Is Difference Between Ones and Zeros in Row and Column easy or hard?

The problem is rated Medium on LeetCode but conceptually straightforward once you recognize that row and column counts can be reused. The main challenge is identifying the formula that combines row and column ones and zeros efficiently without recomputing counts for every cell.

Difference Between Ones and Zeros in Row and Column Python/Java solution

Python and Java implementations follow the same idea: compute rowOnes and colOnes arrays, then build the result matrix using the derived formula. Python typically uses list comprehensions or nested loops, while Java uses integer arrays and standard for-loops. Both implementations run in O(m*n) time and O(m+n) space.

How to solve Difference Between Ones and Zeros in Row and Column in O(n)?

Treat the matrix size as m*n elements and solve it in linear time relative to the grid size. Count ones in each row and column during a traversal, then compute each cell using the formula diff[i][j] = 2*rowOnes[i] + 2*colOnes[j] - m - n. This avoids recalculating counts for every cell and keeps the algorithm O(m*n).

What is the best approach for Difference Between Ones and Zeros in Row and Column?

The best approach precomputes the number of ones in every row and column. After collecting rowOnes and colOnes in O(m*n) time, each cell’s difference can be computed directly using a constant-time formula. This avoids recomputing row and column counts repeatedly and keeps the overall complexity linear in the size of the grid.

Is Difference Between Ones and Zeros in Row and Column asked at Google/Amazon/Meta?

Matrix counting and row/column aggregation problems are common across companies like Google, Amazon, and Meta. Variants that require precomputing row or column statistics appear frequently because they test grid traversal, space optimization, and recognizing reusable computations.

What data structure is used in Difference Between Ones and Zeros in Row and Column?

The solution primarily uses arrays to store row and column counts. Two auxiliary arrays, rowOnes and colOnes, track how many ones appear in each row and column. The final answer is stored in another matrix of the same size as the input grid.

What is the time complexity of Difference Between Ones and Zeros in Row and Column?

The optimal solution runs in O(m*n) time where m is the number of rows and n is the number of columns. One pass gathers row and column counts, and another pass builds the result matrix. Space complexity is O(m+n) for the auxiliary row and column arrays.

Description Hints 2 Solution Video 12:46

2482. Difference Between Ones and Zeros in Row and Column

Medium84.3% acceptance1 company

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 i^th row be onesRow_i.
Let the number of ones in the j^th column be onesCol_j.
Let the number of zeros in the i^th row be zerosRow_i.
Let the number of zeros in the j^th column be zerosCol_j.
diff[i][j] = onesRow_i + onesCol_j - zerosRow_i - zerosCol_j

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.

Asked at 1 company

Amazon

Topics

Hints open one at a time

Hint 1

You need to reuse information about a row or a column many times. Try storing it to avoid computing it multiple times.

Hint 2

Use an array to store the number of 1’s in each row and another array to store the number of 1’s in each column. Once you know the number of 1’s in each row or column, you can also easily calculate the number of 0’s.

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