2309. The best English letters with both appropriates and lowercases One question daily
2309. The best English letters with both appropriates and lowercases
Thought：
- Hash table：The first thing to think of the question is to solve it with a dictionary。But find that you can directly traverse the alphabet，fromZStart looking for，See if the applause exists，Just return directly。
Running time exceeds99.9% 2. Bit operation：We can use two integers mask1 and mask2 Record string separately s 中出现的小写字母and大写字母，in mask1 First i Position representation i Does a lowercase letter appear，and mask2 First i Position representation i Whether an uppercase letter appears。
Then we will mask1 and mask2 Perform and calculate，The results obtained mask First i Position representation i Whether the lethals of the letter appear at the same time。
As long as you get mask The highest level of binary representation 1 s position，Convert it to the corresponding capital letter。If all binary positions are not 1，Explain that there is no letter that appears at the same time,，Return to an empty string。
author：ylb Link：https://leetcode.cn/problems/greatest-english-letter-in-upper-and-lower-case/solutions/2077636/by-lcbin-zbg0/ source：Deduction（LeetCode） 著作权归author所有。商业转载请联系author获得授权，Non -commercial reprint Please indicate the source。
Code：
class Solution:
def greatestLetter(self, s: str) -> str:
for i in range(90, 64, -1):
if chr(i) in s and chr(i+32) in s:
return chr(i)
return ""class Solution {
public:
string greatestLetter(string s) {
unordered_set<char> strin(s.begin(),s.end());
for(char c = 'Z'; c >= 'A'; --c){
if(strin.count(c) && strin.count(char(c+32))){
return string(1,c);
}
}
return "";
}
};class Solution:
def greatestLetter(self, s: str) -> str:
# We can use two integers mask1 and mask2
# Record string separately s 中出现的小写字母and大写字母，in mask1 First i Position representation i Does a lowercase letter appear，
# and mask2 First i Position representation i Whether an uppercase letter appears。
mask1 = mask2 = 0
for i in s:
if i.islower():
mask1 |= 1 << (ord(i) - ord("a"))
else:
mask2 |= 1 << (ord(i) - ord("A"))
mask = mask1 & mask2
return chr(mask.bit_length() - 1 + ord("A")) if mask else ""贡献者
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