HàPhan 河

Perform String Shifts

You are given a string s containing lowercase English letters, and a matrix shift, where `shift[i] = [direction, amount]`:

• `direction` can be `0` (for left shift) or `1` (for right shift).
• `amount` is the amount by which string `s` is to be shifted.
• A left shift by `1` means remove the first character of `s` and append it to the end.
• Similarly, a right shift by `1` means remove the last character of `s` and add it to the beginning.

Return the final string after all operations.

Example 1:

``````Input: s = "abc", shift = [[0,1],[1,2]]
Output: "cab"

Explanation:
[0,1] means shift to left by 1. "abc" -> "bca"
[1,2] means shift to right by 2. "bca" -> "cab"
``````

Example 2:

``````Input: s = "abcdefg", shift = [[1,1],[1,1],[0,2],[1,3]]
Output: "efgabcd"

Explanation:
[1,1] means shift to right by 1. "abcdefg" -> "gabcdef"
[1,1] means shift to right by 1. "gabcdef" -> "fgabcde"
[0,2] means shift to left by 2. "fgabcde" -> "abcdefg"
[1,3] means shift to right by 3. "abcdefg" -> "efgabcd"
``````

Constraints:

`1 <= s.length <= 100`
`s only contains lower case English letters.`
`1 <= shift.length <= 100`
`shift[i].length == 2`
`0 <= shift[i] <= 1`
`0 <= shift[i] <= 100`

Solution

Calculate all rotations to one rotation, then do that only.

``````class Solution {
func stringShift(_ s: String, _ shift: [[Int]]) -> String {
var shifted = shift.reduce(0) { result, next in
if next == 0 {
return result + next
}else{
return result - next
}
} % s.count

var position: Int

if shifted > 0 {
position = shifted
}else{
position = s.count - abs(shifted)
}

let slicePos = s.index(s.startIndex, offsetBy: position)
return String(s[slicePos...]) + String(s[..<slicePos])
}
}
``````