Description
You are given an m x n
binary matrix matrix
.
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 0
to 1
or vice versa).
Return the maximum number of rows that have all values equal after some number of flips.
Example 1:
Input: matrix = [[0,1],[1,1]]
Output: 1
Explanation: After flipping no values, 1 row has all values equal.
Example 2:
Input: matrix = [[0,1],[1,0]]
Output: 2
Explanation: After flipping values in the first column, both rows have equal values.
Example 3:
Input: matrix = [[0,0,0],[0,0,1],[1,1,0]]
Output: 2
Explanation: After flipping values in the first two columns, the last two rows have equal values.
Constraints:
m == matrix.length
n == matrix[i].length
1 <= m, n <= 300
matrix[i][j]
is either 0
or 1
.
Solutions
Solution 1
| class Solution:
def maxEqualRowsAfterFlips(self, matrix: List[List[int]]) -> int:
cnt = Counter()
for row in matrix:
t = tuple(row) if row[0] == 0 else tuple(x ^ 1 for x in row)
cnt[t] += 1
return max(cnt.values())
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14 | class Solution {
public int maxEqualRowsAfterFlips(int[][] matrix) {
Map<String, Integer> cnt = new HashMap<>();
int ans = 0, n = matrix[0].length;
for (var row : matrix) {
char[] cs = new char[n];
for (int i = 0; i < n; ++i) {
cs[i] = (char) (row[0] ^ row[i]);
}
ans = Math.max(ans, cnt.merge(String.valueOf(cs), 1, Integer::sum));
}
return ans;
}
}
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15 | class Solution {
public:
int maxEqualRowsAfterFlips(vector<vector<int>>& matrix) {
unordered_map<string, int> cnt;
int ans = 0;
for (auto& row : matrix) {
string s;
for (int x : row) {
s.push_back('0' + (row[0] == 0 ? x : x ^ 1));
}
ans = max(ans, ++cnt[s]);
}
return ans;
}
};
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16 | func maxEqualRowsAfterFlips(matrix [][]int) (ans int) {
cnt := map[string]int{}
for _, row := range matrix {
s := []byte{}
for _, x := range row {
if row[0] == 1 {
x ^= 1
}
s = append(s, byte(x)+'0')
}
t := string(s)
cnt[t]++
ans = max(ans, cnt[t])
}
return
}
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15 | function maxEqualRowsAfterFlips(matrix: number[][]): number {
const cnt = new Map<string, number>();
let ans = 0;
for (const row of matrix) {
if (row[0] === 1) {
for (let i = 0; i < row.length; i++) {
row[i] ^= 1;
}
}
const s = row.join('');
cnt.set(s, (cnt.get(s) || 0) + 1);
ans = Math.max(ans, cnt.get(s)!);
}
return ans;
}
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