Possibility of a word from a given set of characters

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.


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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP program to check if a query string
// is present is given set.
#include <bits/stdc++.h>
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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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.

chevron_right


Python 3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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.

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program to check if a query string
// is present is given set.
  
function isPresent($s, $q)
{
      
    // Count occurrences of 
    // all characters in s.
    $freq = array(256);
    for ($i = 0; $i < 256; $i++)
        $freq[$i] = 0;
      
    for ($i = 0; $i < strlen($s); $i++)
        $freq[ ord($s[$i]) - ord('a') ]++ ;
          
    // Check if number of occurrences of
    // every character in q is less than
    // or equal to that in s.
    for ($i = 0; $i < strlen($q); $i++)
    {
        $freq[ord($q[$i]) - ord('a')]--;
        if ($freq[ord($q[$i]) - ord('a')] < 0) 
        return false;
    }
  
    return true;
}
  
    // Driver Code
    $s = "abctd";
    $q = "cat";
      
    if (isPresent($s, $q))
        echo "Yes";
    else
        echo "No"
          
// This code is contributed by Sam007
?>

chevron_right



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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up



Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.