Easy
Given two n x n
binary matrices mat
and target
, return true
if it is possible to make mat
equal to target
by rotating mat
in 90-degree increments, or false
otherwise.
Example 1:
Input: mat = [[0,1],[1,0]], target = [[1,0],[0,1]]
Output: true
Explanation: We can rotate mat 90 degrees clockwise to make mat equal target.
Example 2:
Input: mat = [[0,1],[1,1]], target = [[1,0],[0,1]]
Output: false
Explanation: It is impossible to make mat equal to target by rotating mat.
Example 3:
Input: mat = [[0,0,0],[0,1,0],[1,1,1]], target = [[1,1,1],[0,1,0],[0,0,0]]
Output: true
Explanation: We can rotate mat 90 degrees clockwise two times to make mat equal target.
Constraints:
n == mat.length == target.length
n == mat[i].length == target[i].length
1 <= n <= 10
mat[i][j]
and target[i][j]
are either 0
or 1
.class Solution {
fun findRotation(mat: Array<IntArray>, target: Array<IntArray>): Boolean {
for (i in 0..3) {
if (mat.contentDeepEquals(target)) {
return true
}
rotate(mat)
}
return false
}
private fun rotate(mat: Array<IntArray>) {
// Reverse Rows
run {
var i = 0
var j = mat.size - 1
while (i < j) {
val tempRow = mat[i]
mat[i] = mat[j]
mat[j] = tempRow
i++
j--
}
}
// Transpose
for (i in mat.indices) {
for (j in i + 1 until mat.size) {
val temp = mat[i][j]
mat[i][j] = mat[j][i]
mat[j][i] = temp
}
}
}
}