<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> </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> </p> <p><strong>Constraints:</strong></p> <ul> <li><code>1 <= target <= 10<sup>9</sup></code></li> <li><code>1 <= nums.length <= 10<sup>5</sup></code></li> <li><code>1 <= nums[i] <= 10<sup>4</sup></code></li> </ul> <p> </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>.