Set Matrix Zeroes LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
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Given an m x n integer matrix matrix, if an element is 0, set its entire row and column to 0‘s.

You must do it in place.

Example 1:

Input: matrix = [[1,1,1],[1,0,1],[1,1,1]]
Output: [[1,0,1],[0,0,0],[1,0,1]]

Example 2:

Input: matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]
Output: [[0,0,0,0],[0,4,5,0],[0,3,1,0]]

Constraints:

  • m == matrix.length
  • n == matrix[0].length
  • 1 <= m, n <= 200
  • -231 <= matrix[i][j] <= 231 - 1

Follow up:

  • A straightforward solution using O(mn) space is probably a bad idea.
  • A simple improvement uses O(m + n) space, but still not the best solution.
  • Could you devise a constant space solution?

 Set Matrix Zeroes Solutions

Time: O(mn)
Space: O(1)

C++

class Solution {
 public:
  void setZeroes(vector<vector<int>>& matrix) {
    const int m = matrix.size();
    const int n = matrix[0].size();
    bool shouldFillFirstRow = false;
    bool shouldFillFirstCol = false;

    for (int j = 0; j < n; ++j)
      if (matrix[0][j] == 0) {
        shouldFillFirstRow = true;
        break;
      }

    for (int i = 0; i < m; ++i)
      if (matrix[i][0] == 0) {
        shouldFillFirstCol = true;
        break;
      }

    // store the information in the 1st row/col
    for (int i = 1; i < m; ++i)
      for (int j = 1; j < n; ++j)
        if (matrix[i][j] == 0) {
          matrix[i][0] = 0;
          matrix[0][j] = 0;
        }

    // fill 0s for the matrix except the 1st row/col
    for (int i = 1; i < m; ++i)
      for (int j = 1; j < n; ++j)
        if (matrix[i][0] == 0 || matrix[0][j] == 0)
          matrix[i][j] = 0;

    // fill 0s for the 1st row if needed
    if (shouldFillFirstRow)
      for (int j = 0; j < n; ++j)
        matrix[0][j] = 0;

    // fill 0s for the 1st col if needed
    if (shouldFillFirstCol)
      for (int i = 0; i < m; ++i)
        matrix[i][0] = 0;
  }
};

Java

class Solution {
  public void setZeroes(int[][] matrix) {
    final int m = matrix.length;
    final int n = matrix[0].length;
    boolean shouldFillFirstRow = false;
    boolean shouldFillFirstCol = false;

    for (int j = 0; j < n; ++j)
      if (matrix[0][j] == 0) {
        shouldFillFirstRow = true;
        break;
      }

    for (int i = 0; i < m; ++i)
      if (matrix[i][0] == 0) {
        shouldFillFirstCol = true;
        break;
      }

    // store the information in the 1st row/col
    for (int i = 1; i < m; ++i)
      for (int j = 1; j < n; ++j)
        if (matrix[i][j] == 0) {
          matrix[i][0] = 0;
          matrix[0][j] = 0;
        }

    // fill 0s for the matrix except the 1st row/col
    for (int i = 1; i < m; ++i)
      for (int j = 1; j < n; ++j)
        if (matrix[i][0] == 0 || matrix[0][j] == 0)
          matrix[i][j] = 0;

    // fill 0s for the 1st row if needed
    if (shouldFillFirstRow)
      for (int j = 0; j < n; ++j)
        matrix[0][j] = 0;

    // fill 0s for the 1st col if needed
    if (shouldFillFirstCol)
      for (int i = 0; i < m; ++i)
        matrix[i][0] = 0;
  }
}

Python

class Solution:
  def setZeroes(self, matrix: List[List[int]]) -> None:
    m = len(matrix)
    n = len(matrix[0])
    shouldFillFirstRow = 0 in matrix[0]
    shouldFillFirstCol = 0 in list(zip(*matrix))[0]

    # store the information in the 1st row/col
    for i in range(1, m):
      for j in range(1, n):
        if matrix[i][j] == 0:
          matrix[i][0] = 0
          matrix[0][j] = 0

    # fill 0s for the matrix except the 1st row/col
    for i in range(1, m):
      for j in range(1, n):
        if matrix[i][0] == 0 or matrix[0][j] == 0:
          matrix[i][j] = 0

    # fill 0s for the 1st row if needed
    if shouldFillFirstRow:
      matrix[0] = [0] * n

    # fill 0s for the 1st col if needed
    if shouldFillFirstCol:
      for row in matrix:
        row[0] = 0

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