Introduction

In this blog post, we will explore how to solve the LeetCode problem "2245" 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

You are given a 2D integer array grid of size m x n, where each cell contains a positive integer. A cornered path is defined as a set of adjacent cells with at most one turn. More specifically, the path should exclusively move either horizontally or vertically up to the turn (if there is one), without returning to a previously visited cell. After the turn, the path will then move exclusively in the alternate direction: move vertically if it moved horizontally, and vice versa, also without returning to a previously visited cell. The product of a path is defined as the product of all the values in the path. Return the maximum number of trailing zeros in the product of a cornered path found in grid. Note: Horizontal movement means moving in either the left or right direction. Vertical movement means moving in either the up or down direction. Example 1: Input: grid = [[23,17,15,3,20],[8,1,20,27,11],[9,4,6,2,21],[40,9,1,10,6],[22,7,4,5,3]] Output: 3 Explanation: The grid on the left shows a valid cornered path. It has a product of 15 20 6 1 10 = 18000 which has 3 trailing zeros. It can be shown that this is the maximum trailing zeros in the product of a cornered path. The grid in the middle is not a cornered path as it has more than one turn. The grid on the right is not a cornered path as it requires a return to a previously visited cell. Example 2: Input: grid = [[4,3,2],[7,6,1],[8,8,8]] Output: 0 Explanation: The grid is shown in the figure above. There are no cornered paths in the grid that result in a product with a trailing zero. Constraints: m == grid.length n == grid[i].length 1 <= m, n <= 105 1 <= m * n <= 105 1 <= grid[i][j] <= 1000

Explanation

Here's a breakdown of the solution and the code:

• High-Level Approach:

  • Compute prefix sums of twos and fives along rows and columns to efficiently calculate the number of factors 2 and 5 in any straight path.
  • Iterate through each cell as a potential corner. For each cell, consider the four possible corner orientations and calculate the number of trailing zeros by combining the prefix sums of twos and fives along the corresponding rows and columns.
  • Return the maximum number of trailing zeros found.

• Complexity:

  • Runtime Complexity: O(m * n)
  • Storage Complexity: O(m * n)

Code

    def maxTrailingZeros(grid):
    """
    Calculates the maximum number of trailing zeros in the product of a cornered path in the given grid.

    Args:
        grid: A 2D integer array representing the grid.

    Returns:
        The maximum number of trailing zeros found.
    """

    def count_factors(num):
        """Counts the number of factors 2 and 5 in a number."""
        count2 = 0
        count5 = 0
        while num > 0 and num % 2 == 0:
            count2 += 1
            num //= 2
        while num > 0 and num % 5 == 0:
            count5 += 1
            num //= 5
        return count2, count5

    m, n = len(grid), len(grid[0])

    # Precompute prefix sums of 2s and 5s for rows and columns
    row_twos = [[0] * n for _ in range(m)]
    row_fives = [[0] * n for _ in range(m)]
    col_twos = [[0] * m for _ in range(n)]
    col_fives = [[0] * m for _ in range(n)]

    for i in range(m):
        for j in range(n):
            two_count, five_count = count_factors(grid[i][j])
            if j == 0:
                row_twos[i][j] = two_count
                row_fives[i][j] = five_count
            else:
                row_twos[i][j] = row_twos[i][j - 1] + two_count
                row_fives[i][j] = row_fives[i][j - 1] + five_count

            if i == 0:
                col_twos[i][j] = two_count
                col_fives[i][j] = five_count
            else:
                col_twos[i][j] = col_twos[i - 1][j] + two_count
                col_fives[i][j] = col_fives[i - 1][j] + five_count

    max_zeros = 0
    for i in range(m):
        for j in range(n):
            two_count, five_count = count_factors(grid[i][j])

            # Four possible corner orientations:
            # 1. Left and Up
            up_twos = col_twos[i][j]
            up_fives = col_fives[i][j]
            left_twos = row_twos[i][j]
            left_fives = row_fives[i][j]
            max_zeros = max(max_zeros, min(up_twos + left_twos - two_count, up_fives + left_fives - five_count))

            # 2. Left and Down
            down_twos = col_twos[m - 1][j] - col_twos[i][j] + two_count
            down_fives = col_fives[m - 1][j] - col_fives[i][j] + five_count
            max_zeros = max(max_zeros, min(down_twos + left_twos - two_count, down_fives + left_fives - five_count))

            # 3. Right and Up
            right_twos = row_twos[i][n - 1] - row_twos[i][j] + two_count
            right_fives = row_fives[i][n - 1] - row_fives[i][j] + five_count
            max_zeros = max(max_zeros, min(up_twos + right_twos - two_count, up_fives + right_fives - five_count))

            # 4. Right and Down
            max_zeros = max(max_zeros, min(down_twos + right_twos - two_count, down_fives + right_fives - five_count))

    return max_zeros