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 
Explanation:
str2 can only be formed once using characters of str1.

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

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 str1 and str2.
• 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++14

/// C++ program to print the number of times
// str2 can be formed from str1 using the
// characters of str1 only once
#include <bits/stdc++.h>
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[26] = { 0 };
    int freq2[26] = { 0 };

    int l1 = str1.length();
    int l2 = str2.length();
    // iterate and mark the frequencies of
    // all characters in str1
    for (int i = 0; i < l1; i++)
        freq[str1[i] - 'a'] += 1;

  for (int i = 0; i < l2; i++)
        freq2[str2[i] - 'a'] += 1;


    int count = INT_MAX;

    // find the minimum frequency of
    // every character in str1
    for (int i = 0; i < l2; i++)
    {
      if(freq2[str2[i]-'a']!=0)
      count = min(count, freq[str2[i] - 'a']/freq2[str2[i]-'a']);
    }
    return count;
}

// Driver Code
int main()
{
    string str1 = "foreeksgekseg";
    string str2 = "geeks";

    cout << findNumberOfTimes(str1, str2)
         << endl;

    return 0;
}

Java

// 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 freq2[] = 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();
        for (int i = 0; i < l2; i++)
        {
            freq2[str2.charAt(i) - 'a'] += 1;
        }


        int count = Integer.MAX_VALUE;

        // find the minimum frequency of
        // every character in str1
        for (int i = 0; i < l2; i++)
        {
            if(freq2[str2.charAt(i)-'a']!=0)
              count = Math.min(count,
                               freq[str2.charAt(i) - 'a']/freq2[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

# 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)

    freq2= [0] * 26
    l2 = len(str2)

    # iterate and mark the frequencies
    # of all characters in str1
    for i in range(l1):
        freq[ord(str1[i]) - ord("a")] += 1
    for i in range(l2):
        freq2[ord(str2[i]) - ord("a")] += 1

    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')]/freq2[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#

// 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[26];

        int l1 = str1.Length;
        int[] freq2 = new int[26];

        int l2 = str2.Length;

        // iterate and mark the frequencies
        // of all characters in str1
        for (int i = 0; i < l1; i++)
        {
            freq[str1[i] - 'a'] += 1;
        }
        for (int i = 0; i < l2; i++)
        {
            freq2[str2[i] - 'a'] += 1;
        }

        int count = int.MaxValue;

        // find the minimum frequency of
        // every character in str1
        for (int i = 0; i < l2; i++)
        {
            if (freq2[str2[i] - 'a'] != 0)
                count = Math.Min(
                    count, freq[str2[i] - 'a']
                               / freq2[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

PHP

<?php
// PHP program to print the number of times
// str2 can be formed from str1 using the
// characters of str1 only once

// Function to find the number of str2
// that can be formed using characters of str1
function findNumberOfTimes($str1, $str2)
{
    $freq = array_fill(0, 26, NULL);

    $l1 = strlen($str1);


   $freq2 = array_fill(0, 26, NULL);

    $l2 = strlen($str2);

    // iterate and mark the frequencies
    // of all characters in str1
    for ($i = 0; $i < $l1; $i++)
        $freq[ord($str1[$i]) - ord('a')] += 1;

    for ($i = 0; $i < $l2; $i++)
        $freq2[ord($str2[$i]) - ord('a')] += 1;


    $count = PHP_INT_MAX;

    // find the minimum frequency of
    // every character in str1
    for ($i = 0; $i < $l2; $i++)
        $count = min($count, $freq[ord($str2[$i]) -
                                   ord('a')]/$freq2[ord($str2[$i]) -
                                   ord('a')]);

    return $count;
}

// Driver Code
$str1 = "foreeksgekseg";
$str2 = "geeks";

echo findNumberOfTimes($str1, $str2) . "\n";

// This code is contributed by ita_c
?>

Javascript

<script>
// Javascript program to print the number of times
// str2 can be formed from str1 using the
// characters of str1 only once

    // Function to find the number of str2
    // that can be formed using characters of str1
    function findNumberOfTimes(str1, str2)
    {
        let freq = new Array(26);
        let freq2 = new Array(26);
        for(let i = 0; i < 26; i++)
        {
            freq[i] = 0;
            freq2[i] = 0;
        }

        let l1 = str1.length;

        // iterate and mark the frequencies of
        // all characters in str1
        for (let i = 0; i < l1; i++)
        {
            freq[str1[i].charCodeAt(0) - 'a'.charCodeAt(0)] += 1;
        }
            let l2 = str2.length;
        for (let i = 0; i < l2; i++)
        {
            freq2[str2[i].charCodeAt(0) - 'a'.charCodeAt(0)] += 1;
        }

        let count = Number.MAX_VALUE;

        // find the minimum frequency of
        // every character in str1
        for (let i = 0; i < l2; i++)
        {
            if(freq2[str2[i].charCodeAt(0)-'a'.charCodeAt(0)]!=0)
              count = Math.min(count,
                               freq[str2[i].charCodeAt(0) - 'a'.charCodeAt(0)]/freq2[str2[i].charCodeAt(0)-'a'.charCodeAt(0)]);
        }

        return count;
    }

    let str1 = "foreeksgekseg";
    let str2 ="geeks";
    document.write(findNumberOfTimes(str1, str2));


    // This code is contributed by avanitrachhadiya2155
</script>

Output

2

Complexity Analysis:

• Time Complexity: O(l1+l2), where l1 and l2 are length of str1 and str2 respectively. 
• Auxiliary Space: O(1)

Case: Using STL Maps with both uppercase and lowercase characters in str1 and str2.

Approach: 

The operation is similar to the normal hash-array.

Here, the order does not matter. We just need to check if there are sufficient words in str1 to make str2. Traverse both strings str1 and str2 to store characters as key and their frequencies as value in maps freq1 and freq2. Divide the frequencies stored in map freq1 with frequencies that have a matching key in map freq2. 

This will give us the maximum number of cycles of str2 that can be formed. Finally return minimum frequency from map freq1 which will be the final answer.

Implementation:

C++14

#include <bits/stdc++.h>
using namespace std;

int countSubStr(char* str,char* substr)
{
    unordered_map<char,int>freq1;
    unordered_map<char,int>freq2;
    int i,mn=INT_MAX;
    int l1=strlen(str);
    int l2=strlen(substr);

    for(i=0;i<l1;i++)
    freq1[str[i]]++;

    for(i=0;i<l2;i++)
    freq2[substr[i]]++;

    for(auto x:freq2)
    mn=min(mn,freq1[x.first]/x.second);

    return mn;
}

int main()
{
    char str1[]= "arajjhupoot";
    char str2[]="rajput";
    cout<<countSubStr(str1,str2);
    return 0;
}

Java

// java implementation to find sum of
// first n even numbers
import java.io.*;
import java.util.*;

class GFG
{
  public static int countSubStr(String str,String substr)
  {
    HashMap<Character, Integer> freq1 = new HashMap<>();
    HashMap<Character, Integer> freq2 = new HashMap<>();

    int i, mn = Integer.MAX_VALUE;
    int l1 = str.length();
    int l2 = substr.length();

    for(int idx = 0; idx < str.length(); idx++){
      char c = str.charAt(idx);
      if (freq1.containsKey(c)) {

        freq1.put(c, freq1.get(c) + 1);
      }
      else {

        freq1.put(c, 1);
      }
    }

    for(int idx = 0; idx < substr.length(); idx++) {
      char c = substr.charAt(idx);
      if (freq2.containsKey(c)) {

        freq2.put(c, freq2.get(c) + 1);
      }
      else {

        freq2.put(c, 1);
      }
    }

    for (Map.Entry mapElement : freq2.entrySet()) {
      char first = (char)mapElement.getKey();
      int second = (int)mapElement.getValue();
      int second_f1 = freq1.get(first);
      mn = Math.min(mn,second_f1/second);
    }

    return mn;
  }

  // Driver program to test above
  public static void main(String[] args)
  {
    String str1= "arajjhupoot";
    String str2="rajput";
    System.out.println(countSubStr(str1,str2));
  }
}

// This code is contributed by aditya942003patil

Javascript

<script>
function countSubStr(str,substr)
{
    let freq1 = new Map();
    let freq2 = new Map();
    let i,mn=Number.MAX_VALUE;
    let l1 = str.length;
    let l2 = substr.length;

    for(i = 0; i < l1; i++){
        if(freq1.has(str[i]) == true){
            freq1.set(str[i], freq1.get(str[i]) + 1);
        }
        else{
            freq1.set(str[i], 1);
        }
    }

    for(i = 0; i < l2; i++){
        if(freq2.has(substr[i]) == true){
            freq2.set(substr[i], freq1.get(str[i]) + 1);
        }
        else{
            freq2.set(substr[i], 1);
        }
    }

    for(let [x, y] of freq2)
       mn = Math.min(mn,Math.floor(freq1.get(x)/y));

    return mn;
}

// driver code
let str1 = "arajjhupoot";
let str2 ="rajput";
document.write(countSubStr(str1,str2));

// This code is contributed by shinjanpatra

</script>

Python3

from math import floor
import sys
def countSubStr(str,substr):
    freq1 = {}
    freq2 = {}
    mn = sys.maxsize

    l1 = len(str)
    l2 = len(substr)

    for i in range(l1):
        if(str[i] in freq1):
            freq1[str[i]] = freq1[str[i]] + 1

        else:
            freq1[str[i]] = 1

    for i in range(l2):
        if substr[i] in freq2:
            freq2[substr[i]] =  freq1[str[i]] + 1

        else:
            freq2[substr[i]] = 1

    for x, y in freq2.items():
       mn = min(mn,floor(freq1[x]/y))

    return mn

# driver code
str1 = "arajjhupoot"
str2 ="rajput"
print(countSubStr(str1,str2))

# This code is contributed by shinjanpatra

C#

using System;
using System.Collections.Generic;
class GFG {
    static int countSubStr(string str, string substr)
    {
        Dictionary<char, int> freq1
            = new Dictionary<char, int>();
        Dictionary<char, int> freq2
            = new Dictionary<char, int>();
        int mn = Int32.MaxValue;
        int l1 = str.Length;
        int l2 = substr.Length;

        for (int i = 0; i < l1; i++) {
            if (freq1.ContainsKey(str[i])) {
                var val = freq1[str[i]];
                freq1.Remove(str[i]);
                freq1.Add(str[i], val + 1);
            }
            else
                freq1.Add(str[i], 1);
        }

        for (int i = 0; i < l2; i++) {
            if (freq2.ContainsKey(substr[i])) {
                var val = freq2[substr[i]];
                freq2.Remove(substr[i]);
                freq2.Add(substr[i], val + 1);
            }
            else
                freq2.Add(substr[i], 1);
        }
        foreach(KeyValuePair<char, int> x in freq2)
        {
            mn = Math.Min(mn, freq1[x.Key] / x.Value);
        }

        return mn;
    }
    static void Main()
    {
        string str1 = "arajjhupoot";
        string str2 = "rajput";
        Console.Write(countSubStr(str1, str2));
    }
}

Output

1

Complexity Analysis:

• Time Complexity: O(l1+l2), where l1 and l2 are length of str1 and str2 respectively
• Auxiliary Space: O(1) since the total number of distinct characters in both lowercase and uppercase English alphabets is 52 (Constant).