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;` `}` |

## 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` |

## 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` |

## 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 ` |

**Output:**

Yes