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
class Solution {
    public int minSubArrayLen(int target, int[] nums) {
        int n = nums.length;
        long[] s = new long[n + 1];
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + nums[i];
        }
        int ans = n + 1;
        for (int i = 0; i <= n; ++i) {
            int j = search(s, s[i] + target);
            if (j <= n) {
                ans = Math.min(ans, j - i);
            }
        }
        return ans <= n ? ans : 0;
    }

    private int search(long[] nums, long x) {
        int l = 0, r = nums.length;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}