AC Java: 1classSolution {2Map<Character, Character> parent =newHashMap<>();34publicString smallestEquivalentString(String A, String B, String S) {5for(inti = 0; i<A.length(); i++){6chara =A.charAt(i);7charb =B.charAt(i);89if(find(a) !=find(b)){10union(a, b);11}12}1...
Given a string s containing only digits, return the lexicographically smallest string that can be obtained after swapping adjacent digits in s with the same parity at most once. Digits have the same parity if both are odd or both are even. For example, 5 and 9, as well as 2 and 4, h...
classSolution{publicStringsmallestString(String s){int i=0,n=s.length();char[]cs=s.toCharArray();while(i<n&&cs[i]=='a'){i++;}if(i==n){cs[n-1]='z';}while(i<n&&cs[i]!='a'){cs[i]--;i++;}returnString.valueOf(cs);}} Runtime:22 ms, faster than80.00%of Java online...
If you work with Java everyday then the problem might be obvious to you, that to compare strings lexicographically you need to use thecompareTomethod on the first string. And since VTL is implemented in Java, that is also the case here. Instead, the right way to do it is this: I hav...
MonotonicULIDulid=newMonotonicULID();StringulidString=ulid.generate();// Generate ULID in string representationbyte[]ulidBinary=ulid.generateBinary();// Generate ULID in binary representation Develop Please run the following before sending a PR: ...
Note: Two strings are lexicographically equal if they are the same length and contain the same characters in the same positions. Sample Solution: Scala Code: object Scala_String { def test(str1: String, str2: String): String = {
This is a Java implementation ofUniversally Unique Lexicographically Sortable Identifier. In summary: Sorted by generation time; Can be stored as a UUID/GUID; Can be stored as a string of 26 chars; Can be stored as an array of 16 bytes; ...
void permutations(string str) { // Basisfall if (str.length() == 0) { return; } while (1) { // die aktuelle Permutation drucken cout << str << " "; // finde die nächste lexikografisch geordnete Permutation if (!next_permutation(str.begin(), str.end())) { break; } } ...
voidprev_permutation(stringstr) { while(1) { // die aktuelle Permutation drucken cout<<str<<" "; // finde die vorherige lexikografisch geordnete Permutation if(!prev_permutation(str.begin(),str.end())){ break; } } } intmain() ...