Isomorphic Strings Solutions

Question

Given strings s and t, determine if they are isomorphic.

The strings is isomorphic if the characters in s can is replaced to get t.

All occurrences of a character must is replaced with another character while preserving the order of characters. No, characters may map to the same character and a character may map to itself.

For example,
Given "egg" , "add" return True.

Given "foo" , "bar" return FALSE.

Given "paper" , "title" return True.

Solution 1

Use HashMap, remember to check both key and value. Time complexity O (n^2), Space cost O (n).

Hashmap.containsvalue costs O (n)

1  Public classSolution {2      Public Booleanisisomorphic (string s, String t) {3         if(s = =NULL|| t = =NULL)4             return false;5         if(S.length ()! =t.length ())6             return false;7         if(S.length () ==0 && t.length () ==0)8             return true;9         intLength =s.length ();TenMap<character, character> map =NewHashmap<character, character>(); One          for(inti = 0; i < length; i++) { A             CharTMP1 =S.charat (i); -             CharTMP2 =T.charat (i); -             if(Map.containskey (TMP1)) { the                 if(Map.get (TMP1)! =tmp2) -                     return false; -}Else if(Map.containsvalue (TMP2)) { -                 return false; +}Else { - map.put (TMP1, TMP2); +             } A         } at         return true; -     } -}

Solution 2

Use extra space to reduce time complexity.

1  Public classSolution {2      Public Booleanisisomorphic (string s, String t) {3         if(s = =NULL|| t = =NULL)4             return false;5         if(S.length ()! =t.length ())6             return false;7         if(S.length () ==0 && t.length () ==0)8             return true;9         intLength =s.length ();TenMap<character, character> map =NewHashmap<character, character>(); Oneset<character> counts =NewHashset<character>(); A          for(inti = 0; i < length; i++) { -             CharTMP1 =S.charat (i); -             CharTMP2 =T.charat (i); the             if(Map.containskey (TMP1)) { -                 if(Map.get (TMP1)! =tmp2) -                     return false; -}Else if(Counts.contains (TMP2)) { +                 return false; -}Else { + map.put (TMP1, TMP2); A Counts.add (TMP2); at             } -         } -         return true; -     } -}

Isomorphic Strings Solutions