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1427. Perform String Shifts

Description

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

  • directioni can be 0 (for left shift) or 1 (for right shift).
  • amounti 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
  • directioni is either 0 or 1.
  • 0 <= amounti <= 100

Solutions

Solution 1: Simulation

We can denote the length of the string $s$ as $n$. Next, we traverse the array $shift$, accumulate to get the final offset $x$, then take $x$ modulo $n$, the final result is to move the first $n - x$ characters of $s$ to the end.

The time complexity is $O(n + m)$, where $n$ and $m$ are the lengths of the string $s$ and the array $shift$ respectively. The space complexity is $O(1)$.

  • class Solution { public String stringShift(String s, int[][] shift) { int x = 0; for (var e : shift) { if (e[0] == 0) { e[1] *= -1; } x += e[1]; } int n = s.length(); x = (x % n + n) % n; return s.substring(n - x) + s.substring(0, n - x); } } 
  • class Solution { public: string stringShift(string s, vector<vector<int>>& shift) { int x = 0; for (auto& e : shift) { if (e[0] == 0) { e[1] = -e[1]; } x += e[1]; } int n = s.size(); x = (x % n + n) % n; return s.substr(n - x, x) + s.substr(0, n - x); } }; 
  • class Solution: def stringShift(self, s: str, shift: List[List[int]]) -> str: x = sum((b if a else -b) for a, b in shift) x %= len(s) return s[-x:] + s[:-x] 
  • func stringShift(s string, shift [][]int) string { x := 0 for _, e := range shift { if e[0] == 0 { e[1] = -e[1] } x += e[1] } n := len(s) x = (x%n + n) % n return s[n-x:] + s[:n-x] } 
  • function stringShift(s: string, shift: number[][]): string { let x = 0; for (const [a, b] of shift) { x += a === 0 ? -b : b; } x %= s.length; if (x < 0) { x += s.length; } return s.slice(-x) + s.slice(0, -x); }

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