# Count of times second string can be formed from the characters of first string

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

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

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++

 `// C++ implementation of the approach ` `#include ` `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

 `// 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

 `# 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#

 `// 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 `

Output:

```2
```

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