Description
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
Explanation:
The strings s and t can be made identical by:
- Mapping
'e' to 'a'.
- Mapping
'g' to 'd'.
Example 2:
Input: s = "f11", t = "b23"
Output: false
Explanation:
The strings s and t can not be made identical as '1' needs to be mapped to both '2' and '3'.
Example 3:
Input: s = "paper", t = "title"
Output: true
Constraints:
1 <= s.length <= 5 * 104t.length == s.lengths and t consist of any valid ascii character.
Solutions
Solution 1: Hash Table or Array
We can use two hash tables or arrays d_1 and d_2 to record the character mapping relationship between s and t.
Traverse s and t, if the corresponding character mapping relationships in d_1 and d_2 are different, return false, otherwise update the corresponding character mapping relationships in d_1 and d_2. After the traversal is complete, it means that s and t are isomorphic, and return true.
The time complexity is O(n) and the space complexity is O(C). Where n is the length of the string s; and C is the size of the character set, which is C = 256 in this problem.
PythonJavaC++GoTypeScriptRustC#
class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
d1 = {}
d2 = {}
for a, b in zip(s, t):
if (a in d1 and d1[a] != b) or (b in d2 and d2[b] != a):
return False
d1[a] = b
d2[b] = a
return True(code-box)
class Solution {
public boolean isIsomorphic(String s, String t) {
Map<Character, Character> d1 = new HashMap<>();
Map<Character, Character> d2 = new HashMap<>();
int n = s.length();
for (int i = 0; i < n; ++i) {
char a = s.charAt(i), b = t.charAt(i);
if (d1.containsKey(a) && d1.get(a) != b) {
return false;
}
if (d2.containsKey(b) && d2.get(b) != a) {
return false;
}
d1.put(a, b);
d2.put(b, a);
}
return true;
}
}(code-box)
class Solution {
public:
bool isIsomorphic(string s, string t) {
int d1[256]{};
int d2[256]{};
int n = s.size();
for (int i = 0; i < n; ++i) {
char a = s[i], b = t[i];
if (d1[a] != d2[b]) {
return false;
}
d1[a] = d2[b] = i + 1;
}
return true;
}
};(code-box)
func isIsomorphic(s string, t string) bool {
d1 := [256]int{}
d2 := [256]int{}
for i := range s {
if d1[s[i]] != d2[t[i]] {
return false
}
d1[s[i]] = i + 1
d2[t[i]] = i + 1
}
return true
}(code-box)
function isIsomorphic(s: string, t: string): boolean {
const d1: number[] = new Array(256).fill(0);
const d2: number[] = new Array(256).fill(0);
for (let i = 0; i < s.length; ++i) {
const a = s.charCodeAt(i);
const b = t.charCodeAt(i);
if (d1[a] !== d2[b]) {
return false;
}
d1[a] = i + 1;
d2[b] = i + 1;
}
return true;
}(code-box)
use std::collections::HashMap;
impl Solution {
fn help(s: &[u8], t: &[u8]) -> bool {
let mut map = HashMap::new();
for i in 0..s.len() {
if map.contains_key(&s[i]) {
if map.get(&s[i]).unwrap() != &t[i] {
return false;
}
} else {
map.insert(s[i], t[i]);
}
}
true
}
pub fn is_isomorphic(s: String, t: String) -> bool {
let (s, t) = (s.as_bytes(), t.as_bytes());
Self::help(s, t) && Self::help(t, s)
}
}(code-box)
public class Solution {
public bool IsIsomorphic(string s, string t) {
int[] d1 = new int[256];
int[] d2 = new int[256];
for (int i = 0; i < s.Length; ++i) {
var a = s[i];
var b = t[i];
if (d1[a] != d2[b]) {
return false;
}
d1[a] = i + 1;
d2[b] = i + 1;
}
return true;
}
}(code-box)
Solution 2
PythonJava
class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
d1, d2 = [0] * 256, [0] * 256
for i, (a, b) in enumerate(zip(s, t), 1):
a, b = ord(a), ord(b)
if d1[a] != d2[b]:
return False
d1[a] = d2[b] = i
return True(code-box)
class Solution {
public boolean isIsomorphic(String s, String t) {
int[] d1 = new int[256];
int[] d2 = new int[256];
int n = s.length();
for (int i = 0; i < n; ++i) {
char a = s.charAt(i), b = t.charAt(i);
if (d1[a] != d2[b]) {
return false;
}
d1[a] = i + 1;
d2[b] = i + 1;
}
return true;
}
}(code-box)
