# Count ways to place all the characters of two given strings alternately

Given two strings, str1 of length N and str2 of length M of distinct characters, the task is to count the number of ways to place all the characters of str1 and str2 alternatively.

Note: |N – M| ≤ 1

Examples:

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The problem can be solved based on the following observations:

If N != M: Consider only the case of N > M because similarly, it will work for the case N < M.

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!

If N == M:

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!).

Below is the implementation of the above approach:

## 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 `

Output:

```144
```

Time Complexity: O(N + M)
Auxiliary Space: O(1)

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