LeetCode #1072 — MEDIUM

Flip Columns For Maximum Number of Equal Rows

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

The Problem

Problem Statement

You are given an m x n binary matrix matrix.

You can choose any number of columns in the matrix and flip every cell in that column (i.e., Change the value of the cell from 0 to 1 or vice versa).

Return the maximum number of rows that have all values equal after some number of flips.

Example 1:

Input: matrix = [[0,1],[1,1]]
Output: 1
Explanation: After flipping no values, 1 row has all values equal.

Example 2:

Input: matrix = [[0,1],[1,0]]
Output: 2
Explanation: After flipping values in the first column, both rows have equal values.

Example 3:

Input: matrix = [[0,0,0],[0,0,1],[1,1,0]]
Output: 2
Explanation: After flipping values in the first two columns, the last two rows have equal values.

Constraints:

m == matrix.length
n == matrix[i].length
1 <= m, n <= 300
matrix[i][j] is either 0 or 1.

Step 01

Brute Force Baseline

Problem summary: You are given an m x n binary matrix matrix. You can choose any number of columns in the matrix and flip every cell in that column (i.e., Change the value of the cell from 0 to 1 or vice versa). Return the maximum number of rows that have all values equal after some number of flips.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

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

Example 2

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

Example 3

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

Step 02

Core Insight

What unlocks the optimal approach

Flipping a subset of columns is like doing a bitwise XOR of some number K onto each row. We want rows X with X ^ K = all 0s or all 1s. This is the same as X = X^K ^K = (all 0s or all 1s) ^ K, so we want to count rows that have opposite bits set. For example, if K = 1, then we count rows X = (00000...001, or 1111....110).

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

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1072: Flip Columns For Maximum Number of Equal Rows
class Solution {
    public int maxEqualRowsAfterFlips(int[][] matrix) {
        Map<String, Integer> cnt = new HashMap<>();
        int ans = 0, n = matrix[0].length;
        for (var row : matrix) {
            char[] cs = new char[n];
            for (int i = 0; i < n; ++i) {
                cs[i] = (char) (row[0] ^ row[i]);
            }
            ans = Math.max(ans, cnt.merge(String.valueOf(cs), 1, Integer::sum));
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1072: Flip Columns For Maximum Number of Equal Rows
func maxEqualRowsAfterFlips(matrix [][]int) (ans int) {
	cnt := map[string]int{}
	for _, row := range matrix {
		s := []byte{}
		for _, x := range row {
			if row[0] == 1 {
				x ^= 1
			}
			s = append(s, byte(x)+'0')
		}
		t := string(s)
		cnt[t]++
		ans = max(ans, cnt[t])
	}
	return
}

# Accepted solution for LeetCode #1072: Flip Columns For Maximum Number of Equal Rows
class Solution:
    def maxEqualRowsAfterFlips(self, matrix: List[List[int]]) -> int:
        cnt = Counter()
        for row in matrix:
            t = tuple(row) if row[0] == 0 else tuple(x ^ 1 for x in row)
            cnt[t] += 1
        return max(cnt.values())

// Accepted solution for LeetCode #1072: Flip Columns For Maximum Number of Equal Rows
struct Solution;

impl Solution {
    fn max_equal_rows_after_flips(matrix: Vec<Vec<i32>>) -> i32 {
        let n = matrix.len();
        let m = matrix[0].len();
        let mut res = vec![1; n];
        for i in 0..n {
            for j in 0..i {
                let count: usize = matrix[i]
                    .iter()
                    .zip(matrix[j].iter())
                    .map(|(x, y)| if x == y { 1 } else { 0 })
                    .sum();
                if count == 0 || count == m {
                    res[i] += 1;
                    res[j] += 1;
                }
            }
        }
        *res.iter().max().unwrap() as i32
    }
}

#[test]
fn test() {
    let matrix = vec_vec_i32![[0, 1], [1, 1]];
    let res = 1;
    assert_eq!(Solution::max_equal_rows_after_flips(matrix), res);
    let matrix = vec_vec_i32![[0, 1], [1, 0]];
    let res = 2;
    assert_eq!(Solution::max_equal_rows_after_flips(matrix), res);
    let matrix = vec_vec_i32![[0, 0, 0], [0, 0, 1], [1, 1, 0]];
    let res = 2;
    assert_eq!(Solution::max_equal_rows_after_flips(matrix), res);
}

// Accepted solution for LeetCode #1072: Flip Columns For Maximum Number of Equal Rows
function maxEqualRowsAfterFlips(matrix: number[][]): number {
    const cnt = new Map<string, number>();
    let ans = 0;
    for (const row of matrix) {
        if (row[0] === 1) {
            for (let i = 0; i < row.length; i++) {
                row[i] ^= 1;
            }
        }
        const s = row.join('');
        cnt.set(s, (cnt.get(s) || 0) + 1);
        ans = Math.max(ans, cnt.get(s)!);
    }
    return ans;
}

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(n)

Space

O(1)

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.