Minimum Size Subarray Sum
<p>Given an array of positive integers <code>nums</code> and a positive integer <code>target</code>, return <em>the <strong>minimal length</strong> of a </em><span data-keyword="subarray-nonempty"><em>subarray</em></span><em> whose sum is greater than or equal to</em> <code>target</code>. If there is no such subarray, return <code>0</code> instead.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

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

<p><strong class="example">Example 2:</strong></p>

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

<p><strong class="example">Example 3:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= target &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> If you have figured out the <code>O(n)</code> solution, try coding another solution of which the time complexity is <code>O(n log(n))</code>.
Array; Binary Search; Prefix Sum; Sliding Window
impl Solution {
    pub fn min_sub_array_len(target: i32, nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let mut res = n + 1;
        let mut sum = 0;
        let mut i = 0;
        for j in 0..n {
            sum += nums[j];

            while sum >= target {
                res = res.min(j - i + 1);
                sum -= nums[i];
                i += 1;
            }
        }
        if res == n + 1 {
            return 0;
        }
        res as i32
    }
}
