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Difference Between Ones and Zeros in Row and Column

Difficulty: Medium

Problem Description

You are given a 0-indexed m x n binary matrix grid. A 0-indexed m x n difference matrix diff is created based on the number of ones and zeros in each row and column. The value of diff[i][j] is calculated using the formula: diff[i][j] = onesRow[i] + onesCol[j] - zerosRow[i] - zerosCol[j]. Return the difference matrix diff.

Key Insights

  • The number of ones and zeros in each row and column can be precomputed.
  • The formula for diff[i][j] directly utilizes precomputed values, allowing for efficient calculation.
  • The overall complexity can be reduced by using arrays to store counts of ones in rows and columns.

Space and Time Complexity

Time Complexity: O(m * n)
Space Complexity: O(m + n) for storing counts of ones in rows and columns.

Solution

To solve the problem, we will use two arrays to store the count of ones in each row and each column. After counting, we will calculate the difference matrix based on the precomputed values. The algorithm follows these steps:

  1. Initialize arrays to store the count of ones in each row and column.
  2. Traverse the grid to populate these arrays.
  3. Calculate the number of zeros based on the dimensions of the grid and the counts of ones.
  4. Iterate through each cell to compute the value for the difference matrix using the precomputed counts.

Code Solutions

def differenceMatrix(grid):
    m = len(grid)
    n = len(grid[0])
    
    onesRow = [0] * m
    onesCol = [0] * n
    
    # Count the number of ones in each row and column
    for i in range(m):
        for j in range(n):
            if grid[i][j] == 1:
                onesRow[i] += 1
                onesCol[j] += 1
    
    diff = [[0] * n for _ in range(m)]
    
    # Calculate the difference matrix
    for i in range(m):
        for j in range(n):
            zerosRow = n - onesRow[i]
            zerosCol = m - onesCol[j]
            diff[i][j] = onesRow[i] + onesCol[j] - zerosRow - zerosCol
            
    return diff
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