1427. Perform String Shifts 🔒

DifficultyEasy

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 \(\textit{shift}\), 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 \(O(n + m)\), and the space complexity is \(O(1)\), where \(n\) and \(m\) are the lengths of the string \(s\) and the array \(\textit{shift}\) respectively.

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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]

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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);
    }
}

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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);
    }
};

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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]
}

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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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