Hard
In the video game Fallout 4, the quest “Road to Freedom” requires players to reach a metal dial called the “Freedom Trail Ring” and use the dial to spell a specific keyword to open the door.
Given a string ring
that represents the code engraved on the outer ring and another string key
that represents the keyword that needs to be spelled, return the minimum number of steps to spell all the characters in the keyword.
Initially, the first character of the ring is aligned at the "12:00"
direction. You should spell all the characters in key
one by one by rotating ring
clockwise or anticlockwise to make each character of the string key aligned at the "12:00"
direction and then by pressing the center button.
At the stage of rotating the ring to spell the key character key[i]
:
ring
’s characters at the "12:00"
direction, where this character must equal key[i]
.key[i]
has been aligned at the "12:00"
direction, press the center button to spell, which also counts as one step. After the pressing, you could begin to spell the next character in the key (next stage). Otherwise, you have finished all the spelling.Example 1:
Input: ring = “godding”, key = “gd”
Output: 4
Explanation: For the first key character ‘g’, since it is already in place, we just need 1 step to spell this character. For the second key character ‘d’, we need to rotate the ring “godding” anticlockwise by two steps to make it become “ddinggo”. Also, we need 1 more step for spelling. So the final output is 4.
Example 2:
Input: ring = “godding”, key = “godding”
Output: 13
Constraints:
1 <= ring.length, key.length <= 100
ring
and key
consist of only lower case English letters.key
could always be spelled by rotating ring
.class Solution {
fun findRotateSteps(ring: String, key: String): Int {
val indexs: Array<MutableList<Int>?> = arrayOfNulls<MutableList<Int>?>(26)
for (i in 0 until ring.length) {
val num = ring[i].code - 'a'.code
if (indexs[num] == null) {
indexs[num] = ArrayList()
}
indexs[num]!!.add(i)
}
val dp = Array(101) { IntArray(101) }
return find(ring, 0, key, 0, dp, indexs)
}
private fun find(
ring: String,
i: Int,
key: String,
j: Int,
cache: Array<IntArray>,
indexs: Array<MutableList<Int>?>,
): Int {
if (j == key.length) {
return 0
}
if (cache[i][j] != 0) {
return cache[i][j]
}
var ans = Int.MAX_VALUE
val targets: List<Int>? = indexs[key[j].code - 'a'.code]
for (t in targets!!) {
var delta = Math.abs(i - t)
delta = Math.min(delta, Math.abs(ring.length - delta))
ans = Math.min(ans, 1 + delta + find(ring, t, key, j + 1, cache, indexs))
}
cache[i][j] = ans
return ans
}
}