# Count occurrences of a string that can be constructed from another given string

• Difficulty Level : Easy
• Last Updated : 15 Sep, 2022

Given two strings str1 and str2 where str1 being the parent string. The task is to find out the number of string as str2 that can be constructed using letters of str1

Note: All the letters are in lowercase and each character should be used only once.

Examples:

```Input: str1 = "geeksforgeeks", str2 = "geeks"
Output: 2

Input: str1 = "geekgoinggeeky", str2 = "geeks"
Output: 0```

Approach: Store the frequency of characters of str2 in hash2, and do the same for str1 in hash1. Now, find out the minimum value of hash1[i]/hash2[i] for all i where hash2[i]>0.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach``#include ``using` `namespace` `std;` `// Function to find the count``int` `findCount(string str1, string str2)``{``    ``int` `len = str1.size();``    ``int` `len2 = str2.size();``    ``int` `ans = INT_MAX;` `    ``// Initialize hash for both strings``    ``int` `hash1[26] = { 0 }, hash2[26] = { 0 };` `    ``// hash the frequency of letters of str1``    ``for` `(``int` `i = 0; i < len; i++)``        ``hash1[str1[i] - ``'a'``]++;` `    ``// hash the frequency of letters of str2``    ``for` `(``int` `i = 0; i < len2; i++)``        ``hash2[str2[i] - ``'a'``]++;` `    ``// Find the count of str2 constructed from str1``    ``for` `(``int` `i = 0; i < 26; i++)``        ``if` `(hash2[i])``            ``ans = min(ans, hash1[i] / hash2[i]);` `    ``// Return answer``    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``string str1 = ``"geeksclassesatnoida"``;``    ``string str2 = ``"sea"``;``    ``cout << findCount(str1, str2);` `    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``class` `GFG``{``    ``// Function to find the count``    ``static` `int` `findCount(String str1, String str2)``    ``{``        ``int` `len = str1.length();``        ``int` `len2 = str2.length();``        ``int` `ans = Integer.MAX_VALUE;``    ` `        ``// Initialize hash for both strings``        ``int` `[] hash1 = ``new` `int``[``26``];``        ``int` `[] hash2 = ``new` `int``[``26``];``    ` `        ``// hash the frequency of letters of str1``        ``for` `(``int` `i = ``0``; i < len; i++)``            ``hash1[(``int``)(str1.charAt(i) - ``'a'``)]++;``    ` `        ``// hash the frequency of letters of str2``        ``for` `(``int` `i = ``0``; i < len2; i++)``            ``hash2[(``int``)(str2.charAt(i) - ``'a'``)]++;``    ` `        ``// Find the count of str2 constructed from str1``        ``for` `(``int` `i = ``0``; i < ``26``; i++)``            ``if` `(hash2[i] != ``0``)``                ``ans = Math.min(ans, hash1[i] / hash2[i]);``    ` `        ``// Return answer``        ``return` `ans;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main(String []args)``    ``{``        ``String str1 = ``"geeksclassesatnoida"``;``        ``String str2 = ``"sea"``;``        ``System.out.println(findCount(str1, str2));``    ``}``}` `// This code is contributed by ihritik`

## Python3

 `# Python3 implementation of the above approach` `import` `sys` `# Function to find the count``def` `findCount(str1, str2):` `    ``len1 ``=` `len``(str1)``    ``len2 ``=` `len``(str2)``    ``ans ``=` `sys.maxsize` `    ``# Initialize hash for both strings``    ``hash1 ``=` `[``0``] ``*` `26``    ``hash2 ``=` `[``0``] ``*` `26` `    ``# hash the frequency of letters of str1``    ``for` `i ``in` `range``(``0``, len1):``        ``hash1[``ord``(str1[i]) ``-` `97``] ``=` `hash1[``ord``(str1[i]) ``-` `97``] ``+` `1` `    ``# hash the frequency of letters of str2``    ``for` `i ``in` `range``(``0``, len2):``        ``hash2[``ord``(str2[i]) ``-` `97``] ``=` `hash2[``ord``(str2[i]) ``-` `97``] ``+` `1``        ` `    ``# Find the count of str2 constructed from str1``    ``for` `i ``in` `range` `(``0``, ``26``):``        ``if` `(hash2[i] !``=` `0``):``            ``ans ``=` `min``(ans, hash1[i] ``/``/` `hash2[i])` `    ``# Return answer``    ``return` `ans` `    ` `# Driver code``str1 ``=` `"geeksclassesatnoida"``str2 ``=` `"sea"``print``(findCount(str1, str2))` `# This code is contributed by ihritik`

## C#

 `// C# implementation of the above approach``using` `System;` `class` `GFG``{``    ``// Function to find the count``    ``static` `int` `findCount(``string` `str1, ``string` `str2)``    ``{``        ``int` `len = str1.Length;``        ``int` `len2 = str2.Length;``        ``int` `ans = Int32.MaxValue;``    ` `        ``// Initialize hash for both strings``        ``int` `[] hash1 = ``new` `int``[26];``        ``int` `[] hash2 = ``new` `int``[26];``    ` `        ``// hash the frequency of letters of str1``        ``for` `(``int` `i = 0; i < len; i++)``            ``hash1[str1[i] - ``'a'``]++;``    ` `        ``// hash the frequency of letters of str2``        ``for` `(``int` `i = 0; i < len2; i++)``            ``hash2[str2[i] - ``'a'``]++;``    ` `        ``// Find the count of str2 constructed from str1``        ``for` `(``int` `i = 0; i < 26; i++)``            ``if` `(hash2[i] != 0)``                ``ans = Math.Min(ans, hash1[i] / hash2[i]);``    ` `        ``// Return answer``        ``return` `ans;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``string` `str1 = ``"geeksclassesatnoida"``;``        ``string` `str2 = ``"sea"``;``        ``Console.WriteLine(findCount(str1, str2));``    ``}``}` `// This code is contributed by ihritik`

## PHP

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## Javascript

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Output

`3`

Complexity Analysis:

• Time Complexity: O(n)
• Auxiliary Space: O(1)

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