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# Smallest subarray with sum greater than a given value

Given an array arr[] of integers and a number x, the task is to find the smallest subarray with a sum greater than the given value.

Examples:

```arr[] = {1, 4, 45, 6, 0, 19}
x  =  51
Output: 3
Minimum length subarray is {4, 45, 6}

arr[] = {1, 10, 5, 2, 7}
x  = 9
Output: 1
Minimum length subarray is {10}

arr[] = {1, 11, 100, 1, 0, 200, 3, 2, 1, 250}
x = 280
Output: 4
Minimum length subarray is {100, 1, 0, 200}

arr[] = {1, 2, 4}
x = 8
Output : Not Possible
Whole array sum is smaller than 8.```

Naive approach: A simple solution is to use two nested loops. The outer loop picks a starting element, the inner loop considers all elements (on right side of current start) as ending element. Whenever sum of elements between current start and end becomes more than the given number, update the result if current length is smaller than the smallest length so far.

Below is the implementation of the above approach:

## C++

 `# include ``using` `namespace` `std;` `// Returns length of smallest subarray with sum greater than x.``// If there is no subarray with given sum, then returns n+1``int` `smallestSubWithSum(``int` `arr[], ``int` `n, ``int` `x)``{``    ``//  Initialize length of smallest subarray as n+1``     ``int` `min_len = n + 1;` `     ``// Pick every element as starting point``     ``for` `(``int` `start=0; start x) ``return` `1;` `          ``// Try different ending points for current start``          ``for` `(``int` `end=start+1; end x && (end - start + 1) < min_len)``                 ``min_len = (end - start + 1);``          ``}``     ``}``     ``return` `min_len;``}` `/* Driver program to test above function */``int` `main()``{``    ``int` `arr1[] = {1, 4, 45, 6, 10, 19};``    ``int` `x = 51;``    ``int` `n1 = ``sizeof``(arr1)/``sizeof``(arr1);``    ``int` `res1 = smallestSubWithSum(arr1, n1, x);``    ``(res1 == n1+1)? cout << ``"Not possible\n"` `:``                    ``cout << res1 << endl;` `    ``int` `arr2[] = {1, 10, 5, 2, 7};``    ``int` `n2 = ``sizeof``(arr2)/``sizeof``(arr2);``    ``x  = 9;``    ``int` `res2 = smallestSubWithSum(arr2, n2, x);``    ``(res2 == n2+1)? cout << ``"Not possible\n"` `:``                    ``cout << res2 << endl;` `    ``int` `arr3[] = {1, 11, 100, 1, 0, 200, 3, 2, 1, 250};``    ``int` `n3 = ``sizeof``(arr3)/``sizeof``(arr3);``    ``x  = 280;``    ``int` `res3 = smallestSubWithSum(arr3, n3, x);``    ``(res3 == n3+1)? cout << ``"Not possible\n"` `:``                    ``cout << res3 << endl;` `    ``return` `0;``}`

## Java

 `import` `java.io.*;``class` `SmallestSubArraySum``{``    ``// Returns length of smallest subarray with sum greater than x.``    ``// If there is no subarray with given sum, then returns n+1``    ``static` `int` `smallestSubWithSum(``int` `arr[], ``int` `n, ``int` `x)``    ``{``        ``//  Initialize length of smallest subarray as n+1``        ``int` `min_len = n + ``1``;` `        ``// Pick every element as starting point``        ``for` `(``int` `start = ``0``; start < n; start++)``        ``{``            ``// Initialize sum starting with current start``            ``int` `curr_sum = arr[start];` `            ``// If first element itself is greater``            ``if` `(curr_sum > x)``                ``return` `1``;` `            ``// Try different ending points for current start``            ``for` `(``int` `end = start + ``1``; end < n; end++)``            ``{``                ``// add last element to current sum``                ``curr_sum += arr[end];` `                ``// If sum becomes more than x and length of``                ``// this subarray is smaller than current smallest``                ``// length, update the smallest length (or result)``                ``if` `(curr_sum > x && (end - start + ``1``) < min_len)``                    ``min_len = (end - start + ``1``);``            ``}``        ``}``        ``return` `min_len;``    ``}` `    ``// Driver program to test above functions``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr1[] = {``1``, ``4``, ``45``, ``6``, ``10``, ``19``};``        ``int` `x = ``51``;``        ``int` `n1 = arr1.length;``        ``int` `res1 = smallestSubWithSum(arr1, n1, x);``        ``if` `(res1 == n1+``1``)``           ``System.out.println(``"Not Possible"``);``        ``else``           ``System.out.println(res1);`  `        ``int` `arr2[] = {``1``, ``10``, ``5``, ``2``, ``7``};``        ``int` `n2 = arr2.length;``        ``x = ``9``;``        ``int` `res2 = smallestSubWithSum(arr2, n2, x);``        ``if` `(res2 == n2+``1``)``           ``System.out.println(``"Not Possible"``);``        ``else``           ``System.out.println(res2);` `        ``int` `arr3[] = {``1``, ``11``, ``100``, ``1``, ``0``, ``200``, ``3``, ``2``, ``1``, ``250``};``        ``int` `n3 = arr3.length;``        ``x = ``280``;``        ``int` `res3 = smallestSubWithSum(arr3, n3, x);``        ``if` `(res3 == n3+``1``)``           ``System.out.println(``"Not Possible"``);``        ``else``           ``System.out.println(res3);``    ``}``}` `// This code has been contributed by Mayank Jaiswal`

## Python3

 `# Python3 program to find Smallest``# subarray with sum greater``# than a given value` `# Returns length of smallest subarray``# with sum greater than x. If there``# is no subarray with given sum,``# then returns n+1``def` `smallestSubWithSum(arr, n, x):` `    ``# Initialize length of smallest``    ``# subarray as n+1``    ``min_len ``=` `n ``+` `1` `    ``# Pick every element as starting point``    ``for` `start ``in` `range``(``0``,n):``    ` `        ``# Initialize sum starting``        ``# with current start``        ``curr_sum ``=` `arr[start]` `        ``# If first element itself is greater``        ``if` `(curr_sum > x):``            ``return` `1` `        ``# Try different ending points``        ``# for current start``        ``for` `end ``in` `range``(start``+``1``,n):``        ` `            ``# add last element to current sum``            ``curr_sum ``+``=` `arr[end]` `            ``# If sum becomes more than x``            ``# and length of this subarray``            ``# is smaller than current smallest``            ``# length, update the smallest``            ``# length (or result)``            ``if` `curr_sum > x ``and` `(end ``-` `start ``+` `1``) < min_len:``                ``min_len ``=` `(end ``-` `start ``+` `1``)``        ` `    ``return` `min_len;`  `# Driver program to test above function */``arr1 ``=` `[``1``, ``4``, ``45``, ``6``, ``10``, ``19``]``x ``=` `51``n1 ``=` `len``(arr1)``res1 ``=` `smallestSubWithSum(arr1, n1, x);``if` `res1 ``=``=` `n1``+``1``:``    ``print``(``"Not possible"``)``else``:``    ``print``(res1)` `arr2 ``=` `[``1``, ``10``, ``5``, ``2``, ``7``]``n2 ``=` `len``(arr2)``x ``=` `9``res2 ``=` `smallestSubWithSum(arr2, n2, x);``if` `res2 ``=``=` `n2``+``1``:``    ``print``(``"Not possible"``)``else``:``    ``print``(res2)` `arr3 ``=` `[``1``, ``11``, ``100``, ``1``, ``0``, ``200``, ``3``, ``2``, ``1``, ``250``]``n3 ``=` `len``(arr3)``x ``=` `280``res3 ``=` `smallestSubWithSum(arr3, n3, x)``if` `res3 ``=``=` `n3``+``1``:``    ``print``(``"Not possible"``)``else``:``    ``print``(res3)``    ` `# This code is contributed by Smitha Dinesh Semwal`

## C#

 `// C# program to find Smallest``// subarray with sum greater``// than a given value``using` `System;` `class` `GFG``{``    ` `    ``// Returns length of smallest``    ``// subarray with sum greater``    ``// than x. If there is no``    ``// subarray with given sum,``    ``// then returns n+1``    ``static` `int` `smallestSubWithSum(``int` `[]arr,``                                  ``int` `n, ``int` `x)``    ``{``        ``// Initialize length of``        ``// smallest subarray as n+1``        ``int` `min_len = n + 1;` `        ``// Pick every element``        ``// as starting point``        ``for` `(``int` `start = 0; start < n; start++)``        ``{``            ``// Initialize sum starting``            ``// with current start``            ``int` `curr_sum = arr[start];` `            ``// If first element``            ``// itself is greater``            ``if` `(curr_sum > x)``                ``return` `1;` `            ``// Try different ending``            ``// points for current start``            ``for` `(``int` `end = start + 1;``                     ``end < n; end++)``            ``{``                ``// add last element``                ``// to current sum``                ``curr_sum += arr[end];` `                ``// If sum becomes more than``                ``// x and length of this``                ``// subarray is smaller than``                ``// current smallest length,``                ``// update the smallest``                ``// length (or result)``                ``if` `(curr_sum > x &&``                        ``(end - start + 1) < min_len)``                    ``min_len = (end - start + 1);``            ``}``        ``}``        ``return` `min_len;``    ``}` `    ``// Driver Code``    ``static` `public` `void` `Main ()``    ``{``        ``int` `[]arr1 = {1, 4, 45,``                      ``6, 10, 19};``        ``int` `x = 51;``        ``int` `n1 = arr1.Length;``        ``int` `res1 = smallestSubWithSum(arr1,``                                      ``n1, x);``        ``if` `(res1 == n1 + 1)``        ``Console.WriteLine(``"Not Possible"``);``        ``else``        ``Console.WriteLine(res1);`  `        ``int` `[]arr2 = {1, 10, 5, 2, 7};``        ``int` `n2 = arr2.Length;``        ``x = 9;``        ``int` `res2 = smallestSubWithSum(arr2,``                                      ``n2, x);``        ``if` `(res2 == n2 + 1)``        ``Console.WriteLine(``"Not Possible"``);``        ``else``        ``Console.WriteLine(res2);` `        ``int` `[]arr3 = {1, 11, 100, 1, 0,``                      ``200, 3, 2, 1, 250};``        ``int` `n3 = arr3.Length;``        ``x = 280;``        ``int` `res3 = smallestSubWithSum(arr3,``                                      ``n3, x);``        ``if` `(res3 == n3 + 1)``        ``Console.WriteLine(``"Not Possible"``);``        ``else``        ``Console.WriteLine(res3);``    ``}``}` `// This code is contributed by ajit`

## PHP

 ` ``\$x``) ``return` `1;` `        ``// Try different ending``        ``// points for current start``        ``for` `(``\$end``= ``\$start` `+ 1; ``\$end` `< ``\$n``; ``\$end``++)``        ``{``            ``// add last element``            ``// to current sum``            ``\$curr_sum` `+= ``\$arr``[``\$end``];` `            ``// If sum becomes more than``            ``// x and length of this subarray``            ``// is smaller than current``            ``// smallest length, update the``            ``// smallest length (or result)``            ``if` `(``\$curr_sum` `> ``\$x` `&&``                   ``(``\$end` `- ``\$start` `+ 1) < ``\$min_len``)``                ``\$min_len` `= (``\$end` `- ``\$start` `+ 1);``        ``}``    ``}``    ``return` `\$min_len``;``}` `// Driver Code``\$arr1` `= ``array` `(1, 4, 45,``               ``6, 10, 19);``\$x` `= 51;``\$n1` `= sizeof(``\$arr1``);``\$res1` `= smallestSubWithSum(``\$arr1``, ``\$n1``, ``\$x``);` `if` `((``\$res1` `== ``\$n1` `+ 1) == true)``    ``echo` `"Not possible\n"` `;``else``    ``echo` `\$res1` `, ``"\n"``;` `\$arr2` `= ``array``(1, 10, 5, 2, 7);``\$n2` `= sizeof(``\$arr2``);``\$x` `= 9;``\$res2` `= smallestSubWithSum(``\$arr2``, ``\$n2``, ``\$x``);` `if` `((``\$res2` `== ``\$n2` `+ 1) == true)``    ``echo` `"Not possible\n"` `;``else``    ``echo` `\$res2` `, ``"\n"``;` `\$arr3` `= ``array` `(1, 11, 100, 1, 0,``               ``200, 3, 2, 1, 250);``\$n3` `= sizeof(``\$arr3``);``\$x` `= 280;``\$res3` `= smallestSubWithSum(``\$arr3``, ``\$n3``, ``\$x``);``if` `((``\$res3` `== ``\$n3` `+ 1) == true)``    ``echo` `"Not possible\n"` `;``else``    ``echo` `\$res3` `, ``"\n"``;` `// This code is contributed by ajit``?>`

## Javascript

 ``

Output

```3
1
4
```

Time Complexity: O(n2).
Auxiliary Space: O(1)

Efficient Solution: This problem can be solved in O(n) time using the idea used in this post.

## C++14

 `// O(n) solution for finding smallest subarray with sum``// greater than x``#include ``using` `namespace` `std;` `// Returns length of smallest subarray with sum greater than``// x. If there is no subarray with given sum, then returns``// n+1``int` `smallestSubWithSum(``int` `arr[], ``int` `n, ``int` `x)``{``    ``// Initialize current sum and minimum length``    ``int` `curr_sum = 0, min_len = n + 1;` `    ``// Initialize starting and ending indexes``    ``int` `start = 0, end = 0;``    ``while` `(end < n) {``        ``// Keep adding array elements while current sum``        ``// is smaller than or equal to x``        ``while` `(curr_sum <= x && end < n)``            ``curr_sum += arr[end++];` `        ``// If current sum becomes greater than x.``        ``while` `(curr_sum > x && start < n) {``            ``// Update minimum length if needed``            ``if` `(end - start < min_len)``                ``min_len = end - start;` `            ``// remove starting elements``            ``curr_sum -= arr[start++];``        ``}``    ``}``    ``return` `min_len;``}` `/* Driver program to test above function */``int` `main()``{``    ``int` `arr1[] = { 1, 4, 45, 6, 10, 19 };``    ``int` `x = 51;``    ``int` `n1 = ``sizeof``(arr1) / ``sizeof``(arr1);``    ``int` `res1 = smallestSubWithSum(arr1, n1, x);``    ``(res1 == n1 + 1) ? cout << ``"Not possible\n"``                     ``: cout << res1 << endl;` `    ``int` `arr2[] = { 1, 10, 5, 2, 7 };``    ``int` `n2 = ``sizeof``(arr2) / ``sizeof``(arr2);``    ``x = 9;``    ``int` `res2 = smallestSubWithSum(arr2, n2, x);``    ``(res2 == n2 + 1) ? cout << ``"Not possible\n"``                     ``: cout << res2 << endl;` `    ``int` `arr3[] = { 1, 11, 100, 1, 0, 200, 3, 2, 1, 250 };``    ``int` `n3 = ``sizeof``(arr3) / ``sizeof``(arr3);``    ``x = 280;``    ``int` `res3 = smallestSubWithSum(arr3, n3, x);``    ``(res3 == n3 + 1) ? cout << ``"Not possible\n"``                     ``: cout << res3 << endl;` `    ``return` `0;``}`

## Java

 `// O(n) solution for finding smallest subarray with sum``// greater than x``import` `java.io.*;``class` `SmallestSubArraySum {``    ``// Returns length of smallest subarray with sum greater``    ``// than x. If there is no subarray with given sum, then``    ``// returns n+1``    ``static` `int` `smallestSubWithSum(``int` `arr[], ``int` `n, ``int` `x)``    ``{``        ``// Initialize current sum and minimum length``        ``int` `curr_sum = ``0``, min_len = n + ``1``;` `        ``// Initialize starting and ending indexes``        ``int` `start = ``0``, end = ``0``;``        ``while` `(end < n) {``            ``// Keep adding array elements while current sum``            ``// is smaller than or equal to x``            ``while` `(curr_sum <= x && end < n)``                ``curr_sum += arr[end++];` `            ``// If current sum becomes greater than x.``            ``while` `(curr_sum > x && start < n) {``                ``// Update minimum length if needed``                ``if` `(end - start < min_len)``                    ``min_len = end - start;` `                ``// remove starting elements``                ``curr_sum -= arr[start++];``            ``}``        ``}``        ``return` `min_len;``    ``}``    ``// Driver program to test above functions``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr1[] = { ``1``, ``4``, ``45``, ``6``, ``10``, ``19` `};``        ``int` `x = ``51``;``        ``int` `n1 = arr1.length;``        ``int` `res1 = smallestSubWithSum(arr1, n1, x);``        ``if` `(res1 == n1 + ``1``)``            ``System.out.println(``"Not Possible"``);``        ``else``            ``System.out.println(res1);` `        ``int` `arr2[] = { ``1``, ``10``, ``5``, ``2``, ``7` `};``        ``int` `n2 = arr2.length;``        ``x = ``9``;``        ``int` `res2 = smallestSubWithSum(arr2, n2, x);``        ``if` `(res2 == n2 + ``1``)``            ``System.out.println(``"Not Possible"``);``        ``else``            ``System.out.println(res2);` `        ``int` `arr3[]``            ``= { ``1``, ``11``, ``100``, ``1``, ``0``, ``200``, ``3``, ``2``, ``1``, ``250` `};``        ``int` `n3 = arr3.length;``        ``x = ``280``;``        ``int` `res3 = smallestSubWithSum(arr3, n3, x);``        ``if` `(res3 == n3 + ``1``)``            ``System.out.println(``"Not Possible"``);``        ``else``            ``System.out.println(res3);``    ``}``}` `// This code has been contributed by Mayank Jaiswal`

## Python3

 `# O(n) solution for finding smallest``# subarray with sum greater than x` `# Returns length of smallest subarray``# with sum greater than x. If there``# is no subarray with given sum, then``# returns n + 1`  `def` `smallestSubWithSum(arr, n, x):` `    ``# Initialize current sum and minimum length``    ``curr_sum ``=` `0``    ``min_len ``=` `n ``+` `1` `    ``# Initialize starting and ending indexes``    ``start ``=` `0``    ``end ``=` `0``    ``while` `(end < n):` `        ``# Keep adding array elements while current``        ``# sum is smaller than or equal to x``        ``while` `(curr_sum <``=` `x ``and` `end < n):``            ``curr_sum ``+``=` `arr[end]``            ``end ``+``=` `1` `        ``# If current sum becomes greater than x.``        ``while` `(curr_sum > x ``and` `start < n):` `            ``# Update minimum length if needed``            ``if` `(end ``-` `start < min_len):``                ``min_len ``=` `end ``-` `start` `            ``# remove starting elements``            ``curr_sum ``-``=` `arr[start]``            ``start ``+``=` `1` `    ``return` `min_len`  `# Driver program``arr1 ``=` `[``1``, ``4``, ``45``, ``6``, ``10``, ``19``]``x ``=` `51``n1 ``=` `len``(arr1)``res1 ``=` `smallestSubWithSum(arr1, n1, x)``print``(``"Not possible"``) ``if` `(res1 ``=``=` `n1 ``+` `1``) ``else` `print``(res1)` `arr2 ``=` `[``1``, ``10``, ``5``, ``2``, ``7``]``n2 ``=` `len``(arr2)``x ``=` `9``res2 ``=` `smallestSubWithSum(arr2, n2, x)``print``(``"Not possible"``) ``if` `(res2 ``=``=` `n2 ``+` `1``) ``else` `print``(res2)` `arr3 ``=` `[``1``, ``11``, ``100``, ``1``, ``0``, ``200``, ``3``, ``2``, ``1``, ``250``]``n3 ``=` `len``(arr3)``x ``=` `280``res3 ``=` `smallestSubWithSum(arr3, n3, x)``print``(``"Not possible"``) ``if` `(res3 ``=``=` `n3 ``+` `1``) ``else` `print``(res3)` `# This code is contributed by``# Smitha Dinesh Semwal`

## C#

 `// O(n) solution for finding``// smallest subarray with sum``// greater than x``using` `System;` `class` `GFG {` `    ``// Returns length of smallest``    ``// subarray with sum greater``    ``// than x. If there is no``    ``// subarray with given sum,``    ``// then returns n+1``    ``static` `int` `smallestSubWithSum(``int``[] arr, ``int` `n, ``int` `x)``    ``{``        ``// Initialize current``        ``// sum and minimum length``        ``int` `curr_sum = 0, min_len = n + 1;` `        ``// Initialize starting``        ``// and ending indexes``        ``int` `start = 0, end = 0;``        ``while` `(end < n) {``            ``// Keep adding array elements``            ``// while current sum is smaller``            ``// than or equal to x``            ``while` `(curr_sum <= x && end < n)``                ``curr_sum += arr[end++];` `            ``// If current sum becomes``            ``// greater than x.``            ``while` `(curr_sum > x && start < n) {``                ``// Update minimum``                ``// length if needed``                ``if` `(end - start < min_len)``                    ``min_len = end - start;` `                ``// remove starting elements``                ``curr_sum -= arr[start++];``            ``}``        ``}``        ``return` `min_len;``    ``}` `    ``// Driver Code``    ``static` `public` `void` `Main()``    ``{``        ``int``[] arr1 = { 1, 4, 45, 6, 10, 19 };``        ``int` `x = 51;``        ``int` `n1 = arr1.Length;``        ``int` `res1 = smallestSubWithSum(arr1, n1, x);``        ``if` `(res1 == n1 + 1)``            ``Console.WriteLine(``"Not Possible"``);``        ``else``            ``Console.WriteLine(res1);` `        ``int``[] arr2 = { 1, 10, 5, 2, 7 };``        ``int` `n2 = arr2.Length;``        ``x = 9;``        ``int` `res2 = smallestSubWithSum(arr2, n2, x);``        ``if` `(res2 == n2 + 1)``            ``Console.WriteLine(``"Not Possible"``);``        ``else``            ``Console.WriteLine(res2);` `        ``int``[] arr3``            ``= { 1, 11, 100, 1, 0, 200, 3, 2, 1, 250 };``        ``int` `n3 = arr3.Length;``        ``x = 280;``        ``int` `res3 = smallestSubWithSum(arr3, n3, x);``        ``if` `(res3 == n3 + 1)``            ``Console.WriteLine(``"Not Possible"``);``        ``else``            ``Console.WriteLine(res3);``    ``}``}` `// This code is contributed by akt_mit`

## PHP

 ` ``\$x` `&&``               ``\$start` `< ``\$n``)``        ``{``            ``// Update minimum``            ``// length if needed``            ``if` `(``\$end` `- ``\$start` `< ``\$min_len``)``                ``\$min_len` `= ``\$end` `- ``\$start``;` `            ``// remove starting elements``            ``\$curr_sum` `-= ``\$arr``[``\$start``++];``        ``}``    ``}``    ``return` `\$min_len``;``}` `// Driver Code``\$arr1` `= ``array``(1, 4, 45,``              ``6, 10, 19);``\$x` `= 51;``\$n1` `= sizeof(``\$arr1``);``\$res1` `= smallestSubWithSum(``\$arr1``,`` ` `                           ``\$n1``, ``\$x``);``if``(``\$res1` `== ``\$n1` `+ 1)``echo` `"Not possible\n"` `;``else``echo` `\$res1` `,``"\n"``;` `\$arr2` `= ``array``(1, 10, 5, 2, 7);``\$n2` `= sizeof(``\$arr2``);``\$x` `= 9;``\$res2` `= smallestSubWithSum(``\$arr2``,``                           ``\$n2``, ``\$x``);``if``(``\$res2` `== ``\$n2` `+ 1)``echo` `"Not possible\n"` `;``else``echo` `\$res2``,``"\n"``;` `\$arr3` `= ``array``(1, 11, 100, 1, 0,``              ``200, 3, 2, 1, 250);``\$n3` `= sizeof(``\$arr3``);``\$x` `= 280;``\$res3` `= smallestSubWithSum(``\$arr3``,``                           ``\$n3``, ``\$x``);``                           ` `if``(``\$res3` `== ``\$n3` `+ 1)``echo` `"Not possible\n"` `;``else``echo` `\$res3``, ``"\n"``;` `// This code is contributed by ajit``?>`

## Javascript

 ``

Output

```3
1
4
```

Time Complexity: O(n).
Auxiliary Space: O(1)

Another Approach: Binary Search

1. First calculates the cumulative sum of the vector elements and stores them in the sums vector.
2. Then iterates through the sums vector and finds the lower bound of the target sum for each possible subarray.
3. If the lower bound is found and it’s not equal to the target sum (i.e., the subarray sum is greater than the target),
4. Calculates the length of the subarray and updates the ans variable if the length is smaller than the current value.
5. Finally, returns the ans value or 0 if ans was not updated.

## C++

 `// O(n log(n) solution for finding smallest subarray with``// sum greater than x``#include ``using` `namespace` `std;` `int` `smallestSubArrayLen(``int` `target, vector<``int``>& nums)``{``    ``// Get the length of the input vector``    ``int` `n = nums.size();``    ``// If the vector is empty, return 0``    ``if` `(n == 0)``        ``return` `0;``    ``// Initialize the minimum subarray length to INT_MAX-1``    ``int` `ans = INT_MAX - 1;``    ``// Create a new vector "sums" with size n+1, initialized``    ``// to all zeros``    ``vector<``int``> sums(n + 1, 0);``    ``// Compute the running sum of nums and store it in``    ``// "sums"``    ``for` `(``int` `i = 1; i <= n; i++)``        ``sums[i] = sums[i - 1] + nums[i - 1];``    ``// Iterate through each starting index i``    ``for` `(``int` `i = 1; i <= n; i++) {``        ``// Calculate the target sum for the subarray``        ``// starting at index i``        ``int` `to_find = target + sums[i - 1];``        ``// Find the first element in "sums" that is >=``        ``// to_find``        ``auto` `bound = lower_bound(sums.begin(), sums.end(),``                                 ``to_find);``        ``// If such an element is found and it is not equal``        ``// to to_find itself``        ``if` `(bound != sums.end() && *bound != to_find) {``            ``// Compute the length of the subarray and update``            ``// ans if necessary``            ``int` `len = bound - (sums.begin() + i - 1);``            ``ans = min(ans, len);``        ``}``    ``}``    ``// Return ans if it was updated, otherwise return 0``    ``return` `(ans != INT_MAX - 1) ? ans : 0;``}` `/* Driver program to test above function */``int` `main()``{``    ``vector<``int``> arr1 = { 1, 4, 45, 6, 10, 19 };``    ``int` `target1 = 51;``    ``cout << ``"Length of Smallest Subarray :"``         ``<< smallestSubArrayLen(target1, arr1) << endl;` `    ``vector<``int``> arr2 = { 1, 10, 5, 2, 7 };``    ``int` `target2 = 9;``    ``cout << ``"Length of Smallest Subarray :"``         ``<< smallestSubArrayLen(target2, arr2) << endl;` `    ``vector<``int``> arr3 = { 1, 1, 1, 1, 1, 1, 1, 1 };``    ``int` `target3 = 11;``    ``cout << ``"Length of Smallest Subarray :"``         ``<< smallestSubArrayLen(target3, arr3) << endl;` `    ``vector<``int``> arr4``        ``= { 1, 11, 100, 1, 0, 200, 3, 2, 1, 250 };``    ``int` `target4 = 280;``    ``cout << ``"Length of Smallest Subarray :"``         ``<< smallestSubArrayLen(target4, arr4) << endl;` `    ``return` `0;``}`

Output

```Length of Smallest Subarray :3
Length of Smallest Subarray :1
Length of Smallest Subarray :0
Length of Smallest Subarray :4
```

Time Complexity: O (n log(n)).
Auxiliary Space: O(n)

Thanks to Ankit and Nitin for suggesting this optimized solution.

How to handle negative numbers?

The above solution may not work if input array contains negative numbers. For example arr[] = {- 8, 1, 4, 2, -6}. To handle negative numbers, add a condition to ignore subarrays with negative sums. We can use the solution discussed in Find subarray with given sum with negatives allowed in constant space