LeetCode in Kotlin
3240. Minimum Number of Flips to Make Binary Grid Palindromic II
Medium
You are given an m x n binary matrix grid.
A row or column is considered palindromic if its values read the same forward and 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 1’s in grid divisible by 4.
Example 1:
Input: grid = [[1,0,0],[0,1,0],[0,0,1]]
Output: 3
Explanation:
Example 2:
Input: grid = [[0,1],[0,1],[0,0]]
Output: 2
Explanation:
Example 3:
Input: grid = [[1],[1]]
Output: 2
Explanation:
Constraints:
m == grid.lengthn == grid[i].length1 <= m * n <= 2 * 1050 <= grid[i][j] <= 1
Solution
import kotlin.math.min
class Solution {
fun minFlips(grid: Array<IntArray>): Int {
var res = 0
var one = 0
var diff = 0
val m = grid.size
val n = grid[0].size
// Handle quadrants
for (i in 0 until m / 2) {
for (j in 0 until n / 2) {
val v =
(
grid[i][j] +
grid[i][n - 1 - j] +
grid[m - 1 - i][j] +
grid[m - 1 - i][n - 1 - j]
)
res += min(v, (4 - v))
}
}
// Handle middle column
if (n % 2 > 0) {
for (i in 0 until m / 2) {
diff += grid[i][n / 2] xor grid[m - 1 - i][n / 2]
one += grid[i][n / 2] + grid[m - 1 - i][n / 2]
}
}
// Handle middle row
if (m % 2 > 0) {
for (j in 0 until n / 2) {
diff += grid[m / 2][j] xor grid[m / 2][n - 1 - j]
one += grid[m / 2][j] + grid[m / 2][n - 1 - j]
}
}
// Handle center point
if (n % 2 > 0 && m % 2 > 0) {
res += grid[m / 2][n / 2]
}
// Divisible by 4
if (diff == 0 && one % 4 > 0) {
res += 2
}
return res + diff
}
}