Hard
There is a safe protected by a password. The password is a sequence of n
digits where each digit can be in the range [0, k - 1]
.
The safe has a peculiar way of checking the password. When you enter in a sequence, it checks the most recent n
digits that were entered each time you type a digit.
"345"
and you enter in "012345"
:
0
, the most recent 3
digits is "0"
, which is incorrect.1
, the most recent 3
digits is "01"
, which is incorrect.2
, the most recent 3
digits is "012"
, which is incorrect.3
, the most recent 3
digits is "123"
, which is incorrect.4
, the most recent 3
digits is "234"
, which is incorrect.5
, the most recent 3
digits is "345"
, which is correct and the safe unlocks.Return any string of minimum length that will unlock the safe at some point of entering it.
Example 1:
Input: n = 1, k = 2
Output: “10”
Explanation: The password is a single digit, so enter each digit. “01” would also unlock the safe.
Example 2:
Input: n = 2, k = 2
Output: “01100”
Explanation: For each possible password:
“00” is typed in starting from the 4th digit.
“01” is typed in starting from the 1st digit.
“10” is typed in starting from the 3rd digit.
“11” is typed in starting from the 2nd digit.
Thus “01100” will unlock the safe. “01100”, “10011”, and “11001” would also unlock the safe.
Constraints:
1 <= n <= 4
1 <= k <= 10
1 <= kn <= 4096
import kotlin.math.pow
class Solution {
private var foundStr: String? = null
fun crackSafe(n: Int, k: Int): String? {
val targetCnt = k.toDouble().pow(n.toDouble()).toInt()
val visited = BooleanArray(10.0.pow(n.toDouble()).toInt())
visited[0] = true
val visitedCnt = 1
val crackStr = StringBuilder()
for (i in 0 until n) {
crackStr.append('0')
}
dfsAddPwd(n, k, crackStr, 0, visited, visitedCnt, targetCnt)
return foundStr
}
private fun dfsAddPwd(
n: Int,
k: Int,
crackStr: StringBuilder,
prev: Int,
visited: BooleanArray,
visitedCnt: Int,
targetCnt: Int,
) {
if (foundStr != null) {
return
}
if (visitedCnt == targetCnt) {
foundStr = crackStr.toString()
return
}
val root = 10 * prev % 10.0.pow(n.toDouble()).toInt()
for (i in 0 until k) {
val current = root + i
if (!visited[current]) {
crackStr.append(i)
visited[current] = true
dfsAddPwd(n, k, crackStr, current, visited, visitedCnt + 1, targetCnt)
crackStr.setLength(crackStr.length - 1)
visited[current] = false
}
}
}
}