LeetCode #73 — MEDIUM

Set Matrix Zeroes

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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?

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[[1,1,1],[1,0,1],[1,1,1]]

Example 2

[[0,1,2,0],[3,4,5,2],[1,3,1,5]]

Core Insight

What unlocks the optimal approach

If any cell of the matrix has a zero we can record its row and column number using additional memory. But if you don't want to use extra memory then you can manipulate the array instead. i.e. simulating exactly what the question says.
Setting cell values to zero on the fly while iterating might lead to discrepancies. What if you use some other integer value as your marker? There is still a better approach for this problem with O(1) space.
We could have used 2 sets to keep a record of rows/columns which need to be set to zero. But for an O(1) space solution, you can use one of the rows and and one of the columns to keep track of this information.
We can use the first cell of every row and column as a flag. This flag would determine whether a row or column has been set to zero.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

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

        for (int r = 0; r < m; r++) {
            if (matrix[r][0] == 0) firstColZero = true;
            for (int c = 1; c < n; c++) {
                if (matrix[r][c] == 0) {
                    matrix[r][0] = 0;
                    matrix[0][c] = 0;
                }
            }
        }

        for (int r = 1; r < m; r++) {
            for (int c = 1; c < n; c++) {
                if (matrix[r][0] == 0 || matrix[0][c] == 0) matrix[r][c] = 0;
            }
        }

        if (matrix[0][0] == 0) {
            for (int c = 0; c < n; c++) matrix[0][c] = 0;
        }
        if (firstColZero) {
            for (int r = 0; r < m; r++) matrix[r][0] = 0;
        }
    }
}

func setZeroes(matrix [][]int) {
    m, n := len(matrix), len(matrix[0])
    firstColZero := false

    for r := 0; r < m; r++ {
        if matrix[r][0] == 0 {
            firstColZero = true
        }
        for c := 1; c < n; c++ {
            if matrix[r][c] == 0 {
                matrix[r][0] = 0
                matrix[0][c] = 0
            }
        }
    }

    for r := 1; r < m; r++ {
        for c := 1; c < n; c++ {
            if matrix[r][0] == 0 || matrix[0][c] == 0 {
                matrix[r][c] = 0
            }
        }
    }

    if matrix[0][0] == 0 {
        for c := 0; c < n; c++ {
            matrix[0][c] = 0
        }
    }
    if firstColZero {
        for r := 0; r < m; r++ {
            matrix[r][0] = 0
        }
    }
}

class Solution:
    def setZeroes(self, matrix: List[List[int]]) -> None:
        m, n = len(matrix), len(matrix[0])
        first_col_zero = False

        for r in range(m):
            if matrix[r][0] == 0:
                first_col_zero = True
            for c in range(1, n):
                if matrix[r][c] == 0:
                    matrix[r][0] = 0
                    matrix[0][c] = 0

        for r in range(1, m):
            for c in range(1, n):
                if matrix[r][0] == 0 or matrix[0][c] == 0:
                    matrix[r][c] = 0

        if matrix[0][0] == 0:
            for c in range(n):
                matrix[0][c] = 0
        if first_col_zero:
            for r in range(m):
                matrix[r][0] = 0

impl Solution {
    pub fn set_zeroes(matrix: &mut Vec<Vec<i32>>) {
        let m = matrix.len();
        let n = matrix[0].len();
        let mut first_col_zero = false;

        for r in 0..m {
            if matrix[r][0] == 0 {
                first_col_zero = true;
            }
            for c in 1..n {
                if matrix[r][c] == 0 {
                    matrix[r][0] = 0;
                    matrix[0][c] = 0;
                }
            }
        }

        for r in 1..m {
            for c in 1..n {
                if matrix[r][0] == 0 || matrix[0][c] == 0 {
                    matrix[r][c] = 0;
                }
            }
        }

        if matrix[0][0] == 0 {
            for c in 0..n {
                matrix[0][c] = 0;
            }
        }
        if first_col_zero {
            for r in 0..m {
                matrix[r][0] = 0;
            }
        }
    }
}

function setZeroes(matrix: number[][]): void {
  const m = matrix.length;
  const n = matrix[0].length;
  let firstColZero = false;

  for (let r = 0; r < m; r++) {
    if (matrix[r][0] === 0) firstColZero = true;
    for (let c = 1; c < n; c++) {
      if (matrix[r][c] === 0) {
        matrix[r][0] = 0;
        matrix[0][c] = 0;
      }
    }
  }

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

  if (matrix[0][0] === 0) {
    for (let c = 0; c < n; c++) matrix[0][c] = 0;
  }
  if (firstColZero) {
    for (let r = 0; r < m; r++) matrix[r][0] = 0;
  }
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m× n)

Space

O(m+n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.