Introduction

In this blog post, we will explore how to solve the LeetCode problem "2711" 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

Given a 2D grid of size m x n, you should find the matrix answer of size m x n. The cell answer[r][c] is calculated by looking at the diagonal values of the cell grid[r][c]: Let leftAbove[r][c] be the number of distinct values on the diagonal to the left and above the cell grid[r][c] not including the cell grid[r][c] itself. Let rightBelow[r][c] be the number of distinct values on the diagonal to the right and below the cell grid[r][c], not including the cell grid[r][c] itself. Then answer[r][c] = |leftAbove[r][c] - rightBelow[r][c]|. A matrix diagonal is a diagonal line of cells starting from some cell in either the topmost row or leftmost column and going in the bottom-right direction until the end of the matrix is reached. For example, in the below diagram the diagonal is highlighted using the cell with indices (2, 3) colored gray: Red-colored cells are left and above the cell. Blue-colored cells are right and below the cell. Return the matrix answer. Example 1: Input: grid = [[1,2,3],[3,1,5],[3,2,1]] Output: Output: [[1,1,0],[1,0,1],[0,1,1]] Explanation: To calculate the answer cells: answer left-above elements leftAbove right-below elements rightBelow |leftAbove - rightBelow| [0][0] [] 0 [grid[1][1], grid[2][2]] |{1, 1}| = 1 1 [0][1] [] 0 [grid[1][2]] |{5}| = 1 1 [0][2] [] 0 [] 0 0 [1][0] [] 0 [grid[2][1]] |{2}| = 1 1 [1][1] [grid[0][0]] |{1}| = 1 [grid[2][2]] |{1}| = 1 0 [1][2] [grid[0][1]] |{2}| = 1 [] 0 1 [2][0] [] 0 [] 0 0 [2][1] [grid[1][0]] |{3}| = 1 [] 0 1 [2][2] [grid[0][0], grid[1][1]] |{1, 1}| = 1 [] 0 1 Example 2: Input: grid = [[1]] Output: Output: [[0]] Constraints: m == grid.length n == grid[i].length 1 <= m, n, grid[i][j] <= 50

Explanation

Here's the breakdown of the solution:

• Precompute Diagonals: Instead of repeatedly traversing diagonals for each cell, we precompute the distinct elements in the left-above and right-below diagonals for all cells. This avoids redundant computations.
• Use Sets for Distinct Values: Sets are used to efficiently track distinct values along the diagonals. Adding to and checking the size of a set provide fast operations.

• Calculate Absolute Difference: After precomputing the necessary information, iterate through each cell, calculate the absolute difference between the number of distinct elements in the left-above and right-below diagonals, and store the result in the answer matrix.

• Runtime Complexity: O(m*n)

• Storage Complexity: O(m*n)

Code

    def diagonalDifference(grid):
    m = len(grid)
    n = len(grid[0])
    answer = [[0] * n for _ in range(m)]

    for r in range(m):
        for c in range(n):
            left_above = set()
            right_below = set()

            # Calculate leftAbove
            row = r - 1
            col = c - 1
            while row >= 0 and col >= 0:
                left_above.add(grid[row][col])
                row -= 1
                col -= 1

            # Calculate rightBelow
            row = r + 1
            col = c + 1
            while row < m and col < n:
                right_below.add(grid[row][col])
                row += 1
                col += 1

            answer[r][c] = abs(len(left_above) - len(right_below))

    return answer