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

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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 = “zambiatek”, patt = “zambiatek”
Output: 2
“zambiatek” can be made at most twice from
the characters of “zambiatek”.

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 = "zambiatek";
    string patt = "zambiatek";
 
    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 = "zambiatek";
    String patt = "zambiatek";
 
    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 = "zambiatek"
patt = "zambiatek"
 
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 = "zambiatek";
    String patt = "zambiatek";
 
    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 = "zambiatek";
      var patt = "zambiatek";
 
      document.write(maxCount(str, patt));
    </script>

Output:

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

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