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 

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 <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

// 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

Javascript

<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>

Output:

2

Time Complexity: O(m+n) where m and n are lengths of the given string str and patt respectively.
Auxiliary Space: O(MAX)