[LeetCode] Weekly Challenge Perform String Shifts

Posted 2021-03-01 codingEskimo

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][0] <= 1
6. 0 <= shift[i][1] <= 100

这道题的想法比较直接，分析以后就发现不要每次都移动s，因为移动s的时间复杂度是O(L), 减少移动就减少了复杂度。需要做的就是把所有的移动合并，先合并所有的同方向，然后比较左右看需要如何最终移动。有一个点需要注意，最后集合了所有的shift以后，有可能长度超过s，所以要取个模。如果shift的长度为N，那么这道题最终的时间复杂度是O(N+L), 空间复杂度是O(L)。这是copy string需要的。

class Solution:
    def stringShift(self, s: str, shift: List[List[int]]) -> str:
        if not s or not shift: return s
        
        move = [0,0]
        for sh in shift:
            move[sh[0]] += sh[1]
        move = [x % len(s) for x in move]
        
        if move[0] == move[1]:
            return s
        elif move[0] < move[1]:
            r_step = move[1] - move[0]
            return s[-r_step:]+s[0:-r_step]
        else:
            l_step = move[0] - move[1]
            return s[l_step:]+s[0:l_step]