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

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the above approach
#include <bits/stdc++.h>
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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP implementation of the above approach 
  
// Function to find the count 
function findCount($str1, $str2
    $len = strlen($str1) ; 
    $len2 = strlen($str1); 
    $ans = PHP_INT_MAX; 
  
    // Initialize hash for both strings 
    $hash1 = array_fill(0, 26, 0) ;
    $hash2 = array_fill(0, 26, 0); 
  
    // hash the frequency of letters of str1 
    for ($i = 0; $i < $len; $i++) 
        $hash1[ord($str1[$i]) - ord('a')]++; 
  
    // hash the frequency of letters of str2 
    for ($i = 0; $i < $len2; $i++) 
        $hash2[ord($str2[$i]) - ord('a')]++; 
  
    // Find the count of str2 constructed from str1 
    for ($i = 0; $i < 26; $i++) 
        if ($hash2[$i]) 
            $ans = min($ans, $hash1[$i] / $hash2[$i]); 
  
    // Return answer 
    return $ans
  
    // Driver code 
    $str1 = "geeksclassesatnoida"
    $str2 = "sea"
    echo findCount($str1, $str2); 
      
// This code is contributed by Ryuga
?>

chevron_right


Output:

3


My Personal Notes arrow_drop_up

Recommended Posts: