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Diffstat (limited to 'min_num_flip_bin_grid_palin2.py')
-rw-r--r-- | min_num_flip_bin_grid_palin2.py | 42 |
1 files changed, 42 insertions, 0 deletions
diff --git a/min_num_flip_bin_grid_palin2.py b/min_num_flip_bin_grid_palin2.py new file mode 100644 index 0000000..3f196bb --- /dev/null +++ b/min_num_flip_bin_grid_palin2.py @@ -0,0 +1,42 @@ +# 3240. Minimum Number of Flips to Make Binary Grid Palindromic II + +""" +you are given an 'm x n' binary matrix 'grid'. a row or column is considered +palindromic if its values read the same forward as backward. you can flip any +number of cells in 'grid' from 0 to 1 or from 1 to 0. return the minimum +number of cells that need to be flipped to make all rows and columns +palindromic and the total number of 1s in 'grid' divisible by 4. +""" + + +class Solution(object): + def minFlips(self, grid): + """ + :type grid: List[List[int]] + :rtype: int + """ + ans, one, diff, m, n = 0, 0, 0, len(grid), len(grid[0]) + for i in range(m // 2): + for j in range(n // 2): + v = grid[i][j] + grid[i][~j] + grid[~i][j] + grid[~i][~j] + ans += min(v, 4 - v) + if n % 2: + for i in range(m // 2): + diff += grid[i][n // 2] ^ grid[~i][n // 2] + one += grid[i][n // 2] + grid[~i][n // 2] + if m % 2: + for i in range(n // 2): + diff += grid[m // 2][j] ^ grid[m // 2][~j] + one += grid[m // 2][j] + grid[m // 2][~j] + if n % 2 and m % 2: + ans += grid[m // 2][n // 2] + if diff == 0 and one % 4: + ans += 2 + return ans + diff + + +if __name__ == "__main__": + obj = Solution() + print(obj.minFlips(grid=[[1, 0, 0], [0, 1, 0], [0, 0, 1]])) + print(obj.minFlips(grid=[[0, 1], [0, 1], [0, 0]])) + print(obj.minFlips(grid=[[1], [1]])) |