Perform string shifts

Time: O(N+L); Space: O(1); easy

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 <= len(s) <= 100

  • s only contains lower case English letters.

  • 1 <= len(shift) <= 100

  • len(shift[i]) == 2

  • 0 <= shift[i][0] <= 1

  • 0 <= shift[i][1] <= 100

[1]:

class Solution1(object):
    """
    Time: O(N+L)
    Space: O(L)
    """
    def stringShift(self, s, shift):
        """
        :type s: str
        :type shift: List[List[int]]
        :rtype: str
        """
        left_shifts = 0
        for direction, amount in shift:
            if not direction:
                left_shifts += amount
            else:
                left_shifts -= amount
        left_shifts %= len(s)

        return s[left_shifts:] + s[:left_shifts]

[2]:

sol = Solution1()

s = "abc"
shift = [[0,1],[1,2]]
assert sol.stringShift(s, shift) == "cab"

s = "abcdefg"
shift = [[1,1],[1,1],[0,2],[1,3]]
assert sol.stringShift(s, shift) == "efgabcd"