Number of ways to merge two arrays such retaining order

Given two array of size n and m. The task is to find the number of ways we can merge the given arrays into one array such that order of elements of each array doesn’t change.

Examples:

Input : n = 2, m = 2
Output : 6
Let first array of size n = 2 be [1, 2] and second array of size m = 2 be [3, 4].
So, possible merged array of n + m elements can be:
[1, 2, 3, 4]
[1, 3, 2, 4]
[3, 4, 1, 2]
[3, 1, 4, 2]
[1, 3, 4, 2]
[3, 1, 2, 4]
 
Input : n = 4, m = 6
Output : 210

The idea is to use the concept of combinatorics. Suppose we have two array A{a1, a2, …., am} and B{b1, b2, …., bn} having m and n elements respectively and now we have to merge them without losing their order. 

After merging we know that the total number of element will be (m + n) element after merging. So, now we just need the ways to choose m places out of (m + n) where you will place element of array A in its actual order, which is ^{m + n}C_n. 

After placing m element of array A, n spaces will be left, which can be filled by the n elements of B array in its actual order. 

So, total number of ways to merge two array such that their order in merged array is same is ^{m + n}C_n

Below is the implementation of this approach:  

C++

// CPP Program to find number of ways
// to merge two array such that their
// order in merged array is same
#include <bits/stdc++.h>
using namespace std;

// function to find the binomial coefficient
int binomialCoeff(int n, int k)
{
    int C[k + 1];
    memset(C, 0, sizeof(C));

    C[0] = 1; // nC0 is 1

    for (int i = 1; i <= n; i++) {

        // Compute next row of pascal triangle
        // using the previous row
        for (int j = min(i, k); j > 0; j--)
            C[j] = C[j] + C[j - 1];
    }
    return C[k];
}

// function to find number of ways
// to merge two array such that their
// order in merged array is same
int numOfWays(int n, int m)
{
    return binomialCoeff(m + n, m);
}

// Driven Program
int main()
{
    int n = 2, m = 2;
    cout << numOfWays(n, m) << endl;
    return 0;
}

Java

// Java Program to find number of ways
// to merge two array such that their
// order in merged array is same

import java.io.*;

class GFG {

    // function to find the binomial
    // coefficient
    static int binomialCoeff(int n, int k)
    {
        int C[] = new int[k + 1];
        // memset(C, 0, sizeof(C));

        C[0] = 1; // nC0 is 1

        for (int i = 1; i <= n; i++) {

            // Compute next row of pascal
            // triangle using the previous
            // row
            for (int j = Math.min(i, k);
                               j > 0; j--)
                C[j] = C[j] + C[j - 1];
        }

        return C[k];
    }

    // function to find number of ways
    // to merge two array such that their
    // order in merged array is same
    static int numOfWays(int n, int m)
    {
        return binomialCoeff(m + n, m);
    }

    // Driven Program
    public static void main (String[] args)
    {
        int n = 2, m = 2;
        System.out.println(numOfWays(n, m));
    }
}

// This code is contributed by anuj_67.

Python3

# Python 3 Program to find number of ways
# to merge two array such that their
# order in merged array is same

# function to find the binomial coefficient
def binomialCoeff(n, k):
    C = [0 for i in range(k + 1)]

    C[0] = 1

    for i in range(1, n + 1, 1):

        # Compute next row of pascal
        # triangle using the previous row
        j = min(i, k)
        while(j > 0):
            C[j] = C[j] + C[j - 1]
            j -= 1

    return C[k]

# function to find number of ways
# to merge two array such that their
# order in merged array is same
def numOfWays(n, m):
    return binomialCoeff(m + n, m)

# Driver Code
if __name__ == '__main__':
    n = 2
    m = 2
    print(numOfWays(n, m))

# This code is contributed by
# Sahil_shelangia

C#

// C# Program to find number of ways
// to merge two array such that their
// order in merged array is same

using System;

class GFG {

    // function to find the binomial
    // coefficient
    static int binomialCoeff(int n, int k)
    {
        int []C = new int[k + 1];
        // memset(C, 0, sizeof(C));

        C[0] = 1; // nC0 is 1

        for (int i = 1; i <= n; i++) {

            // Compute next row of pascal
            // triangle using the previous
            // row
            for (int j = Math.Min(i, k);
                            j > 0; j--)
                C[j] = C[j] + C[j - 1];
        }

        return C[k];
    }

    // function to find number of ways
    // to merge two array such that their
    // order in merged array is same
    static int numOfWays(int n, int m)
    {
        return binomialCoeff(m + n, m);
    }

    // Driven Program
    public static void Main ()
    {
        int n = 2, m = 2;
        Console.WriteLine(numOfWays(n, m));
    }
}

// This code is contributed by anuj_67.

PHP

<?php
// PHP Program to find number of ways
// to merge two array such that their
// order in merged array is same


// function to find the binomial coefficient
function binomialCoeff($n, $k)
{
     $C = array($k + 1);
      for($i=0; $i < count($C); $i++)
        $C[$i] = 0;

    $C[0] = 1; // nC0 is 1

    for ( $i = 1; $i <= $n; $i++) {

        // Compute next row of pascal triangle
        // using the previous row
        for ( $j = min($i, $k); $j > 0; $j--)
            $C[$j] = $C[$j] + $C[$j - 1 ];
    }
    return $C[$k];
}

// function to find number of ways
// to merge two array such that their
// order in merged array is same
function numOfWays( $n, $m)
{
    return binomialCoeff($m + $n, $m);
}

    $n = 2; $m = 2;
    echo numOfWays($n, $m);
   //This code is contributed by Rajput-Ji.
?>

Javascript

<script>
    // Javascript Program to find number of ways
    // to merge two array such that their
    // order in merged array is same

    // function to find the binomial
    // coefficient
    function binomialCoeff(n, k)
    {
        let C = new Array(k + 1);
        C.fill(0);
        // memset(C, 0, sizeof(C));

        C[0] = 1; // nC0 is 1

        for (let i = 1; i <= n; i++) {

            // Compute next row of pascal
            // triangle using the previous
            // row
            for (let j = Math.min(i, k); j > 0; j--)
                C[j] = C[j] + C[j - 1];
        }

        return C[k];
    }

    // function to find number of ways
    // to merge two array such that their
    // order in merged array is same
    function numOfWays(n, m)
    {
        return binomialCoeff(m + n, m);
    }

    let n = 2, m = 2;
      document.write(numOfWays(n, m));

   // This code is contributed by divyeshrabadiya07.
</script>

Output

6

Time Complexity: O((m+n)*m)
Auxiliary Space: O(m)

We can solve above problem in linear time using linear time implementation of binomial coefficient.