Count possible permutations of given array satisfying the given conditions
Given an array, arr[] consisting of N distinct elements, the task is to count possible permutations of the given array that can be generated which satisfies the following properties:
- The two halves must be sorted.
- arr[i] must be less than arr[N / 2 + i]
Note: N is always even and indexing starts from 0.
Examples:
Input: arr[] = {10, 20, 30, 40}
Output: 2
Explanation:
Possible permutations of the given array that satisfy the given conditions are:{{10, 20, 30, 40}, {10, 30, 20, 40}}.
Therefore, the required output is 2.Input: arr[] = {1, 2}
Output: 1
Approach: Follow the steps below to solve the problem:
- Initialize a variable, say cntPerm to store the count of permutations of the given array that satisfy the given condition.
- Find the value of the binomial coefficient of 2NCN using the following formula:
= [{N × (N – 1) × …………. × (N – R + 1)} / {(R × (R – 1) × ….. × 1)}]
- Finally, calculate catalan number = 2NCN / (N + 1) and print it as the required answer.
Below is the implementation of the above approach:
C++
// C++ Program to implement // the above approach #include <bits/stdc++.h> using namespace std; // Function to get the value // of binomial coefficient int binCoff( int N, int R) { // Stores the value of // binomial coefficient int res = 1; if (R > (N - R)) { // Since C(N, R) // = C(N, N - R) R = (N - R); } // Calculate the value // of C(N, R) for ( int i = 0; i < R; i++) { res *= (N - i); res /= (i + 1); } return res; } // Function to get the count of // permutations of the array // that satisfy the condition int cntPermutation( int N) { // Stores count of permutations // of the array that satisfy // the given condition int cntPerm; // Stores the value of C(2N, N) int C_2N_N = binCoff(2 * N, N); // Stores the value of // catalan number cntPerm = C_2N_N / (N + 1); // Return answer return cntPerm; } // Driver Code int main() { int arr[] = { 1, 2, 3, 4 }; int N = sizeof (arr) / sizeof (arr[0]); cout << cntPermutation(N / 2); return 0; } |
Java
// Java Program to implement // the above approach import java.io.*; class GFG{ // Function to get the value // of binomial coefficient static int binCoff( int N, int R) { // Stores the value of // binomial coefficient int res = 1 ; if (R > (N - R)) { // Since C(N, R) // = C(N, N - R) R = (N - R); } // Calculate the value // of C(N, R) for ( int i = 0 ; i < R; i++) { res *= (N - i); res /= (i + 1 ); } return res; } // Function to get the count of // permutations of the array // that satisfy the condition static int cntPermutation( int N) { // Stores count of permutations // of the array that satisfy // the given condition int cntPerm; // Stores the value of C(2N, N) int C_2N_N = binCoff( 2 * N, N); // Stores the value of // catalan number cntPerm = C_2N_N / (N + 1 ); // Return answer return cntPerm; } // Driver Code public static void main (String[] args) { int arr[] = { 1 , 2 , 3 , 4 }; int N = arr.length; System.out.println(cntPermutation(N / 2 )); } } // This code is contributed by sanjoy_62 |
Python3
# Python3 program to implement # the above approach # Function to get the value # of binomial coefficient def binCoff(N, R): # Stores the value of # binomial coefficient res = 1 if (R > (N - R)): # Since C(N, R) # = C(N, N - R) R = (N - R) # Calculate the value # of C(N, R) for i in range (R): res * = (N - i) res / / = (i + 1 ) return res # Function to get the count of # permutations of the array # that satisfy the condition def cntPermutation(N): # Stores count of permutations # of the array that satisfy # the given condition # Stores the value of C(2N, N) C_2N_N = binCoff( 2 * N, N) # Stores the value of # catalan number cntPerm = C_2N_N / / (N + 1 ) # Return answer return cntPerm # Driver Code if __name__ = = '__main__' : arr = [ 1 , 2 , 3 , 4 ] N = len (arr) print (cntPermutation(N / / 2 )) # This code is contributed by mohit kumar 29 |
C#
// C# Program to implement // the above approach using System; class GFG{ // Function to get the value // of binomial coefficient static int binCoff( int N, int R) { // Stores the value of // binomial coefficient int res = 1; if (R > (N - R)) { // Since C(N, R) // = C(N, N - R) R = (N - R); } // Calculate the value // of C(N, R) for ( int i = 0; i < R; i++) { res *= (N - i); res /= (i + 1); } return res; } // Function to get the count of // permutations of the array // that satisfy the condition static int cntPermutation( int N) { // Stores count of permutations // of the array that satisfy // the given condition int cntPerm; // Stores the value of C(2N, N) int C_2N_N = binCoff(2 * N, N); // Stores the value of // catalan number cntPerm = C_2N_N / (N + 1); // Return answer return cntPerm; } // Driver Code public static void Main(String[] args) { int []arr = {1, 2, 3, 4}; int N = arr.Length; Console.WriteLine(cntPermutation(N / 2)); } } // This code is contributed by shikhasingrajput |
Javascript
<script> // javascript Program to implement // the above approach // Function to get the value // of binomial coefficient function binCoff(N , R) { // Stores the value of // binomial coefficient var res = 1; if (R > (N - R)) { // Since C(N, R) // = C(N, N - R) R = (N - R); } // Calculate the value // of C(N, R) for (i = 0; i < R; i++) { res *= (N - i); res /= (i + 1); } return res; } // Function to get the count of // permutations of the array // that satisfy the condition function cntPermutation(N) { // Stores count of permutations // of the array that satisfy // the given condition var cntPerm; // Stores the value of C(2N, N) var C_2N_N = binCoff(2 * N, N); // Stores the value of // catalan number cntPerm = C_2N_N / (N + 1); // Return answer return cntPerm; } // Driver Code var arr = [ 1, 2, 3, 4 ]; var N = arr.length; document.write(cntPermutation(N / 2)); // This code contributed by umadevi9616 </script> |
Output
2
Time Complexity: O(N)
Auxiliary Space: O(1)
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