What is the brute force solution for Difference Between Ones and Zeros in Row and Column?

The brute force approach for Difference Between Ones and Zeros in Row and Column is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we calculate the number of ones and zeros for each row and column separately for every cell in the matrix. This is straightforward but inefficient as it recalculates values multiple times. The optimal Optimal Solution approach improves this to O(m * n).

What algorithm or data structure is used to solve Difference Between Ones and Zeros in Row and Column?

Difference Between Ones and Zeros in Row and Column uses the following concepts: Array, Matrix, Simulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Difference Between Ones and Zeros in Row and Column?

Explain your thought process clearly while coding. Discuss potential optimizations as you write your brute force solution. Practice similar problems to recognize patterns.

What are common mistakes when solving Difference Between Ones and Zeros in Row and Column?

Not counting zeros correctly after counting ones. Forgetting to initialize arrays for counts.

#2482

Difference Between Ones and Zeros in Row and Column

Medium

ArrayMatrixSimulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(m * n)
Space	O(1)	O(m + n)

💡

Intuition

Time O(m * n)Space O(m + n)

In the optimal solution, we first calculate the number of ones in each row and column in a single pass. This allows us to reuse these counts when calculating the difference matrix, significantly reducing redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Create two arrays to store the count of ones in each row and each column.
2Step 2: Traverse the grid to populate these arrays.
3Step 3: Use the counts from the arrays to compute the diff matrix in a single pass.

solution.py18 lines

1# Full working Python code
2
3def differenceMatrix(grid):
4    m, n = len(grid), len(grid[0])
5    onesRow = [0] * m
6    onesCol = [0] * n
7    diff = [[0] * n for _ in range(m)]
8    for i in range(m):
9        for j in range(n):
10            if grid[i][j] == 1:
11                onesRow[i] += 1
12                onesCol[j] += 1
13    for i in range(m):
14        for j in range(n):
15            zerosRow = n - onesRow[i]
16            zerosCol = m - onesCol[j]
17            diff[i][j] = onesRow[i] + onesCol[j] - zerosRow - zerosCol
18    return diff

ℹ

Complexity note: The time complexity is O(m * n) because we make a single pass through the grid to count ones and then another pass to compute the diff matrix. The space complexity is O(m + n) due to the two arrays used to store counts.

1Reusing computed values can significantly reduce time complexity.
2Understanding how to manipulate counts can simplify complex calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.