Modify the Matrix - Practice Coding Problems

Modify the Matrix - Problem

Array Matrix Easy

Given a 0-indexed m x n integer matrix matrix, create a new 0-indexed matrix called answer. Make answer equal to matrix, then replace each element with the value -1 with the maximum element in its respective column.

Return the matrix answer.

Input & Output

Example 1 — Basic Matrix

$ Input: matrix = [[1,2,-1],[4,-1,6],[7,8,9]]

› Output: [[1,2,9],[4,8,6],[7,8,9]]

💡 Note: Column maximums are [7,8,9]. Replace -1 in position [0,2] with 9, and -1 in position [1,1] with 8.

Example 2 — Multiple -1s in Same Column

$ Input: matrix = [[3,-1],[5,2]]

› Output: [[3,2],[5,2]]

💡 Note: Column maximums are [5,2]. Replace -1 in position [0,1] with maximum of column 1, which is 2.

Example 3 — No -1s

$ Input: matrix = [[1,2,3],[4,5,6]]

› Output: [[1,2,3],[4,5,6]]

💡 Note: No -1 elements to replace, matrix remains unchanged.

Constraints

m == matrix.length
n == matrix[i].length
1 ≤ m, n ≤ 10⁵
1 ≤ m * n ≤ 10⁵
-10⁹ ≤ matrix[i][j] ≤ 10⁹
The input contains at least one non-negative number.

Visualization

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Understanding the Visualization

Input Matrix

Matrix with some -1 elements that need replacement

Find Column Max

Compute maximum value for each column

Replace -1s

Replace each -1 with its column maximum

Key Takeaway

🎯 Key Insight: Pre-compute column maximums to avoid redundant scans when replacing multiple -1s

Asked in

a Amazon 15 G Google 12 M Microsoft 8

The key insight is to avoid scanning columns repeatedly by pre-computing column maximums. Best approach is the two-pass optimization which first finds all column maximums, then replaces all -1s in a second pass. Time: O(m×n), Space: O(n).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Scan Column for Each -1	O(m² × n)	O(1)	For each -1 element, scan its entire column to find the maximum
Two-Pass Optimization	O(m × n)	O(n)	First pass finds column maximums, second pass replaces all -1s

Brute Force - Scan Column for Each -1 — Algorithm Steps

Iterate through each cell in the matrix
When finding -1, scan the entire column to find maximum
Replace -1 with the column maximum

Visualization

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Step-by-Step Walkthrough

Find -1

Locate a -1 element in the matrix

Scan Column

Iterate through entire column to find maximum

Replace

Replace -1 with the column maximum

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>

int** solution(int** matrix, int matrixSize, int* matrixColSize, int* returnSize, int** returnColumnSizes) {
    *returnSize = matrixSize;
    *returnColumnSizes = (int*)malloc(matrixSize * sizeof(int));
    
    for (int i = 0; i < matrixSize; i++) {
        (*returnColumnSizes)[i] = matrixColSize[i];
    }
    
    int m = matrixSize, n = matrixColSize[0];
    
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (matrix[i][j] == -1) {
                // Find maximum in column j
                int colMax = INT_MIN;
                for (int k = 0; k < m; k++) {
                    if (matrix[k][j] != -1) {
                        colMax = (matrix[k][j] > colMax) ? matrix[k][j] : colMax;
                    }
                }
                matrix[i][j] = colMax;
            }
        }
    }
    
    return matrix;
}

int main() {
    // For simplicity, assume small matrix input
    int matrix[3][3] = {{1, 2, -1}, {4, -1, 6}, {7, 8, 9}};
    int* matrixPtr[3];
    int matrixColSize[3] = {3, 3, 3};
    
    for (int i = 0; i < 3; i++) {
        matrixPtr[i] = matrix[i];
    }
    
    int returnSize;
    int* returnColumnSizes;
    
    int** result = solution(matrixPtr, 3, matrixColSize, &returnSize, &returnColumnSizes);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("[");
        for (int j = 0; j < returnColumnSizes[i]; j++) {
            printf("%d", result[i][j]);
            if (j < returnColumnSizes[i] - 1) printf(",");
        }
        printf("]");
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(m² × n)

For each cell (m×n), we might scan entire column (m) to find max

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables, modifying matrix in-place

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Scan Column for Each -1 — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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