LeetCode in Kotlin
2381. Shifting Letters II
Medium
You are given a string s of lowercase English letters and a 2D integer array shifts where shifts[i] = [starti, endi, directioni]. For every i, shift the characters in s from the index starti to the index endi (inclusive) forward if directioni = 1, or shift the characters backward if directioni = 0.
Shifting a character forward means replacing it with the next letter in the alphabet (wrapping around so that 'z' becomes 'a'). Similarly, shifting a character backward means replacing it with the previous letter in the alphabet (wrapping around so that 'a' becomes 'z').
Return the final string after all such shifts to s are applied.
Example 1:
Input: s = “abc”, shifts = [[0,1,0],[1,2,1],[0,2,1]]
Output: “ace”
Explanation: Firstly, shift the characters from index 0 to index 1 backward. Now s = “zac”.
Secondly, shift the characters from index 1 to index 2 forward. Now s = “zbd”.
Finally, shift the characters from index 0 to index 2 forward. Now s = “ace”.
Example 2:
Input: s = “dztz”, shifts = [[0,0,0],[1,1,1]]
Output: “catz”
Explanation: Firstly, shift the characters from index 0 to index 0 backward. Now s = “cztz”.
Finally, shift the characters from index 1 to index 1 forward. Now s = “catz”.
Constraints:
1 <= s.length, shifts.length <= 5 * 104shifts[i].length == 30 <= starti <= endi < s.length0 <= directioni <= 1sconsists of lowercase English letters.
Solution
class Solution {
fun shiftingLetters(s: String, shifts: Array<IntArray>): String {
val diff = IntArray(s.length + 1)
var l: Int
var r: Int
for (shift in shifts) {
l = shift[0]
r = shift[1] + 1
diff[l] += 26
diff[r] += 26
if (shift[2] == 0) {
diff[l]--
diff[r]++
} else {
diff[l]++
diff[r]--
}
diff[l] %= 26
diff[r] %= 26
}
val sb = StringBuilder()
var current = 0
var `val`: Int
for (i in 0 until s.length) {
current += diff[i]
`val` = s[i].code - 'a'.code
`val` += current
`val` %= 26
sb.append(('a'.code + `val`).toChar())
}
return sb.toString()
}
}