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

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

Naive Approach

The idea is to find all subarrays and in that choose those subarrays whose sum is greater than S and whose length is greater than K. After that from those subarrays, choose that subarray whose length is minimum. Then print the length of that subarray.

Steps that were to follow the above approach:

• Make a variable “ans” and initialize it with the maximum value
• After that run two nested loops to find all subarrays
• While finding all subarray calculate their size and sum of all elements of that subarray
• If the sum of all elements is greater than S and its size is greater than K, then update answer with minimum of answer and length of the subarray

Below is the code to implement the above approach:

## C++

 `// C++ program to implement the above approach` `#include ` `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)` `{` `   ``//To store answer` `   ``int` `ans=INT_MAX;` `  `  `  ``//Traverse the array` `   ``for``(``int` `i=0;iK && sum>S){` `               ``if``(size

## Java

 `import` `java.util.*;`   `public` `class` `Main` `{`   `  ``// 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) ` `  ``{`   `    ``// To store answer` `    ``int` `ans = Integer.MAX_VALUE;`   `    ``// Traverse the array` `    ``for` `(``int` `i = ``0``; i < N; i++) {` `      ``// To store size of subarray` `      ``int` `size = ``0``;` `      ``// To store sum of all elements of subarray` `      ``int` `sum = ``0``;` `      ``for` `(``int` `j = i; j < N; j++) {` `        ``size++;` `        ``sum += arr[j];`   `        ``// when size of subarray is greater than k and sum of all element` `        ``// is greater than S then if size is less than previously stored answer` `        ``// then update answer with size` `        ``if` `(size > K && sum > S) {` `          ``if` `(size < ans) {` `            ``ans = size;` `          ``}` `        ``}` `      ``}` `    ``}` `    ``return` `ans;` `  ``}`   `  ``// 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.println(smallestSubarray(K, S, arr, N));` `  ``}` `}`

## Python3

 `# Function to find the length of the smallest subarray of` `# size > K with sum greater than S` `def` `smallestSubarray(K, S, arr, N):` `  `  `    ``# To store answer` `    ``ans ``=` `float``(``'inf'``)`   `    ``# Traverse the array` `    ``for` `i ``in` `range``(N):` `      `  `        ``# To store size of subarray` `        ``size ``=` `0` `        `  `        ``# To store sum of all elements of subarray` `        ``sum` `=` `0` `        ``for` `j ``in` `range``(i, N):` `            ``size ``+``=` `1` `            ``sum` `+``=` `arr[j]`   `            ``# when size of subarray is greater than k and sum of all element` `            ``# is greater than S then if size is less than previously stored answer` `            ``# then update answer with size` `            ``if` `size > K ``and` `sum` `> S:` `                ``if` `size < ans:` `                    ``ans ``=` `size`   `    ``return` `ans`   `# Driver Code` `arr ``=` `[``1``, ``2``, ``3``, ``4``, ``5``]` `K ``=` `1` `S ``=` `8` `N ``=` `len``(arr)` `print``(smallestSubarray(K, S, arr, N))`

## Javascript

 `// Function to find the length of the smallest subarray of` `// size > K with sum greater than S` `function` `smallestSubarray(K, S, arr, N) {` `  ``// To store answer` `  ``let ans = Infinity;`   `  ``// Traverse the array` `  ``for` `(let i = 0; i < N; i++) {` `    ``// To store size of subarray` `    ``let size = 0;`   `    ``// To store sum of all elements of subarray` `    ``let sum = 0;` `    ``for` `(let j = i; j < N; j++) {` `      ``size += 1;` `      ``sum += arr[j];`   `      ``// when size of subarray is greater than k and sum of all element` `      ``// is greater than S then if size is less than previously stored answer` `      ``// then update answer with size` `      ``if` `(size > K && sum > S) {` `        ``if` `(size < ans) {` `          ``ans = size;` `        ``}` `      ``}` `    ``}` `  ``}` `  ``return` `ans;` `}`   `// Driver Code` `let arr = [1, 2, 3, 4, 5];` `let K = 1;` `let S = 8;` `let N = arr.length;` `console.log(smallestSubarray(K, S, arr, N));`

Output-

`2`

Time Complexity: O(N2), because of two nested for loops
Auxiliary Space:O(1) , because no extra space has been used

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 ` `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 current subarray` `    ``int` `end = 0;` `    ``// Store the sum of the current subarray` `    ``int` `currSum = arr;` `    ``// 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];` `        ``}` `        ``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);` `    ``cout << smallestSubarray(K, S, arr, N);` `}`   `// This code is contributed by Sania Kumari Gupta`

## C

 `// C program to implement the above approach` `#include ` `#include `   `// Find minimum between two numbers.` `int` `min(``int` `num1, ``int` `num2)` `{` `    ``return` `(num1 > num2) ? num2 : num1;` `}`   `// 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 current subarray` `    ``int` `end = 0;` `    ``// Store the sum of the current subarray` `    ``int` `currSum = arr;` `    ``// 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];` `        ``}` `        ``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);` `    ``printf``(``"%d"``, smallestSubarray(K, S, arr, N));` `}`   `// This code is contributed by Sania Kumari Gupta`

## 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 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 current subarray` `    ``int` `end = 0;`   `    ``// Store the sum of the` `    ``// current subarray` `    ``int` `currSum = arr;`   `    ``// 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

 ``

Output:

`2`

Time Complexity: O(N)
Auxiliary Space:O(1)

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