What is the time complexity of Score After Flipping Matrix?

The optimal solution has a time complexity of O(m*n).

What data structures are used to solve Score After Flipping Matrix?

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

What are the key insights for solving Score After Flipping Matrix?

Insight 1: We can always ensure the most significant bit (MSB) of each row is 1 by flipping the row if it starts with 0. This is because flipping a row that starts with 0 will always increase the value of that row. Insight 2: After ensuring all rows start with 1, we can decide whether to flip a column based on whether flipping it would increase the overall score. A column should be flipped if it has more 0s than 1s in that column.

Which companies ask LeetCode 861. Score After Flipping Matrix in interviews?

This problem is frequently asked at IIT Bombay.

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Score After Flipping Matrix - LeetCode 861 Solution

Score After Flipping Matrix - Complete Solution Guide

Score After Flipping Matrix is LeetCode problem 861, a Medium 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 m x n binary matrix grid . A move consists of choosing any row or column and toggling each value in that row or column (i.e., changing all 0 's to 1 's, and all 1 's to 0 's). Every row of the matrix is interpreted as a binary number, and the score of the matrix is the sum of these numbers. Return the highest possible score after making any number of moves (including zero moves) . Example 1: Input: grid = [[0,0,1,1],[1,0,1,0],[1,1,0,0]] Output: 39 Explanation: 0b1111 + 0b1001 +

Detailed Explanation

The problem asks us to maximize the score of a binary matrix by flipping rows and columns. Each row represents a binary number, and the score is the sum of these binary numbers. We can flip any row or column as many times as we want. The goal is to find the highest possible score we can achieve after performing any number of flips.

Solution Approach

The solution uses a greedy approach. First, it ensures that the most significant bit (MSB) of each row is 1. This is done by checking the first element of each row. If it's 0, we conceptually flip the row. Then, for each subsequent column, we count the number of 1s in that column after the row flips (if required). If there are more 0s than 1s in a column, we conceptually flip the column. Finally, we calculate the score by summing the binary values of each row.

Step-by-Step Algorithm

Step 1: Calculate the initial score contribution from the first column. Since we'll ensure all rows start with 1, the first column will be all 1s. The contribution is the number of rows multiplied by 2^(n-1), where n is the number of columns.
Step 2: Iterate through the remaining columns (from the second column onwards).
Step 3: For each column, count the number of 1s that would be present after the potential row flips. A row is flipped if its first element is 0. After the flip, grid[i][j] will be 1 if grid[i][j] == grid[i][0] and 0 otherwise.
Step 4: Determine whether flipping the current column would increase the score. This is done by comparing the number of 1s with the number of 0s (which is the number of rows minus the number of 1s).
Step 5: Calculate the maximum number of 1s we can have in the current column by taking the maximum of the number of 1s and the number of 0s.
Step 6: Add the contribution of the current column to the total score. The contribution is the maximum number of 1s multiplied by 2^(n-1-j), where j is the current column index.
Step 7: Return the total score.

Key Insights

Insight 1: We can always ensure the most significant bit (MSB) of each row is 1 by flipping the row if it starts with 0. This is because flipping a row that starts with 0 will always increase the value of that row.
Insight 2: After ensuring all rows start with 1, we can decide whether to flip a column based on whether flipping it would increase the overall score. A column should be flipped if it has more 0s than 1s in that column.
Insight 3: We do not need to actually flip the matrix. We can calculate how many 1s we would have in each column after the optimal row flips and then calculate the score directly.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(1)

Topics

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

Companies

Asked at: IIT Bombay.

Score After Flipping Matrix | LeetCode 861

Medium

Array Greedy Bit Manipulation Matrix

Time: O(m*n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int matrixScore(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;

        // The score contribution from the first column is fixed.
        // After ensuring the most significant bit of each row is 1,
        // the first column will be all 1s.
        // Its contribution to the total score is m * 2^(n-1).
        int score = (1 << (n - 1)) * m;

        // For the remaining columns, we maximize the number of 1s.
        for (int j = 1; j < n; j++) {
            // Count the number of 1s in the current column after potential row flips.
            // A row is flipped if its first element is 0.
            // After flipping, grid[i][j] becomes 1 - grid[i][j].
            // This is equivalent to checking if grid[i][j] == grid[i][0].
            // If they are the same, the resulting bit is 1. If different, it's 0.
            int ones_count = 0;
            for (int i = 0; i < m; i++) {
                if (grid[i][j] == grid[i][0]) {
                    ones_count += 1;
                }
            }
            
            // The number of 0s is m - ones_count.
            // To maximize score, we want as many 1s as possible.
            // We can flip the column if it has more 0s than 1s.
            // The maximum number of 1s is max(ones_count, m - ones_count).
            int max_ones = Math.max(ones_count, m - ones_count);
            
            // Add the contribution of this column to the total score.
            score += max_ones * (1 << (n - 1 - j));
            
        }
        return score;
    }
}

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Problem Info

Description

Problem 861: Score After Flipping Matrix - Detailed explanation and solution approaches.

Topics

Array Greedy Bit Manipulation Matrix

Companies

IIT Bombay

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:861

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

7 steps

1
Step 1: Calculate the initial score contribution from the first column. Since we'll ensure all rows start with 1, the first column will be all 1s. The contribution is the number of rows multiplied by 2^(n-1), where n is the number of columns.
2
Step 2: Iterate through the remaining columns (from the second column onwards).
3
Step 3: For each column, count the number of 1s that would be present after the potential row flips. A row is flipped if its first element is 0. After the flip, grid[i][j] will be 1 if grid[i][j] == grid[i][0] and 0 otherwise.
4
Step 4: Determine whether flipping the current column would increase the score. This is done by comparing the number of 1s with the number of 0s (which is the number of rows minus the number of 1s).
5
Step 5: Calculate the maximum number of 1s we can have in the current column by taking the maximum of the number of 1s and the number of 0s.
6
Step 6: Add the contribution of the current column to the total score. The contribution is the maximum number of 1s multiplied by 2^(n-1-j), where j is the current column index.
7
Step 7: Return the total score.

💡

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(m*n)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The nested loops iterate through each cell in the grid once. The outer loop goes up to 'n' (number of columns), and the inner loop goes up to 'm' (number of rows). All other operations are constant time. Therefore, the overall time complexity is O(m*n). The space complexity is O(1) because we're only using a few integer variables to store intermediate values, which doesn't scale with the input size.

📖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

Actual Flipping

Instead of just calculating the score, we could actually flip the rows and columns. This would involve modifying the original matrix in place.

Time: O(m*n)Space: O(1)

Trade-offs: This approach modifies the input matrix, which might not be desirable. Also, it does the flips, which is more complex than just calculating the result.

✨

Key Insights

1Insight 1: We can always ensure the most significant bit (MSB) of each row is 1 by flipping the row if it starts with 0. This is because flipping a row that starts with 0 will always increase the value of that row.
2Insight 2: After ensuring all rows start with 1, we can decide whether to flip a column based on whether flipping it would increase the overall score. A column should be flipped if it has more 0s than 1s in that column.
3Insight 3: We do not need to actually flip the matrix. We can calculate how many 1s we would have in each column after the optimal row flips and then calculate the score directly.

📚

Key Terms Explained

For Students

Array

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

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: A matrix with only one row or one column. The solution works correctly because the loops will still execute and calculate the appropriate score.
Edge case 2: An empty matrix (m=0 or n=0). The code might need a condition to check for this case to prevent errors (though the problem constraints specify 1 <= m, n <= 20).
Edge case 3: All elements are initially 0. The initial row flips will make the first column all 1's. The column flipping logic will still result in maximizing 1s in the remaining columns.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to consider the score contribution of the first column. The initial step of ensuring all rows start with 1 is crucial and affects the overall score.

✗Common mistake 2: Incorrectly determining when to flip a column. The decision should be based on maximizing the number of 1s in each column *after* the initial row flips.

✗Common mistake 3: Not using bitwise left shift operators efficiently. Left shift is essential for calculating powers of 2 which represent binary place values.

✅

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: Maximizing data transmission efficiency. Imagine each row represents a data packet, and each column represents a bit position. Flipping rows or columns could represent inverting the signal to minimize errors during transmission.

🏦Real-world scenario 2: Image processing. Matrices are used to represent images. Flipping rows and columns can represent transformations. Although this specific problem may not directly apply, the concept of maximizing a certain value through transformations has applications in image enhancement.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: October 20, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

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

@InterviewProJul 7

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

@CodeNinjaJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Score After Flipping Matrix"?

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

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