What is the time complexity of Transform to Chessboard?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Transform to Chessboard?

This problem uses Array, Math, Bit Manipulation, Matrix. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Transform to Chessboard?

Insight 1: If the board can be transformed into a chessboard, then all rows must be either identical or the bitwise inverse of the first row, and similarly for columns. This is because swaps preserve the relative order and adjacency. Insight 2: The number of 1s in each row and each column must be n/2 (if n is even) or either n/2 or n/2 + 1 (if n is odd) for a chessboard to be possible.

Which companies ask LeetCode 782. Transform to Chessboard in interviews?

This problem is frequently asked at Citadel.

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Transform to Chessboard - LeetCode 782 Solution

Transform to Chessboard - Complete Solution Guide

Transform to Chessboard is LeetCode problem 782, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 t

Detailed Explanation

The problem asks us to transform a given n x n binary grid into a chessboard pattern by swapping rows and columns. A chessboard pattern is defined as one where no two adjacent cells (horizontally or vertically) have the same value. We need to find the minimum number of swaps required to achieve this. If it's impossible to transform the board into a chessboard pattern, we should return -1.

Solution Approach

The solution first checks if it's even possible to transform the board into a chessboard. It verifies that all rows and columns are either identical or the bitwise inverse of the first row and column respectively. Also, it checks if the count of 1s in the rows and columns satisfy the condition as mentioned in insight 2. If it's possible, the algorithm calculates the minimum number of swaps needed for rows and columns independently, considering both possible chessboard configurations (starting with 0 or 1). Finally, it returns the sum of row swaps and column swaps.

Step-by-Step Algorithm

Step 1: Check for impossibility: Iterate through the board and verify that `board[0][0] ^ board[i][0] ^ board[0][j] ^ board[i][j]` is always 0. This checks the consistent relative bitwise values and confirms if the grid can potentially form a chessboard.
Step 2: Check row and column sums: Calculate the sum of the first row (row_sum) and the first column (col_sum).
Step 3: Check if row and column sums are valid: Based on whether `n` is even or odd, verify if `row_sum` and `col_sum` satisfy the conditions to ensure it can form a chessboard. Return -1 if invalid.
Step 4: Calculate row and column mismatches: Count the number of mismatches in the first row (row_mismatches) and the first column (col_mismatches) compared to an ideal chessboard pattern (0101... or 1010...).
Step 5: Calculate row and column swaps: If `n` is odd, correct `row_mismatches` and `col_mismatches` to use the valid chessboard pattern with either 0 or 1 starting based on which has fewer mismatches. Calculate row and column swaps by dividing by 2 as each swap correct two mismatches. If `n` is even, choose the pattern which gives fewer swaps based on `min(row_mismatches, n - row_mismatches)` and `min(col_mismatches, n - col_mismatches)`.
Step 6: Return the total swaps: Return the sum of `row_swaps` and `col_swaps`.

Key Insights

Insight 1: If the board can be transformed into a chessboard, then all rows must be either identical or the bitwise inverse of the first row, and similarly for columns. This is because swaps preserve the relative order and adjacency.
Insight 2: The number of 1s in each row and each column must be n/2 (if n is even) or either n/2 or n/2 + 1 (if n is odd) for a chessboard to be possible.
Insight 3: We only need to calculate the number of swaps for rows and columns separately and then add them together. Because row swaps and column swaps are independent.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

This problem involves: Array, Math, Bit Manipulation, Matrix.

Companies

Asked at: Citadel.

Transform to Chessboard | LeetCode 782

Hard

Array Math Bit Manipulation Matrix

Time: O(n^2)Space: O(1)

Solution Code

Available in:

class Solution {
    public int movesToChessboard(int[][] board) {
        int n = board.length;
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if ((board[0][0] ^ board[i][0] ^ board[0][j] ^ board[i][j]) == 1) {
                    return -1;
                }
            }
        }
        
        int row_sum = 0;
        for (int j = 0; j < n; j++) {
            row_sum += board[0][j];
        }
        
        int col_sum = 0;
        for (int i = 0; i < n; i++) {
            col_sum += board[i][0];
        }
        
        if (n % 2 == 1) {
            if (!(row_sum == n / 2 || row_sum == n / 2 + 1)) {
                return -1;
            }
            if (!(col_sum == n / 2 || col_sum == n / 2 + 1)) {
                return -1;
            }
        } else {
            if (row_sum != n / 2) {
                return -1;
            }
            if (col_sum != n / 2) {
                return -1;
            }
        }
        
        int row_mismatches = 0;
        for (int j = 0; j < n; j++) {
            if (board[0][j] != j % 2) {
                row_mismatches++;
            }
        }
        
        int col_mismatches = 0;
        for (int i = 0; i < n; i++) {
            if (board[i][0] != i % 2) {
                col_mismatches++;
            }
        }
        
        int row_swaps = 0;
        int col_swaps = 0;
        
        if (n % 2 == 1) {
            if (row_mismatches % 2 != 0) {
                row_mismatches = n - row_mismatches;
            }
            if (col_mismatches % 2 != 0) {
                col_mismatches = n - col_mismatches;
            }
            
            row_swaps = row_mismatches / 2;
            col_swaps = col_mismatches / 2;
        } else {
            row_swaps = Math.min(row_mismatches, n - row_mismatches) / 2;
            col_swaps = Math.min(col_mismatches, n - col_mismatches) / 2;
        }
        
        return row_swaps + col_swaps;
    }
}

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Problem Info

Description

Problem 782: Transform to Chessboard - Detailed explanation and solution approaches.

Topics

Array Math Bit Manipulation Matrix

Companies

Citadel

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:782

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Check for impossibility: Iterate through the board and verify that `board[0][0] ^ board[i][0] ^ board[0][j] ^ board[i][j]` is always 0. This checks the consistent relative bitwise values and confirms if the grid can potentially form a chessboard.
2
Step 2: Check row and column sums: Calculate the sum of the first row (row_sum) and the first column (col_sum).
3
Step 3: Check if row and column sums are valid: Based on whether `n` is even or odd, verify if `row_sum` and `col_sum` satisfy the conditions to ensure it can form a chessboard. Return -1 if invalid.
4
Step 4: Calculate row and column mismatches: Count the number of mismatches in the first row (row_mismatches) and the first column (col_mismatches) compared to an ideal chessboard pattern (0101... or 1010...).
5
Step 5: Calculate row and column swaps: If `n` is odd, correct `row_mismatches` and `col_mismatches` to use the valid chessboard pattern with either 0 or 1 starting based on which has fewer mismatches. Calculate row and column swaps by dividing by 2 as each swap correct two mismatches. If `n` is even, choose the pattern which gives fewer swaps based on `min(row_mismatches, n - row_mismatches)` and `min(col_mismatches, n - col_mismatches)`.
6
Step 6: Return the total swaps: Return the sum of `row_swaps` and `col_swaps`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The algorithm iterates through the board once to check for impossibility which takes O(n^2). The rest of the operations are O(n) for calculating sums and mismatches. Since, O(n^2) dominates O(n), the time complexity is O(n^2). The algorithm only uses a few integer variables, so the space complexity is O(1).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Try all possible row and column permutations and check if any permutation results in a valid chessboard. This approach is highly inefficient.

Time: O((n!)^2 * n^2)Space: O(n^2)

Trade-offs: This approach is simple to understand but computationally expensive and should be avoided for larger values of n. The main solution offers significantly better performance.

✨

Key Insights

1Insight 1: If the board can be transformed into a chessboard, then all rows must be either identical or the bitwise inverse of the first row, and similarly for columns. This is because swaps preserve the relative order and adjacency.
2Insight 2: The number of 1s in each row and each column must be n/2 (if n is even) or either n/2 or n/2 + 1 (if n is odd) for a chessboard to be possible.
3Insight 3: We only need to calculate the number of swaps for rows and columns separately and then add them together. Because row swaps and column swaps are independent.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: When the board cannot be transformed into a chessboard. The first loop checks for this condition. Specifically `(board[0][0] ^ board[i][0] ^ board[0][j] ^ board[i][j]) == 1` which means there exists no consistent pattern. In this case, -1 is returned.
Edge case 2: When n is odd, the number of 1s in a row/column can be either n/2 or n/2+1. The algorithm explicitly checks for this and adjusts calculations accordingly to correctly handle the situation when row_mismatches and col_mismatches parity do not match the expectations.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check if the board can be transformed into a chessboard at all. Without this check, the algorithm might return incorrect results or run into errors.

✗Common mistake 2: Not considering both possible chessboard patterns (starting with 0 or 1) and choosing the one that requires the minimum number of swaps.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Optimization problems where resources or elements need to be arranged in an alternating pattern, minimizing the number of moves to achieve that pattern.

🏦Real-world scenario 2: Image processing where a checkerboard pattern is used for calibration purposes. If the original image is distorted, this algorithm can be adapted to determine the minimum transformations needed to restore the pattern.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 2, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Transform to Chessboard"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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