Isomorphic Strings LeetCode Solution

Problem – Isomorphic Strings

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

Two strings s and t 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

Constraints:

• 1 <= s.length <= 5 * 104
• t.length == s.length
• s and t consist of any valid ascii character.

Isomorphic Strings LeetCode Solution in Python

class Solution(object):
    def isIsomorphic(self, s, t):
        s2t, t2s = {}, {}
        for i in range(len(s)):
            if s[i] in s2t and s2t[s[i]] != t[i]:
                return False
            if t[i] in t2s and t2s[t[i]] != s[i]:
                return False
            s2t[s[i]] = t[i]
            t2s[t[i]] = s[i]
        return True
    
    def isIsomorphic1(self, s, t):
        d1, d2 = {}, {}
        for i, val in enumerate(s):
            d1[val] = d1.get(val, []) + [i]
        for i, val in enumerate(t):
            d2[val] = d2.get(val, []) + [i]
        return sorted(d1.values()) == sorted(d2.values())
        
    def isIsomorphic2(self, s, t):
        d1, d2 = [[] for _ in xrange(256)], [[] for _ in xrange(256)]
        for i, val in enumerate(s):
            d1[ord(val)].append(i)
        for i, val in enumerate(t):
            d2[ord(val)].append(i)
        return sorted(d1) == sorted(d2)
    
    def isIsomorphic3(self, s, t):
        return len(set(zip(s, t))) == len(set(s)) == len(set(t))
    
    def isIsomorphic4(self, s, t): 
        return [s.find(i) for i in s] == [t.find(j) for j in t]
    
    def isIsomorphic5(self, s, t):
        return map(s.find, s) == map(t.find, t)

    def isIsomorphic6(self, s, t):
        d1, d2 = [0 for _ in xrange(256)], [0 for _ in xrange(256)]
        for i in xrange(len(s)):
            if d1[ord(s[i])] != d2[ord(t[i])]:
                return False
            d1[ord(s[i])] = i+1
            d2[ord(t[i])] = i+1
        return True

Isomorphic Strings LeetCode Solution in Java

public class Solution {
    public boolean isIsomorphic(String s1, String s2) {
        int[] m = new int[512];
        for (int i = 0; i < s1.length(); i++) {
            if (m[s1.charAt(i)] != m[s2.charAt(i)+256]) return false;
            m[s1.charAt(i)] = m[s2.charAt(i)+256] = i+1;
        }
        return true;
    }
}

Isomorphic Strings LeetCode Solution in C++

bool isIsomorphic(string s, string t) {
        char map_s[128] = { 0 };
        char map_t[128] = { 0 };
        int len = s.size();
        for (int i = 0; i < len; ++i)
        {
            if (map_s[s[i]]!=map_t[t[i]]) return false;
            map_s[s[i]] = i+1;
            map_t[t[i]] = i+1;
        }
        return true;    
    }

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