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 arrangedint 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 methodsint 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 constraintimport 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 constraintimport numpy as np# method returns number of ways with # which items can be arrangeddef 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 codeif __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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