LeetCode #782 — HARD

Transform to Chessboard

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

You are given an n x n binary grid board. In each move, you can swap any two rows with each other, or any two columns with each other.

Return the minimum number of moves to transform the board into a chessboard board. If the task is impossible, return -1.

A chessboard board is a board where no 0's and no 1's are 4-directionally adjacent.

Example 1:

Input: board = [[0,1,1,0],[0,1,1,0],[1,0,0,1],[1,0,0,1]]
Output: 2
Explanation: One potential sequence of moves is shown.
The first move swaps the first and second column.
The second move swaps the second and third row.

Example 2:

Input: board = [[0,1],[1,0]]
Output: 0
Explanation: Also note that the board with 0 in the top left corner, is also a valid chessboard.

Example 3:

Input: board = [[1,0],[1,0]]
Output: -1
Explanation: No matter what sequence of moves you make, you cannot end with a valid chessboard.

Constraints:

  • n == board.length
  • n == board[i].length
  • 2 <= n <= 30
  • board[i][j] is either 0 or 1.
Patterns Used
⚡Bit Manipulation→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given an n x n binary grid board. In each move, you can swap any two rows with each other, or any two columns with each other. Return the minimum number of moves to transform the board into a chessboard board. If the task is impossible, return -1. A chessboard board is a board where no 0's and no 1's are 4-directionally adjacent.

Baseline thinking

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

Pattern signal: Array · Math · Bit Manipulation

Example 1

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

Example 2

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

Example 3

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

Related Problems

  • Minimum Moves to Get a Peaceful Board (minimum-moves-to-get-a-peaceful-board)
Step 02

Core Insight

What unlocks the optimal approach

  • No official hints in dataset. Start from constraints and look for a monotonic or reusable state.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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
Largest constraint values
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 #782: Transform to Chessboard
class Solution {
    private int n;

    public int movesToChessboard(int[][] board) {
        n = board.length;
        int mask = (1 << n) - 1;
        int rowMask = 0, colMask = 0;
        for (int i = 0; i < n; ++i) {
            rowMask |= board[0][i] << i;
            colMask |= board[i][0] << i;
        }
        int revRowMask = mask ^ rowMask;
        int revColMask = mask ^ colMask;
        int sameRow = 0, sameCol = 0;
        for (int i = 0; i < n; ++i) {
            int curRowMask = 0, curColMask = 0;
            for (int j = 0; j < n; ++j) {
                curRowMask |= board[i][j] << j;
                curColMask |= board[j][i] << j;
            }
            if (curRowMask != rowMask && curRowMask != revRowMask) {
                return -1;
            }
            if (curColMask != colMask && curColMask != revColMask) {
                return -1;
            }
            sameRow += curRowMask == rowMask ? 1 : 0;
            sameCol += curColMask == colMask ? 1 : 0;
        }
        int t1 = f(rowMask, sameRow);
        int t2 = f(colMask, sameCol);
        return t1 == -1 || t2 == -1 ? -1 : t1 + t2;
    }

    private int f(int mask, int cnt) {
        int ones = Integer.bitCount(mask);
        if (n % 2 == 1) {
            if (Math.abs(n - ones * 2) != 1 || Math.abs(n - cnt * 2) != 1) {
                return -1;
            }
            if (ones == n / 2) {
                return n / 2 - Integer.bitCount(mask & 0xAAAAAAAA);
            }
            return (n / 2 + 1) - Integer.bitCount(mask & 0x55555555);
        } else {
            if (ones != n / 2 || cnt != n / 2) {
                return -1;
            }
            int cnt0 = n / 2 - Integer.bitCount(mask & 0xAAAAAAAA);
            int cnt1 = n / 2 - Integer.bitCount(mask & 0x55555555);
            return Math.min(cnt0, cnt1);
        }
    }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(1)

Approach Breakdown

SORT + SCAN
O(n log n) time
O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION
O(n) time
O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.
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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

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