可以使用每个字符最多一次从另一个字符串形成的字符串计数

给定两个字符串str1 和 str2，任务是打印使用str1 的字符可以形成str2的次数。然而，在STR1的任何索引的字符只能使用一次在STR2的形成中使用。

例子：

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

方法：既然问题对使用str1中的字符只有一次形成STR2限制。如果一个字符已用于形成一个 str2，则它不能用于形成另一个 str2。 str2 的每个字符都必须存在于 str1 中，至少要形成一个 str1。如果STR2的所有字符都在STR1已经存在，那么它在STR1最小发生的字符将可使用一次STR1的字符形成STR2的数量。以下是步骤：

• 创建一个哈希数组，用于存储 str 中每个字符的出现次数。
• 迭代 str2 的所有字符，并找到 str1 中每个字符出现次数最少的一次。
• 返回将作为答案的最小出现次数。

下面是上述方法的实现：

C++

/// C++ program to print the number of times
// str2 can be formed from str1 using the
// characters of str1 only once
#include 
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

Javascript

// 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

输出

2

时间复杂度： O(max(l1,l2))，其中 l1 和 l2 分别是 str1 和 str2 的长度。

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