The Difference Between Ones and Zeros in Row and Column coding problem tasks you with transforming an binary matrix into a new "difference matrix." For each cell in the output matrix, the value is calculated as: onesRow_i + onesCol_j - zerosRow_i - zerosCol_j. Here, onesRow_i is the number of 1s in the -th row, and zerosRow_i is the number of 0s in that row (similar definitions apply for columns).
Companies like Apple and Amazon use this to test basic Matrix interview patterns and simulation efficiency. While the formula seems complex, the problem tests if you can identify redundant calculations. A naive approach would count 1s and 0s for every single cell, leading to complexity. A more optimized approach uses precomputation to achieve .
The pattern used here is Precomputation (Simulation). You perform one pass over the matrix to pre-calculate the counts of 1s for every row and every column. Since the total number of elements in a row or column is fixed, zerosCount can be derived as total - onesCount. After precomputing these four arrays (or just two), you perform a second pass to populate the difference matrix using the provided formula in per cell.
Suppose grid is:
0 1 1
1 0 1
onesRow = [2, 2].onesCol = [1, 1, 2].onesRow[0] = 2, zerosRow[0] = 3 - 2 = 1.onesCol[0] = 1, zerosCol[0] = 2 - 1 = 1.In matrix problems, always ask: "Can I pre-calculate row and column properties?" Most matrix transformation problems can be reduced from to by using linear arrays to cache sums, counts, or flags for each row and column.
| Title | Difficulty | Topics | LeetCode |
|---|---|---|---|
| Spiral Matrix III | Medium | Solve | |
| Spiral Matrix II | Medium | Solve | |
| Count Unguarded Cells in the Grid | Medium | Solve | |
| Game of Life | Medium | Solve | |
| Diagonal Traverse | Medium | Solve |