Map every character of one string to another such that all occurrences are mapped to the same character
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; } |
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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 |
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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 |
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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 |
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Output:
Yes
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