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 



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++ 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;
}

Output:

3

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up

Article Tags :
Practice Tags :



Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.