Isomorphic Strings [Easy] — Two-Way Character Mapping

Problem Statement

Two strings are isomorphic if characters in s can be replaced to get t, with consistent mapping (no two chars map to same, order preserved).

Input: s="egg", t="add" → true  (e→a, g→d)
Input: s="foo", t="bar" → false (o→a and o→r — conflict)
Input: s="paper", t="title" → true

Intuition

Maintain two maps: s→t and t→s. For each pair (sc, tc), check both mappings are consistent.

Solutions

C++

bool isIsomorphic(string s, string t) {
    char ms[256]={}, mt[256]={};
    for(int i=0;i<s.size();i++){
        if(ms[(unsigned char)s[i]]==0&&mt[(unsigned char)t[i]]==0){
            ms[(unsigned char)s[i]]=t[i]; mt[(unsigned char)t[i]]=s[i];
        } else if(ms[(unsigned char)s[i]]!=t[i]||mt[(unsigned char)t[i]]!=s[i]) return false;
    }
    return true;
}

Java

public boolean isIsomorphic(String s, String t) {
    Map<Character,Character> ms=new HashMap<>(), mt=new HashMap<>();
    for(int i=0;i<s.length();i++){
        char sc=s.charAt(i), tc=t.charAt(i);
        if(ms.containsKey(sc)&&ms.get(sc)!=tc) return false;
        if(mt.containsKey(tc)&&mt.get(tc)!=sc) return false;
        ms.put(sc,tc); mt.put(tc,sc);
    }
    return true;
}

Python

def isIsomorphic(s: str, t: str) -> bool:
    st, ts = {}, {}
    for sc, tc in zip(s, t):
        if st.get(sc, tc) != tc or ts.get(tc, sc) != sc:
            return False
        st[sc] = tc; ts[tc] = sc
    return True

JavaScript

var isIsomorphic = function(s, t) {
    const st=new Map(), ts=new Map();
    for(let i=0;i<s.length;i++){
        const sc=s[i], tc=t[i];
        if((st.has(sc)&&st.get(sc)!==tc)||(ts.has(tc)&&ts.get(tc)!==sc)) return false;
        st.set(sc,tc); ts.set(tc,sc);
    }
    return true;
};

Complexity

Time: O(n), Space: O(1) — bounded character set