LeetCode 1427. Perform String Shifts Solution in Java, C++, Python & More | Explanation + Code

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).

PythonJavaC++GoTypeScript

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](code-box)

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); } }(code-box)

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); } };(code-box)

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] }(code-box)

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); }(code-box)