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Easy
1521
Biweekly Contest 46 Q1
Bit Manipulation
Hash Table
String
Divide and Conquer
Sliding Window

中文文档

Description

A string s is nice if, for every letter of the alphabet that s contains, it appears both in uppercase and lowercase. For example, "abABB" is nice because 'A' and 'a' appear, and 'B' and 'b' appear. However, "abA" is not because 'b' appears, but 'B' does not.

Given a string s, return the longest substring of s that is nice. If there are multiple, return the substring of the earliest occurrence. If there are none, return an empty string.

 

Example 1:

Input: s = "YazaAay"
Output: "aAa"
Explanation: "aAa" is a nice string because 'A/a' is the only letter of the alphabet in s, and both 'A' and 'a' appear.
"aAa" is the longest nice substring.

Example 2:

Input: s = "Bb"
Output: "Bb"
Explanation: "Bb" is a nice string because both 'B' and 'b' appear. The whole string is a substring.

Example 3:

Input: s = "c"
Output: ""
Explanation: There are no nice substrings.

 

Constraints:

  • 1 <= s.length <= 100
  • s consists of uppercase and lowercase English letters.

Solutions

Solution 1

Python3

class Solution:
    def longestNiceSubstring(self, s: str) -> str:
        n = len(s)
        ans = ''
        for i in range(n):
            ss = set()
            for j in range(i, n):
                ss.add(s[j])
                if (
                    all(c.lower() in ss and c.upper() in ss for c in ss)
                    and len(ans) < j - i + 1
                ):
                    ans = s[i : j + 1]
        return ans

Java

class Solution {
    public String longestNiceSubstring(String s) {
        int n = s.length();
        int k = -1;
        int mx = 0;
        for (int i = 0; i < n; ++i) {
            Set<Character> ss = new HashSet<>();
            for (int j = i; j < n; ++j) {
                ss.add(s.charAt(j));
                boolean ok = true;
                for (char a : ss) {
                    char b = (char) (a ^ 32);
                    if (!(ss.contains(a) && ss.contains(b))) {
                        ok = false;
                        break;
                    }
                }
                if (ok && mx < j - i + 1) {
                    mx = j - i + 1;
                    k = i;
                }
            }
        }
        return k == -1 ? "" : s.substring(k, k + mx);
    }
}

C++

class Solution {
public:
    string longestNiceSubstring(string s) {
        int n = s.size();
        int k = -1, mx = 0;
        for (int i = 0; i < n; ++i) {
            unordered_set<char> ss;
            for (int j = i; j < n; ++j) {
                ss.insert(s[j]);
                bool ok = true;
                for (auto& a : ss) {
                    char b = a ^ 32;
                    if (!(ss.count(a) && ss.count(b))) {
                        ok = false;
                        break;
                    }
                }
                if (ok && mx < j - i + 1) {
                    mx = j - i + 1;
                    k = i;
                }
            }
        }
        return k == -1 ? "" : s.substr(k, mx);
    }
};

Go

func longestNiceSubstring(s string) string {
	n := len(s)
	k, mx := -1, 0
	for i := 0; i < n; i++ {
		ss := map[byte]bool{}
		for j := i; j < n; j++ {
			ss[s[j]] = true
			ok := true
			for a := range ss {
				b := a ^ 32
				if !(ss[a] && ss[b]) {
					ok = false
					break
				}
			}
			if ok && mx < j-i+1 {
				mx = j - i + 1
				k = i
			}
		}
	}
	if k < 0 {
		return ""
	}
	return s[k : k+mx]
}

TypeScript

function longestNiceSubstring(s: string): string {
    const n = s.length;
    let ans = '';
    for (let i = 0; i < n; i++) {
        let lower = 0,
            upper = 0;
        for (let j = i; j < n; j++) {
            const c = s.charCodeAt(j);
            if (c > 96) {
                lower |= 1 << (c - 97);
            } else {
                upper |= 1 << (c - 65);
            }
            if (lower == upper && j - i + 1 > ans.length) {
                ans = s.substring(i, j + 1);
            }
        }
    }
    return ans;
}

Solution 2

Python3

class Solution:
    def longestNiceSubstring(self, s: str) -> str:
        n = len(s)
        ans = ''
        for i in range(n):
            lower = upper = 0
            for j in range(i, n):
                if s[j].islower():
                    lower |= 1 << (ord(s[j]) - ord('a'))
                else:
                    upper |= 1 << (ord(s[j]) - ord('A'))
                if lower == upper and len(ans) < j - i + 1:
                    ans = s[i : j + 1]
        return ans

Java

class Solution {
    public String longestNiceSubstring(String s) {
        int n = s.length();
        int k = -1;
        int mx = 0;
        for (int i = 0; i < n; ++i) {
            int lower = 0, upper = 0;
            for (int j = i; j < n; ++j) {
                char c = s.charAt(j);
                if (Character.isLowerCase(c)) {
                    lower |= 1 << (c - 'a');
                } else {
                    upper |= 1 << (c - 'A');
                }
                if (lower == upper && mx < j - i + 1) {
                    mx = j - i + 1;
                    k = i;
                }
            }
        }
        return k == -1 ? "" : s.substring(k, k + mx);
    }
}

C++

class Solution {
public:
    string longestNiceSubstring(string s) {
        int n = s.size();
        int k = -1, mx = 0;
        for (int i = 0; i < n; ++i) {
            int lower = 0, upper = 0;
            for (int j = i; j < n; ++j) {
                char c = s[j];
                if (islower(c))
                    lower |= 1 << (c - 'a');
                else
                    upper |= 1 << (c - 'A');
                if (lower == upper && mx < j - i + 1) {
                    mx = j - i + 1;
                    k = i;
                }
            }
        }
        return k == -1 ? "" : s.substr(k, mx);
    }
};

Go

func longestNiceSubstring(s string) string {
	n := len(s)
	k, mx := -1, 0
	for i := 0; i < n; i++ {
		var lower, upper int
		for j := i; j < n; j++ {
			if unicode.IsLower(rune(s[j])) {
				lower |= 1 << (s[j] - 'a')
			} else {
				upper |= 1 << (s[j] - 'A')
			}
			if lower == upper && mx < j-i+1 {
				mx = j - i + 1
				k = i
			}
		}
	}
	if k < 0 {
		return ""
	}
	return s[k : k+mx]
}