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Possibility of a word from a given set of characters
• Difficulty Level : Easy
• Last Updated : 28 Jun, 2018

Given two strings ‘s’ and ‘q’, check if all characters of q are present in ‘s’.

Examples:

```Example:
Input: s = "abctd"
q = "cat"
Output: Yes
Explanation:
All characters of "cat" are
present in "abctd"

Input: s = dog
hod
Output: No
Explanation:
Given query 'hod' hod has the letter
'h' which is not available in set 'dog',
hence the output is no.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to try all characters one by one. Find its number of occurrences in ‘q’, then in ‘s’. Number of occurrences in ‘q’ must be less than or equal to same in ‘s’. If all characters satisfy this condition, return true. Else return false.

An efficient solution is to create a frequency array of length 256 (Number of possible characters) and initialize it to 0. Then we calculate the frequency of each element present in ‘s’. After counting characters in ‘s’, we traverse through ‘q’ and check if frequency of each character is less than its frequency in ‘s’, by reducing its frequency in the frequency array constructed for ‘s’.

Below given is the implementation of the above approach

## C++

 `// CPP program to check if a query string ` `// is present is given set. ` `#include ` `using` `namespace` `std; ` ` `  `const` `int` `MAX_CHAR = 256; ` ` `  `bool` `isPresent(string s, string q) ` `{ ` `    ``// Count occurrences of all characters ` `    ``// in s. ` `    ``int` `freq[MAX_CHAR] = { 0 }; ` `    ``for` `(``int` `i = 0; i < s.length(); i++) ` `        ``freq[s[i]]++; ` ` `  `    ``// Check if number of occurrences of ` `    ``// every character in q is less than ` `    ``// or equal to that in s. ` `    ``for` `(``int` `i = 0; i < q.length(); i++) { ` `        ``freq[q[i]]--; ` `        ``if` `(freq[q[i]] < 0)  ` `           ``return` `false``; ` `    ``} ` ` `  `    ``return` `true``; ` `} ` ` `  `// driver program ` `int` `main() ` `{ ` `    ``string s = ``"abctd"``; ` `    ``string q = ``"cat"``; ` ` `  `    ``if` `(isPresent(s, q)) ` `        ``cout << ``"Yes"``; ` `    ``else` `        ``cout << ``"No"``; ` ` `  `    ``return` `0; ` `} `

## Java

 `// java program to check if a query ` `// string is present is given set. ` `import` `java.io.*; ` ` `  `public` `class` `GFG { ` ` `  `    ``static` `int` `MAX_CHAR = ``256``; ` `     `  `    ``static` `boolean` `isPresent(String s, String q) ` `    ``{ ` `         `  `        ``// Count occurrences of all ` `        ``// characters in s. ` `        ``int` `[]freq = ``new` `int``[MAX_CHAR]; ` `        ``for` `(``int` `i = ``0``; i < s.length(); i++) ` `            ``freq[s.charAt(i)]++; ` `     `  `        ``// Check if number of occurrences of ` `        ``// every character in q is less than ` `        ``// or equal to that in s. ` `        ``for` `(``int` `i = ``0``; i < q.length(); i++) ` `        ``{ ` `            ``freq[q.charAt(i)]--; ` `             `  `            ``if` `(freq[q.charAt(i)] < ``0``)  ` `                ``return` `false``; ` `        ``} ` `     `  `        ``return` `true``; ` `    ``} ` `     `  `    ``// driver program ` `    ``static` `public` `void` `main (String[] args) ` `    ``{ ` `        ``String s = ``"abctd"``; ` `        ``String q = ``"cat"``; ` `     `  `        ``if` `(isPresent(s, q)) ` `            ``System.out.println(``"Yes"``); ` `        ``else` `            ``System.out.println(``"No"``); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## Python 3

 `# Python 3 program to check if a query ` `# string is present is given set. ` `MAX_CHAR ``=` `256` ` `  `def` `isPresent(s, q): ` ` `  `    ``# Count occurrences of all characters ` `    ``# in s. ` `    ``freq ``=` `[``0``] ``*`  `MAX_CHAR  ` `    ``for` `i ``in` `range``(``0` `, ``len``(s)): ` `        ``freq[``ord``(s[i])] ``+``=` `1` ` `  `    ``# Check if number of occurrences of ` `    ``# every character in q is less than ` `    ``# or equal to that in s. ` `    ``for` `i ``in` `range``(``0``, ``len``(q)): ` `        ``freq[``ord``(q[i])] ``-``=` `1` `        ``if` `(freq[``ord``(q[i])] < ``0``):  ` `            ``return` `False` `     `  `    ``return` `True` ` `  `# driver program ` `s ``=` `"abctd"` `q ``=` `"cat"` ` `  `if` `(isPresent(s, q)): ` `    ``print``(``"Yes"``) ` `else``: ` `    ``print``(``"No"``) ` ` `  `# This code is contributed by Smitha `

## C#

 `// C# program to check if a query ` `// string is present is given set. ` `using` `System; ` ` `  `public` `class` `GFG { ` `     `  `    ``static` `int` `MAX_CHAR = 256; ` `     `  `    ``static` `bool` `isPresent(``string` `s, ``string` `q) ` `    ``{ ` ` `  `        ``// Count occurrences of all ` `        ``// characters in s. ` `        ``int` `[]freq = ``new` `int``[MAX_CHAR]; ` `         `  `        ``for` `(``int` `i = 0; i < s.Length; i++) ` `            ``freq[s[i]]++; ` `     `  `        ``// Check if number of occurrences of ` `        ``// every character in q is less than ` `        ``// or equal to that in s. ` `        ``for` `(``int` `i = 0; i < q.Length; i++) ` `        ``{ ` `            ``freq[q[i]]--; ` `             `  `            ``if` `(freq[q[i]] < 0)  ` `                ``return` `false``; ` `        ``} ` `     `  `        ``return` `true``; ` `    ``} ` `     `  `    ``// driver program ` `    ``static` `public` `void` `Main () ` `    ``{ ` `        ``string` `s = ``"abctd"``; ` `        ``string` `q = ``"cat"``; ` `     `  `        ``if` `(isPresent(s, q)) ` `            ``Console.WriteLine(``"Yes"``); ` `        ``else` `            ``Console.WriteLine(``"No"``); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` `

Output:

```Yes
```

Time complexity : O(n)

This article is contributed by Twinkl Bajaj. 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.