计算交替放置两个给定字符串的所有字符的方法

给定两个字符串，长度为N str1和明显不同的字符长度M的STR2，任务是统计的方式发出str1和str2的交替的所有字符数。

注意： | N – M | ≤1

例子：

Input: str1 =“ae”, str2 = “bd”
Output: 8
Explanations:
Possible strings after rearrangements are : {“abed”, “ebad”, “adeb”, “edab”, “bade”, “beda”, “dabe”, “deba”}. Therefore, the required output is 8.

Input: str1= “aegh”, str2=”rsw”
Output: 144

方法：可以根据以下观察结果解决问题：

如果N！= M：仅考虑N> M的情况，因为类似地，它将适用于N 的情况。

交替放置str1 []和str2 []的可能方法

Total number of ways to rearrange all the characters of str1 = N!
Total number of ways to rearrange all the characters of str2 = M!.
Therefore, the total number of ways to place all the characters of str1 and str2 alternatively are = N! * M!

如果N == M：

首先STR1 []，然后的字符放置STR2的字符[]

首先STR2 []，然后的字符放置STR1的字符[]

Total number of ways to rearrange all the characters of str1 = N!
Total number of ways to rearrange all the characters of str2 = M!
Now, 
There are two cases possible here:

• First place the character of str1 and then place the character of str2.
• First place the character of str2 and then place the character of str1.

Therefore, the total number of ways = (2 * N! * M!). 
 

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to get the
// factorial of N
int fact(int n)
{
    int res = 1;
    for (int i = 1; i <= n; i++) {
        res = res * i;
    }
    return res;
}
 
// Function to get the total
// number of  distinct ways
int distinctWays(string str1, string str2)
{
    // Length of str1
    int n = str1.length();
 
    // Length of str2
    int m = str2.length();
 
    // If both strings have equal length
    if (n == m) {
        return 2 * fact(n) * fact(m);
    }
 
    // If both strings do not have
    // equal length
    return fact(n) * fact(m);
}
 
// Driver code
int main()
{
    string str1 = "aegh";
    string str2 = "rsw";
 
    cout << distinctWays(str1, str2);
}

Java

// Java program to implement
// the above approach
import java.io.*;
 
class GFG{
 
// Function to get the
// factorial of N
static int fact(int n)
{
    int res = 1;
    for(int i = 1; i <= n; i++)
    {
        res = res * i;
    }
    return res;
}
 
// Function to get the total
// number of distinct ways
static int distinctWays(String str1,
                        String str2)
{
     
    // Length of str1
    int n = str1.length();
 
    // Length of str2
    int m = str2.length();
 
    // If both strings have equal length
    if (n == m)
    {
        return 2 * fact(n) * fact(m);
    }
 
    // If both strings do not have
    // equal length
    return fact(n) * fact(m);
}
 
// Driver code
public static void main (String[] args)
{
    String str1 = "aegh";
    String str2 = "rsw";
 
    System.out.print(distinctWays(str1, str2));
}
}
 
// This code is contributed by code_hunt

Python3

# Python3 program to implement
# the above approach
 
# Function to get the
# factorial of N
def fact(n):
     
    res = 1
    for i in range(1, n + 1):
        res = res * i
     
    return res
 
# Function to get the total
# number of distinct ways
def distinctWays(str1, str2):
     
    # Length of str1
    n = len(str1)
 
    # Length of str2
    m = len(str2)
 
    # If both strings have equal length
    if (n == m):
        return 2 * fact(n) * fact(m)
     
    # If both strings do not have
    # equal length
    return fact(n) * fact(m)
 
# Driver code
str1 = "aegh"
str2 = "rsw"
 
print(distinctWays(str1, str2))
 
# This code is contributed by code_hunt

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to get the
// factorial of N
static int fact(int n)
{
    int res = 1;
    for(int i = 1; i <= n; i++)
    {
        res = res * i;
    }
    return res;
}
 
// Function to get the total
// number of distinct ways
static int distinctWays(string str1,
                        string str2)
{
     
    // Length of str1
    int n = str1.Length;
 
    // Length of str2
    int m = str2.Length;
 
    // If both strings have equal length
    if (n == m)
    {
        return 2 * fact(n) * fact(m);
    }
 
    // If both strings do not have
    // equal length
    return fact(n) * fact(m);
}
 
// Driver code
public static void Main ()
{
    string str1 = "aegh";
    string str2 = "rsw";
 
    Console.Write(distinctWays(str1, str2));
}
}
 
// This code is contributed by code_hunt

Javascript

 
// JavaScript program to implement
// the above approach
 
    // Function to get the
    // factorial of N
    function fact(n) {
        var res = 1;
        for (i = 1; i <= n; i++) {
            res = res * i;
        }
        return res;
    }
 
    // Function to get the total
    // number of distinct ways
    function distinctWays( str1,  str2) {
 
        // Length of str1
        var n = str1.length;
 
        // Length of str2
        var m = str2.length;
 
        // If both strings have equal length
        if (n == m) {
            return 2 * fact(n) * fact(m);
        }
 
        // If both strings do not have
        // equal length
        return fact(n) * fact(m);
    }
 
    // Driver code
     
        var str1 = "aegh";
        var str2 = "rsw";
 
        document.write(distinctWays(str1, str2));
 
// This code is contributed by todaysgaurav
 

输出：

144

时间复杂度： O(N + M)
辅助空间： O(1)