# Perform String Shifts

#### Created: April 14, 2020 by [lek-tin]

##### Last updated: April 14, 2020

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. `1 <= s.length <= 100`
2. s only contains lower case English letters.
3. `1 <= shift.length <= 100`
4. `shift[i].length == 2`
5. `0 <= shift[i] <= 1`
6. `0 <= shift[i] <= 100`

### Solution

``````class Solution:
def stringShift(self, s: str, shift: List[List[int]]) -> str:
offset = 0
N = len(s)

for sh in shift:
offset += sh if sh > 0 else -sh
offset %= N

res = ""
for i in range(len(s)):
res += s[(-offset+i)%N]

return res
``````