Problem Statement
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.
Example 1:
Input: s ="egg",t ="add"
Output: true
Example 2:
Input: s ="foo",t ="bar"
Output: false
Example 3:
Input: s ="paper",t ="title"
Output: true
Note:
You may assume both s and t have the same length.
Problem link
Video Tutorial
You can find the detailed video tutorial here
Thought Process
A relatively straightforward problem, very similar to Word Pattern. All we have to do is check the one to one mapping from string a to string b, also it needs to maintain a bijection mapping (meaning no two different characters in a should map to the same character in b)
Use bijection mapping
Check character one by one from a and b. If char in a hasn't been seen before, create a one to one mapping between this char in a and the char in b so later if this char in a is seen again, it has to map to b, else we return false. Moreover, need to make sure the char in b is never mapped by a different character.
|
|
|
An explanation int the video |
Solutions
public boolean isIsomorphic(String a, String b) {
if (a == null || b == null || a.length() != b.length()) {
return false;
}
Map<Character, Character> lookup = new HashMap<>();
Set<Character> dupSet = new HashSet<>();
for (int i = 0; i < a.length(); i++) {
char c1 = a.charAt(i);
char c2 = b.charAt(i);
if (lookup.containsKey(c1)) {
if (c2 != lookup.get(c1)) {
return false;
}
} else {
lookup.put(c1, c2);
// this to prevent different c1s map to the same c2, it has to be a bijection mapping
if (dupSet.contains(c2)) {
return false;
}
dupSet.add(c2);
}
}
return true;
}
Time Complexity: O(N), N is the length of string a or string b
Space Complexity: O(N), N is the length of string a or string b because the hashmap and set we use