Solution Explanation:

This problem asks to find the k-th smallest element in an m x n multiplication table. A brute-force approach of generating the entire table and sorting would be highly inefficient, especially for large m and n. Instead, we can leverage binary search for an optimized solution.

Approach:

The core idea is to use binary search on the range of possible values (1 to m*n). For each potential k-th smallest number ( mid in the code), we count how many numbers in the multiplication table are less than or equal to mid. This count is compared with k.

1. Binary Search: We initialize left to 1 and right to m*n, representing the smallest and largest possible values in the table. The binary search iteratively narrows down the search space.

2. Counting Numbers: For a given mid, we iterate through each row (i) of the multiplication table. For each row, we determine the maximum column index (j) such that i*j <= mid. This is efficiently calculated using min(mid/i, n). The total count of numbers less than or equal to mid is accumulated.

3. Comparison and Adjustment:

   • If the cnt (count of numbers <= mid) is greater than or equal to k, it means the k-th smallest number is in the lower half of the search space. We update right to mid to search in the lower half.
   • If cnt is less than k, the k-th smallest number lies in the upper half. We update left to mid + 1.

4. Result: The loop continues until left equals right. At this point, left (or right) represents the k-th smallest number in the multiplication table.

Time Complexity: O(m log(m*n))

The binary search takes O(log(mn)) iterations. In each iteration, the inner loop iterates through 'm' rows, performing a constant-time operation (min(mid/i, n)). Thus the dominant factor in the time complexity is m * log(mn).

Space Complexity: O(1)

The algorithm uses a constant amount of extra space to store variables like left, right, mid, and cnt. It doesn't require additional data structures proportional to the input size.

Code Explanation (Python3 Example):

class Solution:
    def findKthNumber(self, m: int, n: int, k: int) -> int:
        left, right = 1, m * n  # Initialize the search space
        while left < right:     # Binary search loop
            mid = (left + right) >> 1 #Efficient way to calculate average
            cnt = 0
            for i in range(1, m + 1): # Iterate through rows
                cnt += min(mid // i, n) # Count numbers <= mid in each row
 
            if cnt >= k: # If count >= k, kth smallest is in lower half
                right = mid
            else:        # Otherwise, in upper half
                left = mid + 1
        return left # left(or right) is the kth smallest number

The Java, C++, and Go code implementations follow the same logic and algorithmic steps, differing only in syntax. They all achieve the same time and space complexity.