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796. Rotate String

DifficultyEasy

Description

Given two strings s and goal, return true if and only if s can become goal after some number of shifts on s.

A shift on s consists of moving the leftmost character of s to the rightmost position.

  • For example, if s = "abcde", then it will be "bcdea" after one shift.

 

Example 1:

Input: s = "abcde", goal = "cdeab"

Output: true

Explanation:

Rotating s to the left by 2 positions moves "ab" to the end, resulting in "cdeab", which is equal to goal.

Example 2:

Input: s = "abcde", goal = "abced"

Output: false

Explanation:

No sequence of rotations of s can produce "abced". The characters appear in a different relative order, so goal is not a rotation of s.

 

Constraints:

  • 1 <= s.length, goal.length <= 100
  • s and goal consist of lowercase English letters.

Solutions

Solution 1

Thinking

Is \(goal\) a rotation of \(s\)? Unequal lengths fail; otherwise \(goal\) is a substring of \(s+s\).

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class Solution:
    def rotateString(self, s: str, goal: str) -> bool:
        return len(s) == len(goal) and goal in s + s

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class Solution {
    public boolean rotateString(String s, String goal) {
        return s.length() == goal.length() && (s + s).contains(goal);
    }
}

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class Solution {
public:
    bool rotateString(string s, string goal) {
        return s.size() == goal.size() && strstr((s + s).data(), goal.data());
    }
};

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func rotateString(s string, goal string) bool {
    return len(s) == len(goal) && strings.Contains(s+s, goal)
}

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function rotateString(s: string, goal: string): boolean {
    return s.length === goal.length && (goal + goal).includes(s);
}

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impl Solution {
    pub fn rotate_string(s: String, goal: String) -> bool {
        s.len() == goal.len() && (s.clone() + &s).contains(&goal)
    }
}

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class Solution {
    /**
     * @param String $s
     * @param String $goal
     * @return Boolean
     */
    function rotateString($s, $goal) {
        return strlen($goal) === strlen($s) && strpos($s . $s, $goal) !== false;
    }
}

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