<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:
    // observation: the rows that form the answer have the same "symmetry"
    // e.g. 001 and 110 are the same, but 101 and 010 are not the same
    // (2 majority elements and in the same position)
    // count the max freq of majority element and fix it, use ump to count
    
    // since m and n <= 300, cant use number and bitset, use string with '0' and '1'
    // for simplicity, fix first element to be '0'
    int maxEqualRowsAfterFlips(vector<vector<int>>& matrix) {
        int m = matrix.size();
        int n = matrix[0].size();
        unordered_map<string, int> mp;
        int ans = 0;
        string temp;
        temp.reserve(n);
        bool flip;
        for (int i = 0; i < m; ++i) {
            temp.clear();
            flip = (*matrix[i].begin() == 1);
            for (int j = 0; j < n; ++j) {
                if (flip) {
                    temp += '0' + (matrix[i][j] + 1) % 2;
                } else {
                    temp += '0' + matrix[i][j];
                }
            }
            ++mp[temp];
            ans = max(ans, mp[temp]);
            
        }
        return ans;
    }
};
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