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 a 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

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

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

Output:

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My Personal Notes

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

C++

Java

Python3

C#

PHP

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