HackerRank Flipping the Matrix Java Solution Explained

Problem Overview

Original Problem: Flipping the Matrix | HackerRank

Difficulty: Medium

Category: Arrays, Greedy Algorithms

Description: Given a 2n × 2n matrix of integers, maximize the sum of the elements in the n × n upper-left quadrant by reversing any number of rows or columns in the entire matrix. Return the maximum possible sum as an integer.

This problem, part of HackerRank’s One Week Preparation Kit (Day 2 Mock Test), challenges beginners to work with 2D arrays and apply greedy logic to optimize a submatrix sum, a valuable skill for coding interviews.

Understanding the Problem

The Flipping the Matrix problem involves a square matrix of size 2n × 2n, where you can reverse any row or column any number of times to maximize the sum of the n × n upper-left quadrant (i.e., elements matrix[i][j] where 0 ≤ i, j < n). For example, in a 4 × 4 matrix (n = 2), you focus on the top-left 2 × 2 submatrix. By reversing rows or columns, you can move elements within the matrix to increase this quadrant’s sum. The goal is to compute the maximum possible sum of the upper-left n × n submatrix.

Input and Output

Input:
- First line: An integer q, the number of queries.
- For each query:
  - First line: An integer n, half the matrix size .
  - Next 2n lines: Each contains 2n space-separated integers matrix[i][j].
Output:
- For each query, an integer representing the maximum sum of the n × n upper-left quadrant.
Constraints:
- 1 ≤ q ≤ 16
- 1 ≤ n ≤ 128
- 0 ≤ matrix[i][j] < 4069

Initial Reasoning

To solve this, we need to:

Understand that reversing rows or columns allows us to reposition elements within the matrix.
For each position in the upper-left n × n quadrant, determine the maximum possible value by considering all positions affected by row and column flips.
Since the matrix is 2n × 2n, each position in the upper-left quadrant corresponds to a 2 × 2 submatrix in the larger matrix, and flips can move any of these four values to the quadrant position.
Use a greedy approach to select the maximum value for each position in the n × n quadrant.
This approach ensures we maximize the sum efficiently without needing to simulate all possible flips.

Thought Process

Let’s break down the solution step-by-step:

Understand Flipping Mechanics:
- Reversing a row swaps elements left-to-right (e.g., matrix[i][j] ↔ matrix[i][2n-1-j]).
- Reversing a column swaps elements top-to-bottom (e.g., matrix[i][j] ↔ matrix[2n-1-i][j]).
- For any position matrix[i][j] in the upper-left n × n quadrant (0 ≤ i, j < n), flipping rows and columns can place one of four values in that position:
  - matrix[i][j] (original position).
  - matrix[i][2n-1-j] (same row, opposite column).
  - matrix[2n-1-i][j] (opposite row, same column).
  - matrix[2n-1-i][2n-1-j] (opposite row, opposite column).
Greedy Strategy:
- For each position (i, j) in the upper-left n × n quadrant, select the maximum of the four possible values that can occupy that position after any combination of row/column flips.
- Sum these maximum values to get the maximum possible sum for the quadrant.
- Since flips are independent for each 2 × 2 submatrix, choosing the maximum value per position is optimal.
Handle Constraints:
- The matrix size (2n × 2n, n ≤ 128) allows for a straightforward O(n²) solution.
- Use int for calculations (as sums won’t exceed 32-bit limits for n ≤ 128).
- Multiple queries (q ≤ 16) require processing each matrix independently.
Edge Cases:
- If n = 1, the matrix is 2 × 2, and the upper-left quadrant is 1 × 1. The maximum value among the four positions is the answer.
- Non-negative elements (0 ≤ matrix[i][j]) ensure sums are non-negative.
- The constraints guarantee valid inputs, so no additional validation is needed.

Pseudocode

// flippingMatrix method:
Initialize sum for upper-left quadrant
For each row in the upper quadrant
	For each column in the upper quadrant
		Find maximum of possible value for that position
        Add tehe value to the sum 
Return the maximum sum

Solution Code

// Note: This is one of many possible solutions
// Note: Due to HackerRank's test constraints, variable names are not updated in the code below to follow Java naming conventions.
import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;

class Result {

    public static int flippingMatrix(List<List<Integer>> matrix) {
        int quadrantSize = matrix.size() / 2; // n, for the n x n upper-left quadrant
        int maxSum = 0;

        // Iterate over the upper-left n x n quadrant
        for (int i = 0; i < quadrantSize; i++) {
            for (int j = 0; j < quadrantSize; j++) {
                // Find maximum of four possible values for position (i, j)
                int maxValue = Math.max(
                        Math.max(matrix.get(i).get(j), matrix.get(i).get(2 * quadrantSize - 1 - j)),
                        Math.max(matrix.get(2 * quadrantSize - 1 - i).get(j),
                                matrix.get(2 * quadrantSize - 1 - i).get(2 * quadrantSize - 1 - j)));
                maxSum += maxValue;
            }
        }

        return maxSum;
    }

}

public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        int q = Integer.parseInt(bufferedReader.readLine().trim());

        IntStream.range(0, q).forEach(qItr -> {
            try {
                int n = Integer.parseInt(bufferedReader.readLine().trim());

                List<List<Integer>> matrix = new ArrayList<>();

                IntStream.range(0, 2 * n).forEach(i -> {
                    try {
                        matrix.add(
                                Stream.of(bufferedReader.readLine().replaceAll("\\s+$", "").split(" "))
                                        .map(Integer::parseInt)
                                        .collect(toList()));
                    } catch (IOException ex) {
                        throw new RuntimeException(ex);
                    }
                });

                int result = Result.flippingMatrix(matrix);

                bufferedWriter.write(String.valueOf(result));
                bufferedWriter.newLine();
            } catch (IOException ex) {
                throw new RuntimeException(ex);
            }
        });

        bufferedReader.close();
        bufferedWriter.close();
    }
}

Step-by-Step Explanation

Let’s break down the Java solution for beginners:

Input Handling:
- The main method reads:
  - q, the number of queries, using Integer.parseInt(bufferedReader.readLine().trim()).
  - For each query:
    - n, half the matrix size.
    - 2n lines of 2n space-separated integers, processed into a List<List<Integer>> using Stream.of and map(Integer::parseInt).
Initialize Variables:
- In flippingMatrix, calculate quadrantSize = matrix.size() / 2 to get n, the size of the upper-left quadrant.
- Initialize maxSum to 0 to store the sum of the maximum values in the n × n quadrant.
Iterate Over Quadrant:
- Use nested loops to iterate over the upper-left n × n quadrant (i from 0 to n-1, j from 0 to n-1).
- For each position (i, j), compute the maximum of the four possible values:
  - matrix[i][j] (original position).
  - matrix[i][2n-1-j] (same row, flipped column).
  - matrix[2n-1-i][j] (flipped row, same column).
  - matrix[2n-1-i][2n-1-j] (flipped row and column).
- Use nested Math.max calls to find the largest value efficiently.
- Add this maximum value to maxSum.
Return the Result:
- Return maxSum, the maximum possible sum of the n × n quadrant.
- Since elements are non-negative and n ≤ 128, the sum fits within an int.
Output Handling:
- For each query, write the result to the output file using bufferedWriter.write(String.valueOf(result)).
- Add a newline with bufferedWriter.newLine().
- Close bufferedReader and bufferedWriter to prevent resource leaks.

Common Mistakes to Avoid

Here are common pitfalls for beginners:

Simulating All Flips: Attempting to try all possible row and column flips is infeasible due to the exponential number of combinations (2^(2n) possibilities). The greedy approach avoids this.
Incorrect Indices: Miscalculating the indices for flipped positions (e.g., using n-1-j instead of 2n-1-j) can lead to wrong values. Always verify the mirror positions.
Ignoring Quadrant Size: Focusing on the entire 2n × 2n matrix instead of the n × n quadrant can cause confusion. Only sum the upper-left n × n elements.

Real-World Application

The Flipping the Matrix problem mirrors optimization tasks in game theory, image processing, and data analysis. For example, in image processing, manipulating pixel values in a matrix quadrant might optimize visual features. In operations research, similar techniques optimize resource allocation within constraints. This problem teaches 2D array manipulation and greedy algorithms, skills used in software development and algorithmic problem-solving.

FAQs

Q: Why is the greedy approach optimal for this problem?
A: Each position in the n × n quadrant can independently take the maximum of four possible values via row/column flips. Choosing the maximum for each position ensures the highest possible sum without conflicts.

Q: Can I simulate row and column flips?
A: Simulating all flip combinations is computationally expensive (O(2^(2n))). The greedy approach is more efficient, as it directly selects the maximum value per position in O(n²).