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Check whether second string can be formed from characters of first string
• Difficulty Level : Easy
• Last Updated : 12 May, 2021

Given two strings str1 and str2, check if str2 can be formed from str1
Example :

```Input : str1 = geekforgeeks, str2 = geeks
Output : Yes
Here, string2 can be formed from string1.

Input : str1 = geekforgeeks, str2 = and
Output :  No
Here string2 cannot be formed from string1.

Input : str1 = geekforgeeks, str2 = geeeek
Output :  Yes
Here string2 can be formed from string1
as string1 contains 'e' comes 4 times in
string2 which is present in string1. ```

The idea is to count frequencies of characters of str1 in a count array. Then traverse str2 and decrease frequency of characters of str2 in the count array. If frequency of a characters becomes negative at any point, return false.
Below is the implementation of above approach :

C++

 `// CPP program to check whether second string``// can be formed from first string``#include ``using` `namespace` `std;``const` `int` `MAX = 256;` `bool` `canMakeStr2(string str1, string str2)``{``    ``// Create a count array and count frequencies``    ``// characters in str1.``    ``int` `count[MAX] = {0};``    ``for` `(``int` `i = 0; i < str1.length(); i++)``        ``count[str1[i]]++;``        ` `    ``// Now traverse through str2 to check``    ``// if every character has enough counts``    ``for` `(``int` `i = 0; i < str2.length(); i++)``    ``{``        ``if` `(count[str2[i]] == 0)``           ``return` `false``;``        ``count[str2[i]]--;``    ``}``    ``return` `true``;``}` `// Driver Code``int` `main()``{``    ``string str1 = ``"geekforgeeks"``;``    ``string str2 = ``"for"``;``    ``if` `(canMakeStr2(str1, str2))``       ``cout << ``"Yes"``;``    ``else``       ``cout << ``"No"``;``    ``return` `0;``}`

Java

 `// Java program to check whether second string``// can be formed from first string`` ` `class` `GFG {`` ` `    ``static` `int` `MAX = ``256``;`` ` `    ``static` `boolean` `canMakeStr2(String str1, String str2)``    ``{``        ``// Create a count array and count frequencies``        ``// characters in str1.``        ``int``[] count = ``new` `int``[MAX];``        ``char` `[]str3 = str1.toCharArray();``        ``for` `(``int` `i = ``0``; i < str3.length; i++)``            ``count[str3[i]]++;`` ` `        ``// Now traverse through str2 to check``        ``// if every character has enough counts``        ` `        ``char` `[]str4 = str2.toCharArray();``        ``for` `(``int` `i = ``0``; i < str4.length; i++) {``            ``if` `(count[str4[i]] == ``0``)``                ``return` `false``;``            ``count[str4[i]]--;``        ``}``        ``return` `true``;``    ``}`` ` `    ``// Driver Code``    ``static` `public` `void` `main(String []args)``    ``{``        ``String str1 = ``"geekforgeeks"``;``        ``String str2 = ``"for"``;``        ``if` `(canMakeStr2(str1, str2))``            ``System.out.println(``"Yes"``);``        ``else``            ``System.out.println(``"No"``);``    ``}``}`

Python3

 `# Python program to check whether second string``# can be formed from first string``def` `canMakeStr2(s1, s2):` `    ``# Create a count array and count``    ``# frequencies characters in s1``    ``count ``=` `{s1[i] : ``0` `for` `i ``in` `range``(``len``(s1))}``    ` `    ``for` `i ``in` `range``(``len``(s1)):``        ``count[s1[i]] ``+``=` `1``    ` `    ``# Now traverse through str2 to check``    ``# if every character has enough counts``    ``for` `i ``in` `range``(``len``(s2)):``        ``if` `(count.get(s2[i]) ``=``=` `None` `or` `count[s2[i]] ``=``=` `0``):``            ``return` `False``        ``count[s2[i]] ``-``=` `1``    ``return` `True` `# Driver Code``s1 ``=` `"geekforgeeks"``s2 ``=` `"for"` `if` `canMakeStr2(s1, s2):``    ``print``(``"Yes"``)``else``:``    ``print``(``"No"``)`

C#

 `// C# program to check whether second string``// can be formed from first string``using` `System;` `class` `GFG {` `    ``static` `int` `MAX = 256;` `    ``static` `bool` `canMakeStr2(``string` `str1, ``string` `str2)``    ``{``        ``// Create a count array and count frequencies``        ``// characters in str1.``        ``int``[] count = ``new` `int``[MAX];``        ``for` `(``int` `i = 0; i < str1.Length; i++)``            ``count[str1[i]]++;` `        ``// Now traverse through str2 to check``        ``// if every character has enough counts``        ``for` `(``int` `i = 0; i < str2.Length; i++) {``            ``if` `(count[str2[i]] == 0)``                ``return` `false``;``            ``count[str2[i]]--;``        ``}``        ``return` `true``;``    ``}` `    ``// Driver Code``    ``static` `public` `void` `Main()``    ``{``        ``string` `str1 = ``"geekforgeeks"``;``        ``string` `str2 = ``"for"``;``        ``if` `(canMakeStr2(str1, str2))``            ``Console.WriteLine(``"Yes"``);``        ``else``            ``Console.WriteLine(``"No"``);``    ``}``}` `// This code is contributed by vt_m.`

PHP

 ``

Javascript

 ``
Output
`Yes`

Output :

`Yes`

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