Given two strings str and patt, the task is to find the count of times patt can be formed using the characters of str.
Examples:
Input: str = “geeksforgeeks”, patt = “geeks”
Output: 2
“geeks” can be made at most twice from
the characters of “geeksforgeeks”.Input: str = “abcbca”, patt = “aabc”
Output: 1
Approach: Count the frequency of all the characters of str and patt and store them in arrays strFreq[] and pattFreq[] respectively. Now any character ch which appears in patt can be used in a maximum of strFreq[ch] / pattFreq[ch] words and the minimum of this value among all the characters of patt is the required answer.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
const int MAX = 26;
// Function to update the freq[] array // to store the frequencies of // all the characters of str void updateFreq(string str, int freq[])
{ int len = str.length();
// Update the frequency of the characters
for ( int i = 0; i < len; i++) {
freq[str[i] - 'a' ]++;
}
} // Function to return the maximum count // of times patt can be formed // using the characters of str int maxCount(string str, string patt)
{ // To store the frequencies of
// all the characters of str
int strFreq[MAX] = { 0 };
updateFreq(str, strFreq);
// To store the frequencies of
// all the characters of patt
int pattFreq[MAX] = { 0 };
updateFreq(patt, pattFreq);
// To store the result
int ans = INT_MAX;
// For every character
for ( int i = 0; i < MAX; i++) {
// If the current character
// doesn't appear in patt
if (pattFreq[i] == 0)
continue ;
// Update the result
ans = min(ans, strFreq[i] / pattFreq[i]);
}
return ans;
} // Driver code int main()
{ string str = "geeksforgeeks" ;
string patt = "geeks" ;
cout << maxCount(str, patt);
return 0;
} |
// Java implementation of the approach class GFG
{ static int MAX = 26 ;
// Function to update the freq[] array // to store the frequencies of // all the characters of str static void updateFreq(String str, int freq[])
{ int len = str.length();
// Update the frequency of the characters
for ( int i = 0 ; i < len; i++)
{
freq[str.charAt(i) - 'a' ]++;
}
} // Function to return the maximum count // of times patt can be formed // using the characters of str static int maxCount(String str, String patt)
{ // To store the frequencies of
// all the characters of str
int []strFreq = new int [MAX];
updateFreq(str, strFreq);
// To store the frequencies of
// all the characters of patt
int []pattFreq = new int [MAX];
updateFreq(patt, pattFreq);
// To store the result
int ans = Integer.MAX_VALUE;
// For every character
for ( int i = 0 ; i < MAX; i++)
{
// If the current character
// doesn't appear in patt
if (pattFreq[i] == 0 )
continue ;
// Update the result
ans = Math.min(ans, strFreq[i] / pattFreq[i]);
}
return ans;
} // Driver code public static void main(String[] args)
{ String str = "geeksforgeeks" ;
String patt = "geeks" ;
System.out.print(maxCount(str, patt));
} } // This code is contributed by Rajput-Ji |
# Python3 implementation of the approach MAX = 26
# Function to update the freq[] array # to store the frequencies of # all the characters of strr def updateFreq(strr, freq):
lenn = len (strr)
# Update the frequency of the characters
for i in range (lenn):
freq[ ord (strr[i]) - ord ( 'a' )] + = 1
# Function to return the maximum count # of times patt can be formed # using the characters of strr def maxCount(strr, patt):
# To store the frequencies of
# all the characters of strr
strrFreq = [ 0 for i in range ( MAX )]
updateFreq(strr, strrFreq)
# To store the frequencies of
# all the characters of patt
pattFreq = [ 0 for i in range ( MAX )]
updateFreq(patt, pattFreq)
# To store the result
ans = 10 * * 9
# For every character
for i in range ( MAX ):
# If the current character
# doesn't appear in patt
if (pattFreq[i] = = 0 ):
continue
# Update the result
ans = min (ans, strrFreq[i] / / pattFreq[i])
return ans
# Driver code strr = "geeksforgeeks"
patt = "geeks"
print (maxCount(strr, patt))
# This code is contributed by Mohit Kumar |
// C# implementation of the approach using System;
class GFG
{ static int MAX = 26;
// Function to update the []freq array // to store the frequencies of // all the characters of str static void updateFreq(String str, int []freq)
{ int len = str.Length;
// Update the frequency of the characters
for ( int i = 0; i < len; i++)
{
freq[str[i] - 'a' ]++;
}
} // Function to return the maximum count // of times patt can be formed // using the characters of str static int maxCount(String str, String patt)
{ // To store the frequencies of
// all the characters of str
int []strFreq = new int [MAX];
updateFreq(str, strFreq);
// To store the frequencies of
// all the characters of patt
int []pattFreq = new int [MAX];
updateFreq(patt, pattFreq);
// To store the result
int ans = int .MaxValue;
// For every character
for ( int i = 0; i < MAX; i++)
{
// If the current character
// doesn't appear in patt
if (pattFreq[i] == 0)
continue ;
// Update the result
ans = Math.Min(ans, strFreq[i] / pattFreq[i]);
}
return ans;
} // Driver code public static void Main(String[] args)
{ String str = "geeksforgeeks" ;
String patt = "geeks" ;
Console.Write(maxCount(str, patt));
} } // This code is contributed by 29AjayKumar |
<script> // JavaScript implementation of the approach
const MAX = 26;
// Function to update the freq[] array
// to store the frequencies of
// all the characters of str
function updateFreq(str, freq) {
var len = str.length;
// Update the frequency of the characters
for ( var i = 0; i < len; i++) {
freq[str[i].charCodeAt(0) - "a" .charCodeAt(0)]++;
}
}
// Function to return the maximum count
// of times patt can be formed
// using the characters of str
function maxCount(str, patt) {
// To store the frequencies of
// all the characters of str
var strFreq = new Array(MAX).fill(0);
updateFreq(str, strFreq);
// To store the frequencies of
// all the characters of patt
var pattFreq = new Array(MAX).fill(0);
updateFreq(patt, pattFreq);
// To store the result
var ans = 21474836473;
// For every character
for ( var i = 0; i < MAX; i++) {
// If the current character
// doesn't appear in patt
if (pattFreq[i] == 0) continue ;
// Update the result
ans = Math.min(ans, strFreq[i] / pattFreq[i]);
}
return ans;
}
// Driver code
var str = "geeksforgeeks" ;
var patt = "geeks" ;
document.write(maxCount(str, patt));
</script>
|
2
Time Complexity: O(m+n) where m and n are lengths of the given string str and patt respectively.
Auxiliary Space: O(MAX)