left pointer to 0 and sum to 0.nums array with an index i representing the end pointer of the current window.
nums[i] to sum.sum is greater than or equal to target:
ans to be the minimum of ans and the current subarray size i + 1 - left.nums[left] from sum and increment left.ans was updated during the process, return ans. Otherwise, return 0 if no such subarray was found.