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Author: NamVr

1147-flip-columns-for-maximum-number-of-equal-rows/README.md

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+<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>

1147-flip-columns-for-maximum-number-of-equal-rows/solution.cpp

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+class Solution {
+public:
+    int maxEqualRowsAfterFlips(vector<vector<int>>& matrix) {
+        unordered_map<string, int> map;
+
+		for (auto& r : matrix) {
+			// generate the pattern for the current row
+			string s(r.size(), 'T');
+			for (int i = 1; i < r.size(); i++) {
+				if (r[i] != r[0]) s[i] = 'F';
+			}
+			// put the pattern to map
+			map[s]++;
+		}
+
+		// find the highest freq of the pattern
+		int ans = 0;
+		for (auto& p : map)
+			ans = max(ans, p.second);
+		
+		return ans;
+    }
+};