[start]
          start = start + 1
 
  # Check if it is possible to reduce
  # the length of the current window
  while start < N:
      if ((currSum > S) and
         ((end - start + 1) > K)):
          res = min(res, (end - start + 1))
       
      currSum -= arr[start]
      start = start + 1
 
  return res;
 
# Driver Code
if __name__ == "__main__":
     
  arr = [ 1, 2, 3, 4, 5 ]
  K = 1
  S = 8
  N = len(arr)
   
  print(smallestSubarray(K, S, arr, N))
 
# This code is contributed by akhilsaini
C#



// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to find the length of the
// smallest subarray of size > K with
// sum greater than S
static int smallestSubarray(int K, int S,
                            int[] arr, int N)
{
     
    // Store the first index of
    // the current subarray
    int start = 0;
 
    // Store the last index of
    // the the current subarray
    int end = 0;
 
    // Store the sum of the
    // current subarray
    int currSum = arr[0];
 
    // Store the length of
    // the smallest subarray
    int res = int.MaxValue;
 
    while (end < N - 1)
    {
         
        // If sum of the current subarray <= S
        // or length of current subarray <= K
        if (currSum <= S ||
           (end - start + 1) <= K)
        {
             
            // Increase the subarray
            // sum and size
            currSum += arr[++end];
        }
 
        // Otherwise
        else
        {
 
            // Update to store the minimum
            // size of subarray obtained
            res = Math.Min(res, end - start + 1);
 
            // Decrement current subarray
            // size by removing first element
            currSum -= arr[start++];
        }
    }
 
    // Check if it is possible to reduce
    // the length of the current window
    while (start < N)
    {
        if (currSum > S && (end - start + 1) > K)
            res = Math.Min(res, (end - start + 1));
 
        currSum -= arr[start++];
    }
    return res;
}
 
// Driver Code
static public void Main()
{
    int[] arr = { 1, 2, 3, 4, 5 };
    int K = 1, S = 8;
    int N = arr.Length;
     
    Console.Write(smallestSubarray(K, S, arr, N));
}
}
 
// This code is contributed by akhilsaini
Javascript



<script>
// JavaScript program to implement
// the above approach
 
// Function to find the length of the
// smallest subarray of size > K with
// sum greater than S
function smallestSubarray(K, S, arr, N)
{
 
    // Store the first index of
    // the current subarray
    let start = 0;
 
    // Store the last index of
    // the the current subarray
    let end = 0;
 
    // Store the sum of the
    // current subarray
    let currSum = arr[0];
 
    // Store the length of
    // the smallest subarray
    let res = Number.MAX_SAFE_INTEGER;
    while (end < N - 1)
    {
 
        // If sum of the current subarray <= S
        // or length of current subarray <= K
        if (currSum <= S
            || (end - start + 1) <= K)
        {
         
            // Increase the subarray
            // sum and size
            currSum += arr[++end];
        }
 
        // Otherwise
        else {
 
            // Update to store the minimum
            // size of subarray obtained
            res = Math.min(res, end - start + 1);
 
            // Decrement current subarray
            // size by removing first element
            currSum -= arr[start++];
        }
    }
 
    // Check if it is possible to reduce
    // the length of the current window
    while (start < N)
    {
        if (currSum > S
            && (end - start + 1) > K)
            res = Math.min(res, (end - start + 1));
 
        currSum -= arr[start++];
    }
    return res;
}
 
// Driver Code
    let arr = [ 1, 2, 3, 4, 5 ];
    let K = 1, S = 8;
    let N = arr.length;
    document.write(smallestSubarray(K, S, arr, N));
 
// This code is contributed by Surbhi tyagi.
</script>
 
 
Output: 
2
 
Time Complexity: O(N)
Auxiliary Space:O(1) 



My Personal Notes
arrow_drop_up
Previous
Sum of maximum and minimum of Kth subset ordered by increasing subset sum
Next

Print all subsequences of a string
Recommended Articles
Page :
Length of Smallest subarray in range 1 to N with sum greater than a given value
31, Jan 20
Smallest subarray with sum greater than a given value
13, May 14
Length of longest subarray in which elements greater than K are more than elements not greater than K
14, Aug 19
Size of smallest subarray to be removed to make count of array elements greater and smaller than K equal
26, Aug 20
Smallest subarray from a given Array with sum greater than or equal to K
28, Jul 20
Smallest subarray from a given Array with sum greater than or equal to K | Set 2
03, Sep 20
Maximum subarray size, such that all subarrays of that size have sum less than k
30, Dec 16
First subarray having sum at least half the maximum sum of any subarray of size K
09, Sep 20
Minimum cost to convert all elements of a K-size subarray to 0 from given Ternary Array with subarray sum as cost
05, Jul 21
Partition array into two subarrays with every element in the right subarray strictly greater than every element in left subarray
01, Mar 21
Smallest subarray such that all elements are greater than K
19, Nov 18
Smallest subarray having an element with frequency greater than that of other elements
18, Jul 20
Maximum sum subarray of size K with sum less than X
01, Nov 20
Smallest subarray whose sum is multiple of array size
14, Nov 16
Count the subarray with sum strictly greater than the sum of remaining elements
26, Jan 20
Maximize product of min value of subarray and sum of subarray over all subarrays of length K
17, Nov 21
Check if a subarray exists with sum greater than the given Array
16, Apr 20
Highest and Smallest power of K less than and greater than equal to N respectively
02, Mar 20
Smallest pair of indices with product of subarray co-prime with product of the subarray on the left or right
23, Nov 20
Maximum sum subarray having sum less than or equal to given sum using Set
22, Apr 20
Maximum sum subarray having sum less than or equal to given sum
05, Aug 17
Minimize operations to make minimum value of one array greater than maximum value of the other
20, Jul 21
Smallest subarray whose product leaves remainder K when divided by size of the array
18, Jul 20
Minimum value to be assigned to the elements so that sum becomes greater than initial sum
09, Oct 19
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
kholiyagaurav121
@kholiyagaurav121
Vote for difficulty
Current difficulty :
Hard





Improved By :
Article Tags :
Practice Tags :
Report Issue

We use cookies to ensure you have the best browsing experience on our website. By using our site, you
acknowledge that you have read and understood our
Cookie Policy &
        Privacy Policy

Lightbox

Start Your Coding Journey Now!