# Count of strings that can be formed from another string using each character at-most once

Given two strings str1 and str2, the task is to print the number of times str2 can be formed using characters of str1. However, a character at any index of str1 can only be used once in the formation of str2.

Examples:

Input: str1 = “arajjhupoot”, str2 = “rajput”
Output: 1
str2 can only be formed once using characters of str1.

Input: str1 = “foreeksgekseg”, str2 = “geeks”
Output: 2

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

Approach: Since the problem has a restriction on using characters of str1 only once to form str2. If one character has been used to form one str2, it cannot be used in forming another str2. Every character of str2 must be present in str1 at least for the formation of one str1. If all the characters of str2 are already present in str1, then the character which has the minimum occurrence in str1 will be the number of str2’s that can be formed using the characters of str1 once. Below are the steps:

• Create an hash-array which stores the number of occurrences of each character of str.
• Iterate for all the characters of str2, and find the minimum most occurrences of every character in str1.
• Return the minimum occurrence which will be the answer.

Below is the implementation of the above approach:

 `/// C++ program to print the number of times ` `// str2 can be formed from str1 using the ` `// characters of str1 only once ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the number of str2 ` `// that can be formed using characters of str1 ` `int` `findNumberOfTimes(string str1, string str2) ` `{ ` `    ``int` `freq = { 0 }; ` ` `  `    ``int` `l1 = str1.length(); ` ` `  `    ``// iterate and mark the frequencies of ` `    ``// all characters in str1 ` `    ``for` `(``int` `i = 0; i < l1; i++) ` `        ``freq[str1[i] - ``'a'``] += 1; ` ` `  `    ``int` `l2 = str2.length(); ` `    ``int` `count = INT_MAX; ` ` `  `    ``// find the minimum frequency of ` `    ``// every character in str1 ` `    ``for` `(``int` `i = 0; i < l2; i++) ` `        ``count = min(count, freq[str2[i] - ``'a'``]); ` ` `  `    ``return` `count; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``string str1 = ``"foreeksgekseg"``; ` `    ``string str2 = ``"geeks"``; ` ` `  `    ``cout << findNumberOfTimes(str1, str2)  ` `         ``<< endl; ` ` `  `    ``return` `0; ` `} `

 `// Java program to print the number of times  ` `// str2 can be formed from str1 using the  ` `// characters of str1 only once  ` ` `  `class` `GFG { ` ` `  `// Function to find the number of str2  ` `// that can be formed using characters of str1  ` `    ``static` `int` `findNumberOfTimes(String str1, String str2) { ` `        ``int` `freq[] = ``new` `int``[``26``]; ` ` `  `        ``int` `l1 = str1.length(); ` ` `  `        ``// iterate and mark the frequencies of  ` `        ``// all characters in str1  ` `        ``for` `(``int` `i = ``0``; i < l1; i++) { ` `            ``freq[str1.charAt(i) - ``'a'``] += ``1``; ` `        ``} ` ` `  `        ``int` `l2 = str2.length(); ` `        ``int` `count = Integer.MAX_VALUE; ` ` `  `        ``// find the minimum frequency of  ` `        ``// every character in str1  ` `        ``for` `(``int` `i = ``0``; i < l2; i++) { ` `            ``count = Math.min(count, freq[str2.charAt(i) - ``'a'``]); ` `        ``} ` ` `  `        ``return` `count; ` `    ``} ` ` `  `    ``public` `static` `void` `main(String[] args) { ` ` `  `        ``String str1 = ``"foreeksgekseg"``; ` `        ``String str2 = ``"geeks"``; ` `        ``System.out.println(findNumberOfTimes(str1, str2)); ` ` `  `    ``} ` `} ` `/* This code is contributed by 29AjayKumar*/`

 `# Python3 program to print the number of  ` `# times str2 can be formed from str1 using  ` `# the characters of str1 only once ` `import` `sys ` ` `  `# Function to find the number of str2 ` `# that can be formed using characters of str1 ` `def` `findNumberOfTimes(str1, str2): ` `     `  `    ``freq ``=` `[``0``] ``*` `26` `    ``l1 ``=` `len``(str1) ` `     `  `    ``# iterate and mark the frequencies  ` `    ``# of all characters in str1 ` `    ``for` `i ``in` `range``(l1): ` `        ``freq[``ord``(str1[i]) ``-` `ord``(``"a"``)] ``+``=` `1` `    ``l2 ``=` `len``(str2) ` `    ``count ``=` `sys.maxsize ` `     `  `    ``# find the minimum frequency of ` `    ``# every character in str1 ` `    ``for` `i ``in` `range``(l2): ` `        ``count ``=` `min``(count, freq[``ord``(str2[i]) ``-`  `                                ``ord``(``'a'``)]) ` `    ``return` `count ` `     `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``str1 ``=` `"foreeksgekseg"` `    ``str2 ``=` `"geeks"` `    ``print``(findNumberOfTimes(str1, str2)) ` ` `  `# This code is contributed by PrinciRaj1992 `

 `// C# program to print the number of  ` `// times str2 can be formed from str1 ` `// using the characters of str1 only once  ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to find the number of  ` `// str2 that can be formed using ` `// characters of str1  ` `static` `int` `findNumberOfTimes(String str1,  ` `                             ``String str2)  ` `{ ` `    ``int` `[]freq = ``new` `int``; ` ` `  `    ``int` `l1 = str1.Length; ` ` `  `    ``// iterate and mark the frequencies  ` `    ``// of all characters in str1  ` `    ``for` `(``int` `i = 0; i < l1; i++) ` `    ``{ ` `        ``freq[str1[i] - ``'a'``] += 1; ` `    ``} ` ` `  `    ``int` `l2 = str2.Length; ` `    ``int` `count = ``int``.MaxValue; ` ` `  `    ``// find the minimum frequency of  ` `    ``// every character in str1  ` `    ``for` `(``int` `i = 0; i < l2; i++) ` `    ``{ ` `        ``count = Math.Min(count, freq[str2[i] - ``'a'``]); ` `    ``} ` ` `  `    ``return` `count; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main() ` `{ ` `    ``String str1 = ``"foreeksgekseg"``; ` `    ``String str2 = ``"geeks"``; ` `    ``Console.Write(findNumberOfTimes(str1, str2)); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

 ` `

Output:
```2
```

Time Complexity: O(max(l1,l2)), where l1 and l2 are length of str1 and str2 respectively.

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