Given two string s1 and s2, the task is to check if characters of the first string can be mapped with the character of the second string such that if a character ch1 is mapped with some character ch2 then all the occurrences of ch1 will only be mapped with ch2 for both the strings.
Examples:
Input: s1 = “axx”, s2 = “cbc”
Output: Yes
‘a’ in s1 can be mapped to ‘b’ in s2
and ‘x’ in s1 can be mapped to ‘c’ in s2.Input: s1 = “a”, s2 = “df”
Output: No
Approach: If the lengths of both the strings are unequal then the strings cannot be mapped else create two frequency arrays freq1[] and freq2[] which will store the frequencies of all the characters of the given strings s1 and s2 respectively. Now, for every non-zero value in freq1[] find an equal value in freq2[]. If all the non-zero values from freq1[] can be mapped to some value in freq2[] then the answer is possible else not.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; #define MAX 26 // Function that returns true if the mapping is possible bool canBeMapped(string s1, int l1, string s2, int l2) { // Both the strings are of un-equal lengths if (l1 != l2) return false; // To store the frequencies of the // characters in both the string int freq1[MAX] = { 0 }; int freq2[MAX] = { 0 }; // Update frequencies of the characters for (int i = 0; i < l1; i++) freq1[s1[i] - 'a']++; for (int i = 0; i < l2; i++) freq2[s2[i] - 'a']++; // For every character of s1 for (int i = 0; i < MAX; i++) { // If current character is // not present in s1 if (freq1[i] == 0) continue; bool found = false; // Find a character in s2 that has frequency // equal to the current character's // frequency in s1 for (int j = 0; j < MAX; j++) { // If such character is found if (freq1[i] == freq2[j]) { // Set the frequnecy to -1 so that // it doesn't get picked again freq2[j] = -1; // Set found to true found = true; break; } } // If there is no character in s2 // that could be mapped to the // current character in s1 if (!found) return false; } return true; } // Driver code int main() { string s1 = "axx"; string s2 = "cbc"; int l1 = s1.length(); int l2 = s2.length(); if (canBeMapped(s1, l1, s2, l2)) cout << "Yes"; else cout << "No"; return 0; } |
chevron_right
filter_none
Java
// Java implementation of the approach class GFG { static int MAX = 26; // Function that returns true if the mapping is possible public static boolean canBeMapped(String s1, int l1, String s2, int l2) { // Both the strings are of un-equal lengths if (l1 != l2) return false; // To store the frequencies of the // characters in both the string int[] freq1 = new int[MAX]; int[] freq2 = new int[MAX]; // Update frequencies of the characters for (int i = 0; i < l1; i++) freq1[s1.charAt(i) - 'a']++; for (int i = 0; i < l2; i++) freq2[s2.charAt(i) - 'a']++; // For every character of s1 for (int i = 0; i < MAX; i++) { // If current character is // not present in s1 if (freq1[i] == 0) continue; boolean found = false; // Find a character in s2 that has frequency // equal to the current character's // frequency in s1 for (int j = 0; j < MAX; j++) { // If such character is found if (freq1[i] == freq2[j]) { // Set the frequnecy to -1 so that // it doesn't get picked again freq2[j] = -1; // Set found to true found = true; break; } } // If there is no character in s2 // that could be mapped to the // current character in s1 if (!found) return false; } return true; } // Driver code public static void main(String[] args) { String s1 = "axx"; String s2 = "cbc"; int l1 = s1.length(); int l2 = s2.length(); if (canBeMapped(s1, l1, s2, l2)) System.out.println("Yes"); else System.out.println("No"); } } // This code is contributed by // sanjeev2552 |
chevron_right
filter_none
Python3
# Python 3 implementation of the approach MAX = 26 # Function that returns true if the mapping is possible def canBeMapped(s1, l1, s2, l2): # Both the strings are of un-equal lengths if (l1 != l2): return False # To store the frequencies of the # characters in both the string freq1 = [0 for i in range(MAX)] freq2 = [0 for i in range(MAX)] # Update frequencies of the characters for i in range(l1): freq1[ord(s1[i]) - ord('a')] += 1 for i in range(l2): freq2[ord(s2[i]) - ord('a')] += 1 # For every character of s1 for i in range(MAX): # If current character is # not present in s1 if (freq1[i] == 0): continue found = False # Find a character in s2 that has frequency # equal to the current character's # frequency in s1 for j in range(MAX): # If such character is found if (freq1[i] == freq2[j]): # Set the frequnecy to -1 so that # it doesn't get picked again freq2[j] = -1 # Set found to true found = True break # If there is no character in s2 # that could be mapped to the # current character in s1 if (found==False): return False return True # Driver code if __name__ == '__main__': s1 = "axx" s2 = "cbc" l1 = len(s1) l2 = len(s2) if (canBeMapped(s1, l1, s2, l2)): print("Yes") else: print("No") # This code is contributed by # Surendra_Gangwar |
chevron_right
filter_none
C#
// C# implementation of the approach using System; class GFG { static int MAX = 26; // Function that returns true // if the mapping is possible public static Boolean canBeMapped(String s1, int l1, String s2, int l2) { // Both the strings are of un-equal lengths if (l1 != l2) return false; // To store the frequencies of the // characters in both the string int[] freq1 = new int[MAX]; int[] freq2 = new int[MAX]; // Update frequencies of the characters for (int i = 0; i < l1; i++) freq1[s1[i] - 'a']++; for (int i = 0; i < l2; i++) freq2[s2[i] - 'a']++; // For every character of s1 for (int i = 0; i < MAX; i++) { // If current character is // not present in s1 if (freq1[i] == 0) continue; Boolean found = false; // Find a character in s2 that has frequency // equal to the current character's // frequency in s1 for (int j = 0; j < MAX; j++) { // If such character is found if (freq1[i] == freq2[j]) { // Set the frequnecy to -1 so that // it doesn't get picked again freq2[j] = -1; // Set found to true found = true; break; } } // If there is no character in s2 // that could be mapped to the // current character in s1 if (!found) return false; } return true; } // Driver code public static void Main(String[] args) { String s1 = "axx"; String s2 = "cbc"; int l1 = s1.Length; int l2 = s2.Length; if (canBeMapped(s1, l1, s2, l2)) Console.WriteLine("Yes"); else Console.WriteLine("No"); } } // This code is contributed // by PrinciRaj1992 |
chevron_right
filter_none
Output:
Yes
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.
Recommended Posts:
- Find all strings formed from characters mapped to digits of a number
- Check if frequency of character in one string is a factor or multiple of frequency of same character in other string
- Replace all occurrences of character X with character Y in given string
- Minimize ASCII values sum after removing all occurrences of one character
- Count occurrences of a string that can be constructed from another given string
- Find a string such that every character is lexicographically greater than its immediate next character
- Modify the string such that every character gets replaced with the next character in the keyboard
- Sum of all distances between occurrences of same characters in a given string
- Minimum string such that every adjacent character of given string is still adjacent
- Remove all occurrences of a character in a string
- Number of permutations of a string in which all the occurrences of a given character occurs together
- Remove all occurrences of a character in a string | Recursive approach
- Rearrange the given string such that all prime multiple indexes have same character
- Longest subsequence with at least one character appearing in every string
- Lengths of maximized partitions of a string such that each character of the string appears in one substring
- Maximized partitions of a string such that each character of the string appears in one substring
- Check if max occurring character of one string appears same no. of times in other
- Find length of longest subsequence of one string which is substring of another string
- Replace every character of string by character whose ASCII value is K times more than it
- Replace every character of a string by a different character
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
