Difference Between Ones and Zeros in Row and Column - Video Solutions
Difference Between Ones and Zeros in Row and Column | Intuition | Leetcode-2482
Difference Between Ones and Zeros in Row and Column - Video Solution
Watch 10 video solutions for Difference Between Ones and Zeros in Row and Column, a medium level problem involving Array, Matrix, Simulation. This walkthrough by codestorywithMIK has 3,893 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
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
ithrow beonesRowi. - Let the number of ones in the
jthcolumn beonesColj. - Let the number of zeros in the
ithrow bezerosRowi. - Let the number of zeros in the
jthcolumn bezerosColj. 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.lengthn == grid[i].length1 <= m, n <= 1051 <= m * n <= 105grid[i][j]is either0or1.
Approach Overview
Problem Overview: You are given a binary m x n grid. For each cell (i, j), compute the difference between the number of ones and zeros in its row and column combined. Formally, diff[i][j] = onesRow[i] + onesCol[j] - zerosRow[i] - zerosCol[j]. The task is to build the resulting matrix efficiently.
Approach 1: Precompute Row and Column Counts (Time: O(m*n), Space: O(m+n))
The straightforward strategy is to count how many 1s appear in every row and every column first. Iterate through the grid once to populate two arrays: rowOnes[m] and colOnes[n]. Since each row has n elements and each column has m, you can derive zeros using zerosRow = n - rowOnes[i] and zerosCol = m - colOnes[j]. Then iterate through the grid again and compute the difference for each cell using the formula. This approach is easy to reason about and keeps the logic clean because counting and result computation are separated. It’s a common technique when working with matrix problems that require repeated row/column statistics.
Approach 2: Optimized Counting in Single Pass (Time: O(m*n), Space: O(m+n))
You can slightly streamline the counting by collecting row and column counts during a single traversal of the grid. While iterating through every cell, increment rowOnes[i] and colOnes[j] whenever a 1 is encountered. After this pass, compute each result cell using a simplified formula: diff[i][j] = 2 * rowOnes[i] + 2 * colOnes[j] - m - n. This works because zerosRow = n - rowOnes and zerosCol = m - colOnes. The formula avoids explicitly computing zeros and reduces arithmetic per cell. The approach still runs in linear time relative to the grid size and is typical for problems involving aggregated counts in arrays and simulation style transformations.
Recommended for interviews: The precompute row/column counts approach is what most interviewers expect. It clearly demonstrates that you recognize repeated computations and replace them with cached counts. The optimized formula version shows deeper understanding of the relationship between ones and zeros and reduces redundant calculations while keeping the same O(m*n) complexity.
Complexity Analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Precompute Row and Column Counts | O(m*n) | O(m+n) | Best general solution. Easy to implement and clearly separates counting and result construction. |
| Optimized Counting with Derived Formula | O(m*n) | O(m+n) | Preferred when simplifying arithmetic and reducing repeated calculations using the derived formula. |