<p>You are given an <code>m x n</code> binary matrix <code>matrix</code>.</p>

<p>You can choose any number of columns in the matrix and flip every cell in that column (i.e., Change the value of the cell from <code>0</code> to <code>1</code> or vice versa).</p>

<p>Return <em>the maximum number of rows that have all values equal after some number of flips</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[0,1],[1,1]]
<strong>Output:</strong> 1
<strong>Explanation:</strong> After flipping no values, 1 row has all values equal.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[0,1],[1,0]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> After flipping values in the first column, both rows have equal values.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[0,0,0],[0,0,1],[1,1,0]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> After flipping values in the first two columns, the last two rows have equal values.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == matrix.length</code></li>
	<li><code>n == matrix[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 300</code></li>
	<li><code>matrix[i][j]</code> is either&nbsp;<code>0</code> or <code>1</code>.</li>
</ul>

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

        // Helper function to get the pattern string of a row
        auto getPattern = [](const vector<int>& row) {
            string pattern(row.size(), '0');
            for (int i = 1; i < row.size(); ++i) {
                pattern[i] = (row[i] != row[i - 1]) ? '1' : '0';
            }
            return pattern;
        };

        int maxCount = 0;

        for (const auto& row : matrix) {
            // Get the pattern of the row
            string pattern = getPattern(row);
            // Create a flipped pattern (flip '0' <-> '1')
            string flippedPattern(row.size(), '0');
            for (int i = 0; i < row.size(); ++i) {
                flippedPattern[i] = (pattern[i] == '0') ? '1' : '0';
            }

            // Count occurrences of both the pattern and the flipped pattern
            countMap[pattern]++;
            countMap[flippedPattern]++;

            // Update maxCount based on the highest frequency of either pattern
            maxCount = max(maxCount, countMap[pattern]);
            maxCount = max(maxCount, countMap[flippedPattern]);
        }

        return maxCount;
    }
};

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