Solution Explanation:

This problem asks to find the size of any subarray within nums of length k where every element is greater than threshold / k. Two approaches are presented, both leveraging the concept of finding the longest subarray fulfilling the condition.

Solution 1: Union Find

This approach cleverly uses a Union-Find data structure to efficiently merge adjacent subarrays.

1. Initialization:

   • p: A parent array for the Union-Find, initially setting each element to itself (representing individual subarrays).
   • size: An array storing the size of each subarray (initially 1).
   • arr: A sorted array of pairs (num, index), sorted in descending order of num. This allows us to process the largest elements first.

2. Iteration and Merging:

   • The code iterates through arr (largest elements first).
   • For each element (v, i):
     • It checks if the left and right neighbors (i-1 and i+1) are already visited (vis). If so, it merges their subarrays using the merge function.
     • The merge function performs the Union-Find operation, updating the parent and size arrays.
     • If v > threshold / size[find(i)] (the current element is greater than the threshold divided by its subarray size), then it means a valid subarray is found, and its size is returned.
   • Finally, if no such subarray is found, it returns -1.

3. Union Find Functions:

   • find(x): Finds the root parent of element x (representing its subarray).
   • merge(a, b): Merges the subarrays containing elements a and b.

Time Complexity Analysis (Solution 1):

• Sorting arr takes O(N log N) time.
• The find and merge operations in Union Find take (amortized) O(α(N)) time, where α(N) is the inverse Ackermann function, which is practically constant for all realistic input sizes.
• The main loop iterates N times.
• Therefore, the overall time complexity is dominated by the sorting step: O(N log N).

Space Complexity Analysis (Solution 1):

• The space complexity is O(N) due to the arrays p, size, arr, and vis.

Solution 2: Monotonic Stack

This approach utilizes two monotonic stacks to efficiently find the left and right boundaries of the subarray for each element.

1. Left Boundary:

   • A monotonic decreasing stack stk is used.
   • The stack maintains indices of elements in descending order.
   • When an element v is processed, the stack is popped until the top element is smaller than v.
   • The left[i] array stores the index of the leftmost element that is smaller than nums[i].

2. Right Boundary:

   • A similar process is applied to find the right[i] array, using a monotonic decreasing stack iterating from right to left.

3. Verification:

   • The code iterates through the array, checking the condition v > threshold / k for each element, where k = right[i] - left[i] - 1. If this is true, it means a valid subarray of length k is found.

Time Complexity Analysis (Solution 2):

• Each element is pushed and popped from the stack at most once, resulting in a linear time complexity O(N) for each stack.
• Therefore, the overall time complexity is O(N).

Space Complexity Analysis (Solution 2):

• The space complexity is O(N) for the arrays left, right, and the stacks.

Conclusion:

Both solutions correctly solve the problem. Solution 2 (Monotonic Stack) is generally preferred because of its superior O(N) time complexity compared to Solution 1's O(N log N) time complexity. However, Solution 1's Union Find approach is a creative and elegant method that could be advantageous in slightly different variations of the problem. The choice depends on the specific constraints and performance requirements.