#2482

Medium

LeetCode Problem Workspace

Difference Between Ones and Zeros in Row and Column

This problem requires computing a difference matrix based on the ones and zeros in each row and column of a binary matrix.

array matrix simulation

Problem Statement

You are given a 0-indexed m x n binary matrix, grid. Your task is to construct a new matrix diff with the same dimensions, where each element in diff[i][j] is calculated based on the number of ones and zeros in the ith row and jth column of grid. The formula for diff[i][j] is: onesRow[i] + onesCol[j] - zerosRow[i] - zerosCol[j], where onesRow[i] represents the number of ones in row i, onesCol[j] represents the number of ones in column j, zerosRow[i] represents the number of zeros in row i, and zerosCol[j] represents the number of zeros in column j.

Return the matrix diff. You can assume that both m and n are between 1 and 10^5, and the total number of elements in the grid does not exceed 10^5.

Examples

Example 1

Input: grid = [[0,1,1],[1,0,1],[0,0,1]]

Output: [[0,0,4],[0,0,4],[-2,-2,2]]

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]]

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.

Solution Approach

Efficient Row and Column Counting

First, iterate through the grid to calculate the number of ones and zeros for each row and column. Store these values separately in arrays so you don't have to compute them multiple times for each element in the result matrix.

Construct the Difference Matrix

For each element diff[i][j], calculate its value by applying the formula: onesRow[i] + onesCol[j] - zerosRow[i] - zerosCol[j]. This approach leverages the precomputed row and column data, which optimizes the solution.

Time and Space Complexity Optimization

Ensure that the approach maintains an overall time complexity of O(M * N), where M is the number of rows and N is the number of columns. By storing row and column sums, the space complexity is reduced to O(M + N), which is manageable within the problem's constraints.

Complexity Analysis

Metric	Value
Time	O(M * N)
Space	O(M + N)

The time complexity is O(M * N) due to the need to process each element in the matrix. The space complexity is O(M + N) since we store the row and column sums for ones and zeros separately.

What Interviewers Usually Probe

Ability to optimize calculations by reusing row and column data.
Skill in handling large matrix sizes efficiently.
Understanding of how to manage time and space complexity constraints.

Common Pitfalls or Variants

Common pitfalls

Not reusing row and column sums, leading to redundant calculations and inefficient solutions.
Incorrectly handling edge cases where either m or n is 1.
Overcomplicating the problem by not recognizing the need for precomputing sums.

Follow-up variants

Alter the formula to consider other metrics such as sum of all elements in a row and column.
Extend the problem to work with larger matrices and additional conditions on elements.
Introduce dynamic changes to the matrix after the initial construction of the difference matrix.

FAQ

What is the core idea behind solving the Difference Between Ones and Zeros in Row and Column problem?

The core idea is to efficiently compute a difference matrix by reusing precomputed row and column sums of ones and zeros.

How do I optimize my solution to handle large grid sizes in the Difference Between Ones and Zeros problem?

You can optimize your solution by precomputing row and column sums and storing them, avoiding repetitive calculations.

Can I solve the Difference Between Ones and Zeros problem without using extra space?

No, because you need to store row and column sums to efficiently compute the difference matrix. However, the space complexity can be minimized to O(M + N).

What common mistakes should I avoid in the Difference Between Ones and Zeros problem?

Avoid recalculating row and column sums for every element in the matrix. Additionally, make sure to handle edge cases like very small grids properly.

How does the solution handle edge cases in the Difference Between Ones and Zeros problem?

The solution handles edge cases by ensuring that row and column sums are computed efficiently, and the formula applies correctly even for the smallest grid sizes.

terminal

Solution

Solution 1: Simulation

We can solve this problem by simulating the process as described in the problem statement.

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class Solution:
    def onesMinusZeros(self, grid: List[List[int]]) -> List[List[int]]:
        m, n = len(grid), len(grid[0])
        rows = [0] * m
        cols = [0] * n
        for i, row in enumerate(grid):
            for j, v in enumerate(row):
                rows[i] += v
                cols[j] += v
        diff = [[0] * n for _ in range(m)]
        for i, r in enumerate(rows):
            for j, c in enumerate(cols):
                diff[i][j] = r + c - (n - r) - (m - c)
        return diff

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