Number of ways to arrange N items under given constraints

We are given N items which are of total K different colors. Items of the same color are indistinguishable and colors can be numbered from 1 to K and count of items of each color is also given as k1, k2, and so on. Now we need to arrange these items one by one under the constraint that the last item of color i comes before the last item of color (i + 1) for all possible colors. Our goal is to find out how many ways this can be achieved.

Examples:

Input : N = 3        
        k1 = 1    k2 = 2
Output : 2
Explanation :
Possible ways to arrange are,
k1, k2, k2
k2, k1, k2 

Input : N = 4        
        k1 = 2    k2 = 2
Output : 3
Explanation :
Possible ways to arrange are,
k1, k2, k1, k2
k1, k1, k2, k2
k2, k1, k1, k2 

Method 1: 

We can solve this problem using dynamic programming. Let dp[i] stores the number of ways to arrange first i colored items. For one colored item answer will be one because there is only one way. Now Let’s assume all items are in a sequence. Now, to go from dp[i] to dp[i + 1], we need to put at least one item of color (i + 1) at the very end, but the other items of color (i + 1) can go anywhere in the sequence. The number of ways to arrange the item of color (i + 1) is combination of (k1 + k2 .. + ki + k(i + 1) – 1) over (k(i + 1) – 1) which can be represented as (k1 + k2 .. + ki + k(i + 1) – 1)C(k(i + 1) – 1). In this expression we subtracted one because we need to put one item at the very end. 

In the below code, first, we have calculated the combination values, you can read more about that from here. After that we looped over all the different colors and calculated the final value using the above relation.

C++

// C++ program to find number of ways to arrange
// items under given constraint
#include <bits/stdc++.h>
using namespace std;

// method returns number of ways with which items
// can be arranged
int waysToArrange(int N, int K, int k[])
{
    int C[N + 1][N + 1];
    int i, j;

    // Calculate value of Binomial Coefficient in
    // bottom up manner
    for (i = 0; i <= N; i++) {
        for (j = 0; j <= i; j++) {

            // Base Cases
            if (j == 0 || j == i)
                C[i][j] = 1;

            // Calculate value using previously
            // stored values
            else
                C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]);
        }
    }

    // declare dp array to store result up to ith
    // colored item
    int dp[K];

    // variable to keep track of count of items
    // considered till now
    int count = 0;

    dp[0] = 1;

    // loop over all different colors
    for (int i = 0; i < K; i++) {

        // populate next value using current value
        // and stated relation
        dp[i + 1] = (dp[i] * C[count + k[i] - 1][k[i] - 1]);
        count += k[i];
    }

    // return value stored at last index
    return dp[K];
}

// Driver code to test above methods
int main()
{
    int N = 4;
    int k[] = { 2, 2 };

    int K = sizeof(k) / sizeof(int);
    cout << waysToArrange(N, K, k) << endl;
    return 0;
}

Java

// Java program to find number of ways to arrange
// items under given constraint
import java.io.*;
class GFG {

    // method returns number of ways with which items
    // can be arranged
    static int waysToArrange(int N, int K, int[] k)
    {
        int[][] C = new int[N + 1][N + 1];
        int i, j;

        // Calculate value of Binomial Coefficient in
        // bottom up manner
        for (i = 0; i <= N; i++) {
            for (j = 0; j <= i; j++) {

                // Base Cases
                if (j == 0 || j == i) {
                    C[i][j] = 1;
                }

                // Calculate value using previously
                // stored values
                else {
                    C[i][j]
                        = (C[i - 1][j - 1] + C[i - 1][j]);
                }
            }
        }

        // declare dp array to store result up to ith
        // colored item
        int[] dp = new int[K + 1];

        // variable to keep track of count of items
        // considered till now
        int count = 0;

        dp[0] = 1;

        // loop over all different colors
        for (i = 0; i < K; i++) {

            // populate next value using current value
            // and stated relation
            dp[i + 1]
                = (dp[i] * C[count + k[i] - 1][k[i] - 1]);
            count += k[i];
        }

        // return value stored at last index
        return dp[K];
    }

    // Driver code
    public static void main(String[] args)
    {
        int N = 4;
        int[] k = new int[] { 2, 2 };
        int K = k.length;
        System.out.println(waysToArrange(N, K, k));
    }
}

// This code has been contributed by 29AjayKumar

Python3

# Python3 program to find number of ways
# to arrange items under given constraint
import numpy as np

# method returns number of ways with
# which items can be arranged
def waysToArrange(N, K, k) :

    C = np.zeros((N + 1, N + 1))

    # Calculate value of Binomial
    # Coefficient in bottom up manner
    for i in range(N + 1) :
        for j in range(i + 1) :

            # Base Cases
            if (j == 0 or j == i) :
                C[i][j] = 1

            # Calculate value using previously
            # stored values
            else :
                C[i][j] = (C[i - 1][j - 1] +
                           C[i - 1][j])

    # declare dp array to store result
    # up to ith colored item
    dp = np.zeros((K + 1))

    # variable to keep track of count
    # of items considered till now
    count = 0

    dp[0] = 1

    # loop over all different colors
    for i in range(K) :

        # populate next value using current
        # value and stated relation
        dp[i + 1] = (dp[i] * C[count + k[i] - 1][k[i] - 1])
        count += k[i]

    # return value stored at last index
    return dp[K]

# Driver code
if __name__ == "__main__" :

    N = 4
    k = [ 2, 2 ]

    K = len(k)
    print(int(waysToArrange(N, K, k)))

# This code is contributed by Ryuga

C#

// C# program to find number of ways to arrange
// items under given constraint
using System;

class GFG
{
    // method returns number of ways with which items
    // can be arranged
    static int waysToArrange(int N, int K, int[] k)
    {
        int[,] C = new int[N + 1, N + 1];
        int i, j;

        // Calculate value of Binomial Coefficient in
        // bottom up manner
        for (i = 0; i <= N; i++)
        {
            for (j = 0; j <= i; j++)
            {

                // Base Cases
                if (j == 0 || j == i)
                    C[i, j] = 1;

                // Calculate value using previously
                // stored values
                else
                    C[i, j] = (C[i - 1, j - 1] + C[i - 1,  j]);
            }
        }

        // declare dp array to store result up to ith
        // colored item
        int[] dp = new int[K + 1];

        // variable to keep track of count of items
        // considered till now
        int count = 0;

        dp[0] = 1;

        // loop over all different colors
        for (i = 0; i < K; i++) {

            // populate next value using current value
            // and stated relation
            dp[i + 1] = (dp[i] * C[count + k[i] - 1, k[i] - 1]);
            count += k[i];
        }

        // return value stored at last index
        return dp[K];
    }

    // Driver code
    static void Main()
    {
        int N = 4;
        int[] k = new int[]{ 2, 2 };
        int K = k.Length;
        Console.Write(waysToArrange(N, K, k));
    }
}

// This code is contributed by DrRoot_

PHP

<?php
// PHP program to find number of
// ways to arrange items under
// given constraint

// method returns number of ways
// with which items can be arranged
function waysToArrange($N, $K, $k)
{
    $C[$N + 1][$N + 1] = array(array());

    // Calculate value of Binomial
    // Coefficient in bottom up manner
    for ($i = 0; $i <= $N; $i++)
    {
        for ($j = 0; $j <= $i; $j++)
        {

            // Base Cases
            if ($j == 0 || $j == $i)
                $C[$i][$j] = 1;

            // Calculate value using
            // previously stored values
            else
                $C[$i][$j] = ($C[$i - 1][$j - 1] +
                              $C[$i - 1][$j]);
        }
    }

    // declare dp array to store
    // result up to ith colored item
    $dp[$K] = array();

    // variable to keep track of count
    // of items considered till now
    $count = 0;

    $dp[0] = 1;

    // loop over all different colors
    for ($i = 0; $i < $K; $i++)
    {

        // populate next value using
        // current value and stated relation
        $dp[$i + 1] = ($dp[$i] * $C[$count +
                        $k[$i] - 1][$k[$i] - 1]);
        $count += $k[$i];
    }

    // return value stored at
    // last index
    return $dp[$K];
}

// Driver code
$N = 4;
$k = array( 2, 2 );

$K = sizeof($k);
echo waysToArrange($N, $K, $k),"\n";

// This code is contributed by jit_t
?>

Javascript

<script>
// Javascript program to find number of ways to arrange
// items under given constraint.

    // method returns number of ways with which items
    // can be arranged
    function waysToArrange(N, K, k)
    {
        let C = new Array(N + 1);
        // Loop to create 2D array using 1D array
        for (let i = 0; i < C.length; i++) {
            C[i] = new Array(2);
        }

        let i, j;

        // Calculate value of Binomial Coefficient in
        // bottom up manner
        for (i = 0; i <= N; i++)
        {
            for (j = 0; j <= i; j++)
            {

                // Base Cases
                if (j == 0 || j == i)
                {
                    C[i][j] = 1;
                }

                // Calculate value using previously
                // stored values
                else
                {
                    C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]);
                }
            }
        }

        // declare dp array to store result up to ith
        // colored item
        let dp = Array.from({length: K+1}, (_, i) => 0);

        // variable to keep track of count of items
        // considered till now
        let count = 0;

        dp[0] = 1;

        // loop over all different colors
        for (i = 0; i < K; i++)
        {

            // populate next value using current value
            // and stated relation
            dp[i + 1] = (dp[i] * C[count + k[i] - 1][k[i] - 1]);
            count += k[i];
        }

        // return value stored at last index
        return dp[K];
    }

// driver function

        let N = 4;
        let k = [2, 2];
        let K = k.length;
        document.write(waysToArrange(N, K, k));

</script>

Output

3

Time Complexity: O(N²)
Auxiliary space: O(N*N)

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