[LeetCode] 744. Find Smallest Letter Greater Than Target_Easy tag: Binary Search

Given a list of sorted characters letters containing only lowercase letters, and given a target letter target, find the smallest element in the list that is larger than the given target.

Letters also wrap around. For example, if the target is target = 'z' and letters = ['a', 'b'], the answer is 'a'.

Examples:

Input:
letters = ["c", "f", "j"]
target = "a"
Output: "c"

Input:
letters = ["c", "f", "j"]
target = "c"
Output: "f"

Input:
letters = ["c", "f", "j"]
target = "d"
Output: "f"

Input:
letters = ["c", "f", "j"]
target = "g"
Output: "j"

Input:
letters = ["c", "f", "j"]
target = "j"
Output: "c"

Input:
letters = ["c", "f", "j"]
target = "k"
Output: "c"

 

Note:

1. letters has a length in range [2, 10000].
2. letters consists of lowercase letters, and contains at least 2 unique letters.
3. target is a lowercase letter.

思路for loop letters, 一旦有ord(c) > ord(target), 返回c, 否则返回letter c. 因为是non- decreasing order

O(n)

 

Code

class Solution:
    def nextGreatestLetter(self, letters: List[str], target: str) -> str:
        for c in letters:
            if ord(c) > ord(target):
                return c
        return letters[0]

 

Binary search, change letter into index using ord() method, also deal with edge case first 如果ord(target) 比第一个小或者跟最后一个相等或者更大，返回letters[0], 其他的就找first index s.t ord(letters[index]) > ord(target)

O( lg n)

class Solution:
    def nextGreatestLetter(self, letters: List[str], target: str) -> str:
        index_target = ord(target)
        if index_target >= ord(letters[-1])  or index_target < ord(letters[0]):
            return letters[0]
        l, r = 0, len(letters) - 1
        while l + 1 < r:
            mid = l + (r - l)//2
            index = ord(letters[mid])
            if index > index_target:
                r = mid
            else:
                l = mid
        return letters[r] # 因为letters[l] 总会<=target, edge case 也排除了