Matrix Zeroing

mediumcodingAlgorithmsMatrixArrays~25 min

Asked at Google, Meta

Overview

Given an m × n integer matrix, if an element is 0, set its entire row and column to 0. Do it in place.

Constraints

1 <= m, n <= 200
-2^31 <= matrix[i][j] <= 2^31 - 1
Use O(1) extra space (use the first row/column as markers).

Examples

setZeroes([
  [1, 1, 1],
  [1, 0, 1],
  [1, 1, 1]
]);
// => [[1,0,1],[0,0,0],[1,0,1]]

setZeroes([
  [0, 1, 2, 0],
  [3, 4, 5, 2],
  [1, 3, 1, 5]
]);
// => [[0,0,0,0],[0,4,5,0],[0,3,1,0]]

Solution

Reveal solution

function setZeroes(matrix) {
  const m = matrix.length, n = matrix[0].length;
  let firstRowZero = false, firstColZero = false;

  for (let j = 0; j < n; j++) if (matrix[0][j] === 0) firstRowZero = true;
  for (let i = 0; i < m; i++) if (matrix[i][0] === 0) firstColZero = true;

  for (let i = 1; i < m; i++) {
    for (let j = 1; j < n; j++) {
      if (matrix[i][j] === 0) {
        matrix[i][0] = 0;
        matrix[0][j] = 0;
      }
    }
  }

  for (let i = 1; i < m; i++) {
    for (let j = 1; j < n; j++) {
      if (matrix[i][0] === 0 || matrix[0][j] === 0) {
        matrix[i][j] = 0;
      }
    }
  }

  if (firstRowZero) for (let j = 0; j < n; j++) matrix[0][j] = 0;
  if (firstColZero) for (let i = 0; i < m; i++) matrix[i][0] = 0;

  return matrix;
}

Use the first row and first column as markers. Track whether they themselves need zeroing with two booleans. O(1) extra space.

Resources

matrix-zeroing.js

Matrix Zeroing

mediumcodingAlgorithmsMatrix

Overview

Given an m × n integer matrix, if an element is 0, set its entire row and column to 0. Do it in place.

Constraints

1 <= m, n <= 200
-2^31 <= matrix[i][j] <= 2^31 - 1
Use O(1) extra space (use the first row/column as markers).

Examples

setZeroes([
  [1, 1, 1],
  [1, 0, 1],
  [1, 1, 1]
]);
// => [[1,0,1],[0,0,0],[1,0,1]]

setZeroes([
  [0, 1, 2, 0],
  [3, 4, 5, 2],
  [1, 3, 1, 5]
]);
// => [[0,0,0,0],[0,4,5,0],[0,3,1,0]]

Solution

Reveal solution

function setZeroes(matrix) {
  const m = matrix.length, n = matrix[0].length;
  let firstRowZero = false, firstColZero = false;

  for (let j = 0; j < n; j++) if (matrix[0][j] === 0) firstRowZero = true;
  for (let i = 0; i < m; i++) if (matrix[i][0] === 0) firstColZero = true;

  for (let i = 1; i < m; i++) {
    for (let j = 1; j < n; j++) {
      if (matrix[i][j] === 0) {
        matrix[i][0] = 0;
        matrix[0][j] = 0;
      }
    }
  }

  for (let i = 1; i < m; i++) {
    for (let j = 1; j < n; j++) {
      if (matrix[i][0] === 0 || matrix[0][j] === 0) {
        matrix[i][j] = 0;
      }
    }
  }

  if (firstRowZero) for (let j = 0; j < n; j++) matrix[0][j] = 0;
  if (firstColZero) for (let i = 0; i < m; i++) matrix[i][0] = 0;

  return matrix;
}

Use the first row and first column as markers. Track whether they themselves need zeroing with two booleans. O(1) extra space.

Resources

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