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744. Find Smallest Letter Greater Than Target

Description

You are given an array of characters letters that is sorted in non-decreasing order, and a character target. There are at least two different characters in letters.

Return the smallest character in letters that is lexicographically greater than target. If such a character does not exist, return the first character in letters.

 

Example 1:

Input: letters = ["c","f","j"], target = "a"
Output: "c"
Explanation: The smallest character that is lexicographically greater than 'a' in letters is 'c'.

Example 2:

Input: letters = ["c","f","j"], target = "c"
Output: "f"
Explanation: The smallest character that is lexicographically greater than 'c' in letters is 'f'.

Example 3:

Input: letters = ["x","x","y","y"], target = "z"
Output: "x"
Explanation: There are no characters in letters that is lexicographically greater than 'z' so we return letters[0].

 

Constraints:

  • 2 <= letters.length <= 104
  • letters[i] is a lowercase English letter.
  • letters is sorted in non-decreasing order.
  • letters contains at least two different characters.
  • target is a lowercase English letter.

Solutions

Since letters is sorted in non-decreasing order, we can use binary search to find the smallest character that is larger than target.

We define the left boundary of the binary search as $l = 0$, and the right boundary as $r = n$. For each binary search, we calculate the middle position $mid = (l + r) / 2$. If $letters[mid] > \textit{target}$, it means we need to continue searching in the left half, so we set $r = mid$. Otherwise, we need to continue searching in the right half, so we set $l = mid + 1$.

Finally, we return $letters[l \mod n]$.

The time complexity is $O(\log n)$, where $n$ is the length of letters. The space complexity is $O(1)$.

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class Solution:
    def nextGreatestLetter(self, letters: List[str], target: str) -> str:
        i = bisect_right(letters, ord(target), key=lambda c: ord(c))
        return letters[i % len(letters)]

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class Solution {
    public char nextGreatestLetter(char[] letters, char target) {
        int i = Arrays.binarySearch(letters, (char) (target + 1));
        i = i < 0 ? -i - 1 : i;
        return letters[i % letters.length];
    }
}

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class Solution {
public:
    char nextGreatestLetter(vector<char>& letters, char target) {
        int i = upper_bound(letters.begin(), letters.end(), target) - letters.begin();
        return letters[i % letters.size()];
    }
};

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func nextGreatestLetter(letters []byte, target byte) byte {
    i := sort.Search(len(letters), func(i int) bool { return letters[i] > target })
    return letters[i%len(letters)]
}

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function nextGreatestLetter(letters: string[], target: string): string {
    let [l, r] = [0, letters.length];
    while (l < r) {
        const mid = (l + r) >> 1;
        if (letters[mid] > target) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return letters[l % letters.length];
}

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impl Solution {
    pub fn next_greatest_letter(letters: Vec<char>, target: char) -> char {
        let mut l = 0;
        let mut r = letters.len();
        while l < r {
            let mid = l + (r - l) / 2;
            if letters[mid] > target {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        letters[l % letters.len()]
    }
}

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class Solution {
    /**
     * @param String[] $letters
     * @param String $target
     * @return String
     */
    function nextGreatestLetter($letters, $target) {
        $l = 0;
        $r = count($letters);
        while ($l < $r) {
            $mid = $l + $r >> 1;
            if ($letters[$mid] > $target) {
                $r = $mid;
            } else {
                $l = $mid + 1;
            }
        }
        return $letters[$l % count($letters)];
    }
}

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