// 3240. Minimum Number of Flips to Make Binary Grid Palindromic II #include "leetcode.h" /* * 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. */ int minFlips(int **grid, int gridSize, int *gridColSize) { int ans = 0, one = 0, diff = 0, m = gridSize, n = gridColSize[0]; for (int i = 0; i < m / 2; ++i) for (int j = 0; j < n / 2; ++j) { int v = grid[i][j] + grid[i][n - 1 - j] + grid[m - 1 - i][j] + grid[m - 1 - i][n - 1 - j]; ans += fmin(v, 4 - v); } if (n % 2) for (int i = 0; i < m / 2; ++i) { diff += grid[i][n / 2] != grid[m - 1 - i][n / 2]; one += grid[i][n / 2] + grid[m - 1 - i][n / 2]; } if (m % 2) for (int j = 0; j < n / 2; ++j) { diff += grid[m / 2][j] != grid[m / 2][n - 1 - j]; one += grid[m / 2][j] + grid[m / 2][n - 1 - j]; } if (n % 2 && m % 2) ans += grid[m / 2][n / 2]; if (!diff && one % 4) ans += 2; return ans + diff; } int main() { int g1i[3][3] = {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}, g2i[3][2] = {{0, 1}, {0, 1}, {0, 0}}, g3i[2][1] = {{1}, {1}}; struct two_d_arr g1, g2, g3; two_d_arr_init(&g1, ARRAY_SIZE(g1i), ARRAY_SIZE(g1i[0]), g1i); two_d_arr_init(&g2, ARRAY_SIZE(g2i), ARRAY_SIZE(g2i[0]), g2i); two_d_arr_init(&g3, ARRAY_SIZE(g3i), ARRAY_SIZE(g3i[0]), g3i); printf("%d\n", minFlips(g1.arr, g1.row_size, g1.col_size)); // expect: 3 printf("%d\n", minFlips(g2.arr, g2.row_size, g2.col_size)); // expect: 2 printf("%d\n", minFlips(g3.arr, g3.row_size, g3.col_size)); // expect: 2 two_d_arr_free(&g1); two_d_arr_free(&g2); two_d_arr_free(&g3); }