Given an array **A[]** consisting of **N** integers and an integer **K**, the task is to find the length of the smallest subarray with sum greater than or equal to **K**. If no such subarray exists, print **-1**.

**Examples:**

Input:A[] = {2, -1, 2}, K = 3Output:3Explanation:

Sum of the given array is 3.

Hence, the smallest possible subarray satisfying the required condition is the entire array.

Therefore, the length is 3.

Input:A[] = {2, 1, 1, -4, 3, 1, -1, 2}, K = 5Output:4

**Naive Approach:**

The simplest approach to solve the problem is to generate all possible subarrays of the given array and check which subarray sum is greater than or equal to **K**. Among all such subarrays satisfying the condition, print the subarray having minimum length.**Time Complexity:**O(N^{2})**Auxiliary Space:** O(1)

**Efficient Approach:**

The above approach can be further optimized using Prefix Sum Array and Binary search. Follow the steps below:

- Initialize an array to store the
**Prefix sum**of the original array. - Hash the prefix sum array with the indices using a Map.
- If a
*greater sum with a lesser index*is already found, then there is no point of hashing a prefix sum which is smaller than the largest prefix sum obtained till now. Therefore, hash the*increasing order*of prefix sum. - Traversing the array and if any element is greater than or equal to
**K**, return 1 as the answer. - Otherwise, for every element, perform
**Binary Search**over the indices**(i, n-1)**in the**prefix sum array**to find the first index with sum at least**K**. - Return the minimum length subarray obtained from the above steps.

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 perform Binary Search` `// and return the smallest index with` `// sum greater than value` `int` `binary_search(map<` `int` `, vector<` `int` `> >& m,` ` ` `int` `value, ` `int` `index)` `{` ` ` `// Search the value in map` ` ` `auto` `it = m.lower_bound(value);` ` ` ` ` `// If all keys in the map` ` ` `// are less then value` ` ` `if` `(it == m.end())` ` ` `return` `0;` ` ` ` ` `// Check if the sum is found` ` ` `// at a greater index` ` ` `auto` `it1` ` ` `= lower_bound(it->second.begin(),` ` ` `it->second.end(), index);` ` ` ` ` `if` `((it1 - it->second.begin())` ` ` `!= it->second.size())` ` ` `return` `*it1;` ` ` ` ` `return` `0;` `}` ` ` `// Function to find the smallest subarray` `// with sum greater than equal to K` `int` `findSubarray(` `int` `arr[], ` `int` `n, ` `int` `k)` `{` ` ` ` ` `// Prefix sum array` ` ` `int` `pre_array[n];` ` ` ` ` `// Stores the hashes to prefix sum` ` ` `map<` `int` `, vector<` `int` `> > m;` ` ` ` ` `pre_array[0] = arr[0];` ` ` `m[pre_array[0]].push_back(0);` ` ` ` ` `// If any array element is` ` ` `// greater than equal to k` ` ` `if` `(arr[0] >= k)` ` ` `return` `1;` ` ` ` ` `int` `ans = INT_MAX;` ` ` `for` `(` `int` `i = 1; i < n; i++) {` ` ` ` ` `pre_array[i]` ` ` `= arr[i] + pre_array[i - 1];` ` ` ` ` `// If prefix sum exceeds K` ` ` `if` `(pre_array[i] >= k)` ` ` ` ` `// Update size of subarray` ` ` `ans = min(ans, i + 1);` ` ` ` ` `auto` `it = m.rbegin();` ` ` ` ` `// Hash prefix sum in` ` ` `// increasing order` ` ` `if` `(pre_array[i] >= it->first)` ` ` `m[pre_array[i]].push_back(i);` ` ` `}` ` ` ` ` `for` `(` `int` `i = 1; i < n; i++) {` ` ` ` ` `int` `temp` ` ` `= binary_search(m,` ` ` `pre_array[i - 1] + k,` ` ` `i);` ` ` `if` `(temp == 0)` ` ` `continue` `;` ` ` ` ` `// Update size of subarray` ` ` `ans = min(ans, temp - i + 1);` ` ` `}` ` ` ` ` `// If any subarray is found` ` ` `if` `(ans <= n)` ` ` `return` `ans;` ` ` ` ` `// If no such subarray exists` ` ` `return` `-1;` `}` ` ` `// Driver Code` `int` `main()` `{` ` ` `int` `arr[] = { 2, 1, 1, -4, 3, 1, -1, 2 };` ` ` ` ` `int` `k = 5;` ` ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` ` ` `cout << findSubarray(arr, n, k) << endl;` ` ` ` ` `return` `0;` `}` |

**Output:**

4

**Time Complexity:*** O(NlogN)***Auxiliary Space:*** O(N)*

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