Skip to content
Related Articles

Related Articles

Improve Article
C Program to check if two given strings are isomorphic to each other
  • Difficulty Level : Medium
  • Last Updated : 03 Nov, 2020

Given two strings str1 and str2, the task is to check if the two given strings are isomorphic to each other or not.

Two strings are said to be isomorphic if there is a one to one mapping possible for every character of str1 to every character of str2 and all occurrences of every character in str1 map to same character in str2.

Examples:

Input: str1 = “egg”, str2 = “add”
Output: Yes
Explanation:
‘e’ in str1 with ASCII value 101 is mapped to ‘a’ in str2 with ASCII value 97.
‘g’ in str1 with ASCII value 103 is mapped to ‘d’ in str2 with ASCII value 100.

Input: str1 = “eggs”, str2 = “addd”
Output: No



Hashing Approach: Refer to the previous post for the Hashmap based approach.
Time Complexity: O(N)
Auxiliary Space: O(256)

ASCII-value based Approach: The idea is similar to that of the above approach. Follow the steps below to solve the problem:

  1. Initialize two arrays of size 256.
  2. Iterate through characters of the given strings and increment the index equal to the ASCII value of the character at ith position.
  3. If the are no conflicts in the mapping of the characters, print Yes. Otherwise, print No.

Below is the implementation of the above approach:

C




// C Program to implement
// the above approach
  
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
  
// Function to check and return if strings
// str1 and str2 are ismorphic
bool areIsomorphic(char *str1, char *str2)
{
    // If the length of the strings
    // are not equal
    if (strlen(str1) != strlen(str2)) {
        return false;
    }
  
    // Initialise two arrays
    int arr1[256] = { 0 }, arr2[256] = { 0 };
  
    // Travsersing both the strings
    for (int i = 0; i < strlen(str1); i++) {
  
        // If current characters don't map
        if (arr1[(int)str1[i]] 
        != arr2[(int)str2[i]]) {
            return false;
        }
  
        // Increment the count of characters
        // at their respective ASCII indices
        arr1[(int)str1[i]]++;
        arr2[(int)str2[i]]++;
    }
    return true;
}
  
// Driver Code
int main()
{
    char s1[] = "aab", s2[] = "xxy";
  
    if (areIsomorphic(s1, s2))
        printf("Yes\n");
    else
        printf("No\n");
  
    return 0;
}
Output:
Yes

Time Complexity: O(N)
Auxiliary Space: O(256)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live 




My Personal Notes arrow_drop_up
Recommended Articles
Page :