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## 1. Overview

Matrices are fundamental data structures used in various computer science, mathematics, and engineering fields. In some scenarios, we might need to set certain matrix elements to zero based on specific conditions or requirements. In this tutorial, we’ll discuss various approaches to accomplish this task efficiently in Java.

## 2. Understanding the Problem

Given a matrix, we aim to set the entire row and column of every zero element in the matrix to zero. This operation effectively “zeroes out” rows and columns that contain at least one zero element.

For example, consider the following matrix:

[1, 2, 3]
[4, 0, 6]
[7, 8, 9]

After applying the transformation, the matrix becomes:

[1, 0, 3]
[0, 0, 0]
[7, 0, 9]

## 3. The Naive Solution

A frequently employed strategy for obtaining the desired outcome is to utilize a simple method of problem-solving, often without emphasizing optimization or efficiency considerations. It’s usually the most obvious way to tackle a problem, but it may not be the most effective or efficient in terms of performance or resource usage.

To solve our problem using a naive approach, we can start by duplicating the input matrix to preserve the original values and then traverse it to detect zero elements.

Upon encountering a zero element, we proceed to zero out the entire corresponding row and column in the copied matrix. Finally, we update the original matrix with the modified values obtained from the copy:

public class SetMatrixToZero {
static void setZeroesByNaiveApproach(int[][] matrix) {
int row = matrix.length;
int col = matrix[0].length;
int [][] result = new int[row][col];

for (int i = 0; i<row; i++) {
for (int j = 0; j < col; j++) {
result[i][j] = matrix[i][j];
}
}

for (int i = 0; i < row; i++) {
for (int j = 0; j < col; j++) {
if (matrix[i][j] == 0) {
for (int k = 0; k < col; k++) {
result[i][k] = 0;
}
for (int k = 0; k < row; k++) {
result[k][j] = 0;
}
}
}
}

for (int i = 0; i < row; i++) {
for (int j = 0; j < col; j++) {
matrix[i][j] = result[i][j];
}
}
}
}

While this approach is straightforward, it has a time complexity of , where m is the number of rows and n is the number of columns. Its space complexity is O(m*n). For large matrices, this approach may not be efficient.

Let’s test this approach by executing the below test:

@Test
void givenMatrix_whenUsingSetZeroesByNaiveApproach_thenSetZeroes() {
int[][] matrix = {
{1, 2, 3},
{4, 0, 6},
{7, 8, 9}
};
int[][] expected = {
{1, 0, 3},
{0, 0, 0},
{7, 0, 9}
};
SetMatrixToZero.setZeroesByNaiveApproach(matrix);
assertArrayEquals(expected, matrix);
}

## 4. The Time Optimized Approach

This approach focuses on improving the time complexity of the solution by using an additional space to store the indices of zero elements in the matrix.

It first traverses the matrix to identify zero elements and stores their row and column indices in separate HashSet. Then, it traverses the matrix again and sets the corresponding rows and columns to zero based on the stored indices:

static void setZeroesByTimeOptimizedApproach(int[][] matrix) {
int rows = matrix.length;
int cols = matrix[0].length;

Set<Integer> rowSet = new HashSet<>();
Set<Integer> colSet = new HashSet<>();

for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (matrix[i][j] == 0) {
}
}
}

for (int row: rowSet) {
for (int j = 0; j < cols; j++) {
matrix[row][j] = 0;
}
}

for (int col: colSet) {
for (int i = 0; i < rows; i++) {
matrix[i][col] = 0;
}
}
}

This approach has a time complexity of O(m * n) and requires additional space of O(m + n) for storing indices, which can be inefficient for large matrices.

We can validate the above approach through the below test:

@Test
void givenMatrix_whenUsingSetZeroesByTimeOptimizedApproach_thenSetZeroes() {
int[][] matrix = {
{1, 2, 3},
{4, 0, 6},
{7, 8, 9}
};
int[][] expected = {
{1, 0, 3},
{0, 0, 0},
{7, 0, 9}
};
SetMatrixToZero.setZeroesByTimeOptimizedApproach(matrix);
assertArrayEquals(expected, matrix);
}


## 5. The Optimal Approach

This approach optimizes space complexity by modifying the original matrix in place without using additional space.

It utilizes the matrix’s first row and first column to store information about zero elements. The algorithm first checks if the first row and first column contain any zeros.

Then, it marks zero elements in the first row and first column by setting the corresponding elements to zero. Next, it traverses the matrix (excluding the first row and first column) to mark zero elements in the first row and first column.

Finally, it traverses the first row and first column to update the entire row or column to zero if needed.

We’ll break the above steps into separate helper methods. The below code checks if there is at least one zero in the first row of the matrix:

static boolean hasZeroInFirstRow(int[][] matrix, int cols) {
for (int j = 0; j < cols; j++) {
if (matrix[0][j] == 0) {
return true;
}
}
return false;
}

Similarly, we check if there is at least one zero in the first column of the matrix:

static boolean hasZeroInFirstCol(int[][] matrix, int rows) {
for (int i = 0; i < rows; i++) {
if (matrix[i][0] == 0) {
return true;
}
}
return false;
}

We mark the positions in the matrix where zeros are present by setting corresponding elements in the first row and first column to zero:

static void markZeroesInMatrix(int[][] matrix, int rows, int cols) {
for (int i = 1; i < rows; i++) {
for (int j = 1; j < cols; j++) {
if (matrix[i][j] == 0) {
matrix[i][0] = 0;
matrix[0][j] = 0;
}
}
}
}

Next, we set zeroes in rows based on the markings made in the first column of the matrix:

static void setZeroesInRows(int[][] matrix, int rows, int cols) {
for (int i = 1; i < rows; i++) {
if (matrix[i][0] == 0) {
for (int j = 1; j < cols; j++) {
matrix[i][j] = 0;
}
}
}
}

Similarly, we set zeroes in columns based on the markings made in the first row of the matrix:

static void setZeroesInCols(int[][] matrix, int rows, int cols) {
for (int j = 1; j < cols; j++) {
if (matrix[0][j] == 0) {
for (int i = 1; i < rows; i++) {
matrix[i][j] = 0;
}
}
}
}


Now, we set all elements in the first row to zero:

static void setZeroesInFirstRow(int[][] matrix, int cols) {
for (int j = 0; j < cols; j++) {
matrix[0][j] = 0;
}
}

We follow a similar approach for the first column:

static void setZeroesInFirstCol(int[][] matrix, int rows) {
for (int i = 0; i < rows; i++) {
matrix[i][0] = 0;
}
}


We merge all the preceding steps and execute them within this method:

static void setZeroesByOptimalApproach(int[][] matrix) {
int rows = matrix.length;
int cols = matrix[0].length;

boolean firstRowZero = hasZeroInFirstRow(matrix, cols);
boolean firstColZero = hasZeroInFirstCol(matrix, rows);

markZeroesInMatrix(matrix, rows, cols);

setZeroesInRows(matrix, rows, cols);
setZeroesInCols(matrix, rows, cols);

if (firstRowZero) {
setZeroesInFirstRow(matrix, cols);
}
if (firstColZero) {
setZeroesInFirstCol(matrix, rows);
}
}

This approach eliminates the need for additional space, resulting in an additional space complexity of O(1). The overall space complexity remains O(m * n).

It accomplishes the intended outcome within the existing matrix and maintains a time complexity of O(m * n), where n represents the number of rows and n signifies the number of columns. This efficiency makes it suitable for handling large matrices.

Let’s write the test to assess this approach:

@Test
void givenMatrix_whenUsingSetZeroesByOptimalApproach_thenSetZeroes() {
int[][] matrix = {
{1, 2, 3},
{4, 0, 6},
{7, 8, 9}
};
int[][] expected = {
{1, 0, 3},
{0, 0, 0},
{7, 0, 9}
};
SetMatrixToZero.setZeroesByOptimalApproach(matrix);
assertArrayEquals(expected, matrix);
}

## 6. Conclusion

In this tutorial, we explored various methods for setting elements to zero in a matrix. The naive approach aims to achieve the desired outcome without prioritizing optimization or performance. In contrast, the time-optimized approach reduces time complexity by utilizing extra space.

However, the optimal approach enhances both time complexity and doesn’t necessitate any additional space, making it the better choice out of all the above approaches. It’s important to acknowledge that there could be several other optimal solutions to this problem.

As always, the full source code is available over on GitHub.

### Get started with Spring and Spring Boot, through the Learn Spring course:

>> CHECK OUT THE COURSE
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