Flip Columns For Maximum Number of Equal Rows

class Solution {
public:
    int maxRowsEqualAfterFlips(vector<vector<int>>& matrix) {
        unordered_map<string, int> rowPatterns;

        for (auto& row : matrix) {
            string convertedPattern = "";

            for (int j = 0; j < row.size(); j++) {
                convertedPattern += (row[0] == row[j]) ? "T" : "F";
            }

            rowPatterns[convertedPattern]++;
        }

        int highestCount = 0;
        for (auto& freq : rowPatterns) {
            highestCount = max(freq.second, highestCount);
        }

        return highestCount;
    }
};

The provided C++ code defines a solution to the problem of finding the maximum number of rows in a matrix that can be made equal by potentially flipping columns. The class Solution has a method named maxRowsEqualAfterFlips which takes a matrix of integers as input and returns the maximum number of rows that can be made identical through column flipping.

• The code uses a hash map (unordered_map) to count different row patterns.
• Each row of the matrix is converted to a 'T'/'F' pattern string based on the comparison of each element with the first element of the row. 'T' represents a match with the first element, and 'F' represents no match.
• This pattern string is then used as a key in the hash map to count how frequently each pattern appears in the matrix.
• After processing all rows and storing their patterns, the highest frequency (highest count of similar rows that can be made equal by flips) is determined by iterating through the hash map.

Execute the implementation as follows:

1. Initialize rowPatterns to store each converted pattern as a key and its occurrence count as a value.
2. Loop through each row in the matrix, converting it into a 'T'/'F' string based on its similarity with the first element of the row.
3. Update the occurrence count of each pattern in the rowPatterns.
4. Determine and return the maximum number amongst the stored pattern frequencies, representing the highest number of rows that can be made the same by flipping.

This solution effectively uses hashing and string manipulation to categorize and count row equivalence possibilities after potential flips, yielding the maximum group of identical rows.

class Solution {

    public int maximumEqualRowsAfterOperations(int[][] inputMatrix) {
        Map<String, Integer> rowPatterns = new HashMap<>();

        for (int[] row : inputMatrix) {
            StringBuilder rowPattern = new StringBuilder();

            for (int elementIndex = 0; elementIndex < row.length; elementIndex++) {
                if (row[0] == row[elementIndex]) {
                    rowPattern.append("T");
                } else {
                    rowPattern.append("F");
                }
            }

            String pattern = rowPattern.toString();
            rowPatterns.put(pattern, rowPatterns.getOrDefault(pattern, 0) + 1);
        }

        int maximumCount = 0;
        for (int count : rowPatterns.values()) {
            maximumCount = Math.max(count, maximumCount);
        }

        return maximumCount;
    }
}

This Java solution addresses the problem of finding the maximum number of rows that can be made identical through column flips. The solution uses a hashmap to track frequency of each unique pattern of rows, where:

• A row pattern is determined by comparing each element with the first element of the row. If they are the same, "T" is appended to the pattern; otherwise, "F" is appended.
• Each pattern is stored in the hashmap with its corresponding frequency.

The procedure follows these steps:

1. Initialize a hashmap rowPatterns to store each unique row pattern and its frequency.
2. Iterate through each row of the input matrix. For each row, compute its pattern and update the hashmap:
   1. For each element in the row, compare with the first element. Append "T" if they are equal, otherwise append "F".
   2. Convert the pattern to string and update the hashmap with the frequency of this pattern.
3. Initialize maximumCount to zero. After all rows are processed, determine the maximum frequency of any pattern in the hashmap.
4. Return maximumCount, which reflects the maximum number of rows that can be made identical after flipping columns as required.

This method efficiently captures row patterns and uses them to find the optimal number of flips needed to maximize identical rows.

class Solution:
    def maxEqualRowsAfterFlips(self, matrix: List[List[int]]) -> int:
        pattern_count = {}

        for row in matrix:
            pattern = "".join("T" if x == row[0] else "F" for x in row)
            pattern_count[pattern] = pattern_count.get(pattern, 0) + 1

        return max(pattern_count.values(), default=0)

This Python solution addresses the problem of maximizing the number of equal rows in a matrix after potentially flipping any number of columns. The solution involves transformation and counting of row patterns to determine the highest number of rows that can be made identical through flips.

• Start by initializing a dictionary pattern_count to keep track of how often each transformed row pattern appears.
• Iterate over each row in the matrix. For each row, construct a pattern string. In this pattern, represent each element as "T" if it matches the first element of the row and "F" otherwise.
• For each generated pattern, update the pattern_count dictionary, increasing the count of the current pattern.
• Return the maximum value found in pattern_count, which represents the largest number of rows that can be made identical through flips. Use the default=0 parameter with the max function to handle the case when the dictionary might be empty.

By implementing this approach, efficiently determine the maximum number of rows that can be made the same through potential column flips.