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
// 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;
} |
// 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 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# 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 // 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 ?> |
<script> // Javascript program to check if a query
// string is present is given set.
let MAX_CHAR = 256;
function isPresent(s, q)
{
// Count occurrences of all
// characters in s.
let freq = new Array(MAX_CHAR);
freq.fill(0);
for (let 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 (let i = 0; i < q.length; i++)
{
freq[q[i]]--;
if (freq[q[i]] < 0)
return false ;
}
return true ;
}
let s = "abctd" ;
let q = "cat" ;
if (isPresent(s, q))
document.write( "Yes" );
else
document.write( "No" );
</script> |
Yes
Time complexity : O(n)
Auxiliary Space: O(256)