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📜  给定字符的所有出现同时出现的字符串的排列数

📅  最后修改于: 2021-10-27 08:24:04             🧑  作者: Mango

给定一个字符串’s’ 和一个字符’c’,任务是找到字符串的排列数,其中所有出现的字符’c’ 将在一起(一个接一个)。
例子 :

天真的方法:

  • 生成给定字符串 的所有排列。
  • 对于每个排列,检查字符的所有出现是否一起出现。
  • 上一步的排列数就是答案。

有效的方法:让字符串的长度为’l’,字符串字符的频率为’n’。

  • 存储字符串中每个字符的频率。
  • 如果字符中不存在该字符串,则输出将为“0”。
  • 将所有出现的字符视为一个组合的单个字符。
  • 因此,’l’ 将变为 ‘l – occ + 1’,其中 ‘occ’ 是给定字符的总出现次数,而 ‘l’ 是字符串的新长度。
  • 返回((l 的阶乘)/(除给定字符外每个字符出现次数的阶乘乘积))

下面是上述方法的实现:

C++
// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return factorial
// of the number passed as argument
long long int fact(int n)
{
    long long result = 1;
    for (int i = 1; i <= n; i++)
        result *= i;
    return result;
}
 
// Function to get the total permutations
// which satisfy the given condition
int getResult(string str, char ch)
{
    // Create has to store count
    // of each character
    int has[26] = { 0 };
 
    // Store character occurrences
    for (int i = 0; i < str.length(); i++)
        has[str[i] - 'A']++;
 
    // Count number of times
    // Particular character comes
    int particular = has[ch - 'A'];
 
    // If particular character isn't
    // present in the string then return 0
    if (particular == 0)
        return 0;
 
    // Remove count of particular character
    has[ch - 'A'] = 0;
 
    // Total length
    // of the string
    int total = str.length();
 
    // Assume all occurrences of
    // particular character as a
    // single character.
    total = total - particular + 1;
 
    // Compute factorial of the length
    long long int result = fact(total);
 
    // Divide by the factorials of
    // the no. of occurrences of all
    // the characters.
    for (int i = 0; i < 26; i++) {
        if (has[i] > 1) {
            result = result / fact(has[i]);
        }
    }
 
    // return the result
    return result;
}
 
// Driver Code
int main()
{
    string str = "MISSISSIPPI";
 
    // Assuming the string and the character
    // are all in uppercase
    cout << getResult(str, 'S') << endl;
 
  return 0;
}


Java
// Java implementation of above approach
import java.util.*;
class solution
{
 
// Function to return factorial
// of the number passed as argument
 static int fact(int n)
{
     int result = 1;
    for (int i = 1; i <= n; i++)
        result *= i;
    return result;
}
 
// Function to get the total permutations
// which satisfy the given condition
static int getResult(String str, char ch)
{
    // Create has to store count
    // of each character
    int has[] = new int[26];
     
    for(int i=0;i<26;i++)
    has[i]=0;
 
    // Store character occurrences
    for (int i = 0; i < str.length(); i++)
        has[str.charAt(i) - 'A']++;
 
    // Count number of times
    // Particular character comes
    int particular = has[ch - 'A'];
 
    // If particular character isn't
    // present in the string then return 0
    if (particular == 0)
        return 0;
 
    // Remove count of particular character
    has[ch - 'A'] = 0;
 
    // Total length
    // of the string
    int total = str.length();
 
    // Assume all occurrences of
    // particular character as a
    // single character.
    total = total - particular + 1;
 
    // Compute factorial of the length
     int result = fact(total);
 
    // Divide by the factorials of
    // the no. of occurrences of all
    // the characters.
    for (int i = 0; i < 26; i++) {
        if (has[i] > 1) {
            result = result / fact(has[i]);
        }
    }
 
    // return the result
    return result;
}
 
// Driver Code
public static void main(String args[])
{
    String str = "MISSISSIPPI";
 
    // Assuming the string and the character
    // are all in uppercase
    System.out.println( getResult(str, 'S') );
 
}
}
//contributed by Arnab Kundu


Python 3
# Python3 implementation of
# the approach
 
# Function to return factorial
# of the number passed as argument
def fact(n) :
 
    result = 1
 
    for i in range(1, n + 1) :
        result *= i
 
    return result
 
# Function to get the total permutations
# which satisfy the given condition
def getResult(string, ch):
 
    # Create has to store count
    # of each character
    has = [0] * 26
 
    # Store character occurrences
    for i in range(len(string)) :
        has[ord(string[i]) - ord('A')] += 1
 
    # Count number of times
    # Particular character comes
    particular = has[ord(ch) - ord('A')]
 
    # If particular character isn't
    # present in the string then return 0
    if particular == 0 :
        return 0
 
    # Remove count of particular character
    has[ord(ch) - ord('A')] = 0
 
    # Total length
    # of the string
    total = len(string)
 
    # Assume all occurrences of
    # particular character as a
    # single character.
    total = total - particular + 1
 
    # Compute factorial of the length
    result = fact(total)
 
    # Divide by the factorials of
    # the no. of occurrences of all
    # the characters.
    for i in range(26) :
 
        if has[i] > 1 :
            result /= fact(has[i])
 
    # return the result
    return result
 
 
# Driver code
if __name__ == "__main__" :
 
    string = "MISSISSIPPI"
 
    # Assuming the string and the character
    # are all in uppercase
    print(getResult(string,'S'))
 
# This code is contributed
# by ANKITRAI1


C#
// C# implementation of above approach
using System;
 
class GFG
{
 
// Function to return factorial
// of the number passed as argument
static int fact(int n)
{
    int result = 1;
    for (int i = 1; i <= n; i++)
        result *= i;
    return result;
}
 
// Function to get the total permutations
// which satisfy the given condition
static int getResult(string str, char ch)
{
    // Create has to store count
    // of each character
    int []has = new int[26];
     
    for(int i = 0; i < 26; i++)
    has[i] = 0;
 
    // Store character occurrences
    for (int i = 0; i < str.Length; i++)
        has[str[i] - 'A']++;
 
    // Count number of times
    // Particular character comes
    int particular = has[ch - 'A'];
 
    // If particular character
    // isn't present in the string
    // then return 0
    if (particular == 0)
        return 0;
 
    // Remove count of particular character
    has[ch - 'A'] = 0;
 
    // Total length of the string
    int total = str.Length;
 
    // Assume all occurrences of
    // particular character as a
    // single character.
    total = total - particular + 1;
 
    // Compute factorial of the length
    int result = fact(total);
 
    // Divide by the factorials of
    // the no. of occurrences of all
    // the characters.
    for (int i = 0; i < 26; i++)
    {
        if (has[i] > 1)
        {
            result = result / fact(has[i]);
        }
    }
 
    // return the result
    return result;
}
 
// Driver Code
public static void Main()
{
    string str = "MISSISSIPPI";
 
    // Assuming the string and the
    // character are all in uppercase
    Console.WriteLine(getResult(str, 'S') );
}
}
 
// This code is contributed by anuj_67


PHP
 1)
        {
            $result = $result / fact($has[$i]);
        }
    }
 
    // return the result
    return $result;
}
 
// Driver Code
$str = "MISSISSIPPI";
 
// Assuming the string and the character
// are all in uppercase
echo getResult($str, 'S')."\n" ;
 
// This code is contributed by ita_c
?>


Javascript


输出:
840

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