Introduction

In this blog post, we will explore how to solve the LeetCode problem "668" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Nearly everyone has used the Multiplication Table. The multiplication table of size m x n is an integer matrix mat where mat[i][j] == i j (1-indexed). Given three integers m, n, and k, return the kth smallest element in the m x n multiplication table. Example 1: Input: m = 3, n = 3, k = 5 Output: 3 Explanation: The 5th smallest number is 3. Example 2: Input: m = 2, n = 3, k = 6 Output: 6 Explanation: The 6th smallest number is 6. Constraints: 1 <= m, n <= 3 104 1 <= k <= m * n

Explanation

Here's a breakdown of the solution approach, followed by the Python code:

• Binary Search: The core idea is to use binary search to efficiently find the kth smallest element. We search within the range [1, m * n], which represents the possible values in the multiplication table.
• Counting Elements: For each mid-value during binary search, we count how many elements in the multiplication table are less than or equal to the mid-value.

• Adjusting Search Range: Based on the count, we adjust the binary search range (left or right) to narrow down the search until we find the kth smallest element.

• Runtime Complexity: O(min(m, n) log(m n)). Storage Complexity: O(1).

Code

    def findKthNumber(m: int, n: int, k: int) -> int:
    """
    Finds the kth smallest number in an m x n multiplication table.

    Args:
        m: The number of rows in the multiplication table.
        n: The number of columns in the multiplication table.
        k: The desired rank of the element.

    Returns:
        The kth smallest element in the multiplication table.
    """

    def count_less_equal(x):
        """Counts the number of elements <= x in the multiplication table."""
        count = 0
        for i in range(1, m + 1):
            count += min(x // i, n)
        return count

    left, right = 1, m * n
    while left < right:
        mid = left + (right - left) // 2
        if count_less_equal(mid) < k:
            left = mid + 1
        else:
            right = mid
    return left