Question
Given two strings s and t, determine if they are isomorphic.
Two strings are isomorphic if the characters in s can be replaced to get t.
All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character but 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 class Solution { 2 public boolean isIsomorphic(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 int length = s.length(); 10 Map<Character, Character> map = new HashMap<Character, Character>(); 11 for (int i = 0; i < length; i++) { 12 char tmp1 = s.charAt(i); 13 char tmp2 = t.charAt(i); 14 if (map.containsKey(tmp1)) { 15 if (map.get(tmp1) != tmp2) 16 return false; 17 } else if (map.containsValue(tmp2)) { 18 return false; 19 } else { 20 map.put(tmp1, tmp2); 21 } 22 } 23 return true; 24 } 25 }
Solution 2
Use extra space to reduce time complexity.
1 public class Solution { 2 public boolean isIsomorphic(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 int length = s.length(); 10 Map<Character, Character> map = new HashMap<Character, Character>(); 11 Set<Character> counts = new HashSet<Character>(); 12 for (int i = 0; i < length; i++) { 13 char tmp1 = s.charAt(i); 14 char tmp2 = t.charAt(i); 15 if (map.containsKey(tmp1)) { 16 if (map.get(tmp1) != tmp2) 17 return false; 18 } else if (counts.contains(tmp2)) { 19 return false; 20 } else { 21 map.put(tmp1, tmp2); 22 counts.add(tmp2); 23 } 24 } 25 return true; 26 } 27 }