You are given an integer array nums and an integer threshold.
Find any subarray of nums of length k such that every element in the subarray is greater than threshold / k.
Return the size of any such subarray. If there is no such subarray, return -1.
A subarray is a contiguous non-empty sequence of elements within an array.
Input: nums = [1,3,4,3,1], threshold = 6 Output: 3 Explanation: The subarray [3,4,3] has a size of 3, and every element is greater than 6 / 3 = 2. Note that this is the only valid subarray.
Input: nums = [6,5,6,5,8], threshold = 7 Output: 1 Explanation: The subarray [8] has a size of 1, and 8 > 7 / 1 = 7. So 1 is returned. Note that the subarray [6,5] has a size of 2, and every element is greater than 7 / 2 = 3.5. Similarly, the subarrays [6,5,6], [6,5,6,5], [6,5,6,5,8] also satisfy the given conditions. Therefore, 2, 3, 4, or 5 may also be returned.
1 <= nums.length <= 1051 <= nums[i], threshold <= 109
implSolution{pubfnvalid_subarray_size(nums:Vec<i32>,threshold:i32) -> i32{letmut sublens = vec![0; nums.len()];letmut asc_stack = vec![];for i in0..nums.len(){while asc_stack.last().unwrap_or(&(-1,0)).1 >= nums[i]{ asc_stack.pop();} sublens[i] = i asi32 - asc_stack.last().unwrap_or(&(-1,0)).0; asc_stack.push((i asi32, nums[i]));} asc_stack.clear();for i in(0..nums.len()).rev(){while asc_stack.last().unwrap_or(&(nums.len()asi32,0)).1 >= nums[i]{ asc_stack.pop();} sublens[i] += asc_stack.last().unwrap_or(&(nums.len()asi32,0)).0 - i asi32 - 1; asc_stack.push((i asi32, nums[i]));if nums[i] > threshold / sublens[i]{return sublens[i];}} -1}}