LeetCode in Kotlin
821. Shortest Distance to a Character
Easy
Given a string s and a character c that occurs in s, return an array of integers answer where answer.length == s.length and answer[i] is the distance from index i to the closest occurrence of character c in s.
The distance between two indices i and j is abs(i - j), where abs is the absolute value function.
Example 1:
Input: s = “loveleetcode”, c = “e”
Output: [3,2,1,0,1,0,0,1,2,2,1,0]
Explanation: The character ‘e’ appears at indices 3, 5, 6, and 11 (0-indexed).
The closest occurrence of ‘e’ for index 0 is at index 3, so the distance is abs(0 - 3) = 3.
The closest occurrence of ‘e’ for index 1 is at index 3, so the distance is abs(1 - 3) = 2.
For index 4, there is a tie between the ‘e’ at index 3 and the ‘e’ at index 5, but the distance is still the same: abs(4 - 3) == abs(4 - 5) = 1.
The closest occurrence of ‘e’ for index 8 is at index 6, so the distance is abs(8 - 6) = 2.
Example 2:
Input: s = “aaab”, c = “b”
Output: [3,2,1,0]
Constraints:
1 <= s.length <= 104s[i]andcare lowercase English letters.- It is guaranteed that
coccurs at least once ins.
Solution
class Solution {
fun shortestToChar(s: String, c: Char): IntArray {
val result = IntArray(s.length)
result.fill(Int.MAX_VALUE)
for (i in s.indices) {
if (s[i] == c) {
result[i] = 0
}
}
for (i in s.indices) {
if (result[i] != 0) {
var j = i - 1
while (j >= 0 && result[j] != 0) {
j--
}
if (j >= 0) {
result[i] = i - j
}
j = i + 1
while (j < s.length && result[j] != 0) {
j++
}
if (j < s.length) {
result[i] = result[i].coerceAtMost(j - i)
}
}
}
return result
}
}