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Day 14 of 30 days Leetcode challenge - Perform String Shifts

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][0] <= 1
0 <= shift[i][1] <= 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] == 0 {
                return result + next[1]
            }else{
                return result - next[1]
            }
        } % 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])
    }
}

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