# 209. Minimum Size Subarray Sum

**209。Minimum Size Subarray Sum**

Given an array of positive integers `nums` and a positive integer `target`, return *the **minimal length** of a* 

*subarray whose sum is greater than or equal to* `target`. If there is no such subarray, return `0` instead.

**Example 1:**

<pre><code><strong>Input: target = 7, nums = [2,3,1,2,4,3]
</strong><strong>Output: 2
</strong><strong>Explanation: The subarray [4,3] has the minimal length under the problem constraint.
</strong></code></pre>

**Example 2:**

<pre><code><strong>Input: target = 4, nums = [1,4,4]
</strong><strong>Output: 1
</strong></code></pre>

**Example 3:**

<pre><code><strong>Input: target = 11, nums = [1,1,1,1,1,1,1,1]
</strong><strong>Output: 0
</strong></code></pre>

**Constraints:**

* `1 <= target <= 109`
* `1 <= nums.length <= 105`
* `1 <= nums[i] <= 104`

**Follow up:** If you have figured out the `O(n)` solution, try coding another solution of which the time complexity is `O(n log(n))`.

***My Solutions：***

```
class Solution {
    public int minSubArrayLen(int target, int[] nums) {
        int min = nums.length; 
        boolean existsMin = false;
        int l = 0, sum = 0;
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            while (sum >= target) {
                min = Math.min(min, i - l + 1);
                existsMin = true;
                sum -= nums[l];
                l++;
            }
        }

        return existsMin ? min : 0;
    }
}
```