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 <= 100s consists of uppercase and lowercase English letters.
Solutions#
Solution 1#
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| 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
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| 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);
}
}
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| 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);
}
};
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| 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]
}
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| 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#
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| 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
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| 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);
}
}
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| 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);
}
};
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| 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]
}
|
