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2309. Greatest English Letter in Upper and Lower Case — Daily Problem

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2309. Greatest English Letter in Upper and Lower Case

Approach

  1. Hash table: The first approach that comes to mind is using a dictionary. But it turns out we can directly traverse the alphabet from Z downward, check if both cases exist, and return immediately.

Runtime beats 99.9%

  1. Bit manipulation: We use two integers mask1 and mask2 to record which lowercase and uppercase letters appear in string s. In mask1, bit i represents whether lowercase letter i appears; in mask2, bit i represents whether uppercase letter i appears.

We then perform a bitwise AND of mask1 and mask2. The result mask has bit i set only if both cases of letter i appear.

We find the highest set bit in mask and convert it to the corresponding uppercase letter. If no bit is set, return an empty string.

Author: ylb
Link: https://leetcode.cn/problems/greatest-english-letter-in-upper-and-lower-case/solutions/2077636/by-lcbin-zbg0/
Source: LeetCode
All rights reserved by the author. Commercial use requires authorization; non-commercial use must cite 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:
        # Use two integers mask1 and mask2 to record lowercase and uppercase letters in s
        # In mask1, bit i represents whether lowercase letter i appears;
        # In mask2, bit i represents whether uppercase letter i 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 ""

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