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

Last Updated : 07 Apr, 2021

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

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.

Possible ways to place str1[] and str2[] alternatively

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:

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[]

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 <bits/stdc++.h>
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

<script>
 
// 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 
 
</script>

Output:

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

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

C++

Java

Python3

C#

Javascript

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