# Subarray Sum

Given an array of positive integers `nums`

and a positive integer `target`

, return the minimal length of a **contiguous subarray**`[nums`

of which the sum is greater than or equal to _{l}, nums_{l+1}, ..., nums_{r-1}, nums_{r}]`target`

. If there is no such subarray, return `0`

instead.

```
Example 1:
Input: target = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: The subarray [4,3] has the minimal length under the problem constraint.
Example 2:
Input: target = 4, nums = [1,4,4]
Output: 1
Example 3:
Input: target = 11, nums = [1,1,1,1,1,1,1,1]
Output: 0
Constraints:
1 <= target <= 109
1 <= nums.length <= 105
1 <= nums[i] <= 105
```

## Solution

```
fun minSubArrayLen(target: Int, nums: IntArray): Int {
var minLen = Int.MAX_VALUE
var windowStart = 0
var windowSum = 0
for (windowEnd in nums.indices) {
windowSum += nums[windowEnd]
while (windowSum >= target) {
minLen = kotlin.math.min(minLen, windowEnd + 1 - windowStart)
windowSum -= nums[windowStart++]
}
}
return minLen.takeIf { minLen != Int.MAX_VALUE } ?: 0
}
```

- First, we will add-up elements from the beginning of the array until their sum becomes greater than or equal to
`target`

. - These elements will constitute our sliding window. We are asked to find the smallest such window having a sum greater than or equal to
`target`

. We will remember the length of this window as the smallest window so far. - After this, we will keep adding one element in the sliding window (i.e. slide the window ahead) in a stepwise fashion.
- In each step, we will also try to shrink the window from the beginning. We will shrink the window until the windowâ€™s sum is smaller than
`target`

again. This is needed as we intend to find the smallest window. This shrinking will also happen in multiple steps; in each step, we will do two things:

- Check if the current window length is the smallest so far, and if so, remember its length.
- Subtract the first element of the window from the running sum to shrink the sliding window.