Introduction

In this blog post, we will explore how to solve the LeetCode problem "2482" 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 0-indexed m x n binary matrix grid. A 0-indexed m x n difference matrix diff is created with the following procedure: Let the number of ones in the ith row be onesRowi. Let the number of ones in the jth column be onesColj. Let the number of zeros in the ith row be zerosRowi. Let the number of zeros in the jth column be zerosColj. diff[i][j] = onesRowi + onesColj - zerosRowi - zerosColj Return the difference matrix diff. Example 1: Input: grid = [[0,1,1],[1,0,1],[0,0,1]] Output: [[0,0,4],[0,0,4],[-2,-2,2]] Explanation: - diff[0][0] = onesRow0 + onesCol0 - zerosRow0 - zerosCol0 = 2 + 1 - 1 - 2 = 0 - diff[0][1] = onesRow0 + onesCol1 - zerosRow0 - zerosCol1 = 2 + 1 - 1 - 2 = 0 - diff[0][2] = onesRow0 + onesCol2 - zerosRow0 - zerosCol2 = 2 + 3 - 1 - 0 = 4 - diff[1][0] = onesRow1 + onesCol0 - zerosRow1 - zerosCol0 = 2 + 1 - 1 - 2 = 0 - diff[1][1] = onesRow1 + onesCol1 - zerosRow1 - zerosCol1 = 2 + 1 - 1 - 2 = 0 - diff[1][2] = onesRow1 + onesCol2 - zerosRow1 - zerosCol2 = 2 + 3 - 1 - 0 = 4 - diff[2][0] = onesRow2 + onesCol0 - zerosRow2 - zerosCol0 = 1 + 1 - 2 - 2 = -2 - diff[2][1] = onesRow2 + onesCol1 - zerosRow2 - zerosCol1 = 1 + 1 - 2 - 2 = -2 - diff[2][2] = onesRow2 + onesCol2 - zerosRow2 - zerosCol2 = 1 + 3 - 2 - 0 = 2 Example 2: Input: grid = [[1,1,1],[1,1,1]] Output: [[5,5,5],[5,5,5]] Explanation: - diff[0][0] = onesRow0 + onesCol0 - zerosRow0 - zerosCol0 = 3 + 2 - 0 - 0 = 5 - diff[0][1] = onesRow0 + onesCol1 - zerosRow0 - zerosCol1 = 3 + 2 - 0 - 0 = 5 - diff[0][2] = onesRow0 + onesCol2 - zerosRow0 - zerosCol2 = 3 + 2 - 0 - 0 = 5 - diff[1][0] = onesRow1 + onesCol0 - zerosRow1 - zerosCol0 = 3 + 2 - 0 - 0 = 5 - diff[1][1] = onesRow1 + onesCol1 - zerosRow1 - zerosCol1 = 3 + 2 - 0 - 0 = 5 - diff[1][2] = onesRow1 + onesCol2 - zerosRow1 - zerosCol2 = 3 + 2 - 0 - 0 = 5 Constraints: m == grid.length n == grid[i].length 1 <= m, n <= 105 1 <= m * n <= 105 grid[i][j] is either 0 or 1.

Explanation

Here's the breakdown of the solution:

• Precompute Row and Column Sums: Calculate the number of ones in each row (onesRow) and each column (onesCol) beforehand. This avoids redundant counting within the nested loops.
• Calculate Zeros from Ones: Derive the number of zeros in each row (zerosRow) and each column (zerosCol) using the row/column length and the precomputed onesRow and onesCol values.

• Compute the Difference Matrix: Iterate through the grid and apply the formula diff[i][j] = onesRow[i] + onesCol[j] - zerosRow[i] - zerosCol[j] to construct the diff matrix.

• Runtime Complexity: O(m * n), where m is the number of rows and n is the number of columns.

• Storage Complexity: O(m + n) due to the onesRow, onesCol arrays. The diff matrix itself contributes O(m * n), but since the problem requires returning it, it's considered part of the problem's inherent output size rather than auxiliary space.

Code

    def differenceOfDistinctValues(grid):
    m = len(grid)
    n = len(grid[0])

    onesRow = [0] * m
    onesCol = [0] * n

    for i in range(m):
        for j in range(n):
            if grid[i][j] == 1:
                onesRow[i] += 1
                onesCol[j] += 1

    zerosRow = [n - x for x in onesRow]
    zerosCol = [m - x for x in onesCol]

    diff = [[0] * n for _ in range(m)]
    for i in range(m):
        for j in range(n):
            diff[i][j] = onesRow[i] + onesCol[j] - zerosRow[i] - zerosCol[j]

    return diff