Count permutation such that sequence is non decreasing

Given an array arr[] of integers, the task is to find the count of permutation of the array such that the permutation is in increasing order i.e. arr[0] ≤ arr[1] ≤ arr[2] ≤ … ≤ arr[n – 1].

Examples:

Input: arr[] = {1, 2, 1}
Output: 2
1, 1, 2 and 1, 1, 2 are the only valid permutations.



Input: arr[] = {5, 4, 4, 5}
Output: 4

Approach: The sequence should be non-descending i.e. arr[0] ≤ arr[1] ≤ arr[2] ≤ … ≤ arr[n – 1].
First, sort the array and then focus on the block where all elements are equal as these elements can be rearranged in P! ways where P is the size of that block.
Permuting that block will not violate the given condition. Now, find all the blocks where all elements are equal and multiply the answer of that individual block to the final answer to get the total count of possible permutations.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
#define N 20
  
// To store the factorials
int fact[N];
  
// Function to update fact[] array
// such that fact[i] = i!
void pre()
{
  
    // 0! = 1
    fact[0] = 1;
    for (int i = 1; i < N; i++) {
  
        // i! = i * (i - 1)!
        fact[i] = i * fact[i - 1];
    }
}
  
// Function to return the count
// of possible permutations
int CountPermutation(int a[], int n)
{
  
    // To store the result
    int ways = 1;
  
    // Sort the array
    sort(a, a + n);
  
    // Initial size of the block
    int size = 1;
    for (int i = 1; i < n; i++) {
  
        // Increase the size of block
        if (a[i] == a[i - 1]) {
            size++;
        }
        else {
  
            // Update the result for
            // the previous block
            ways *= fact[size];
  
            // Reset the size to 1
            size = 1;
        }
    }
  
    // Update the result for
    // the last block
    ways *= fact[size];
  
    return ways;
}
  
// Driver code
int main()
{
  
    int a[] = { 1, 2, 4, 4, 2, 4 };
    int n = sizeof(a) / sizeof(a[0]);
  
    // Pre-calculating factorials
    pre();
  
    cout << CountPermutation(a, n);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

//Java implementation of the approach
import java.util.Arrays; 
import java.io.*;
  
class GFG 
{
static int N = 20;
  
// To store the factorials
static int []fact=new int[N];
  
// Function to update fact[] array
// such that fact[i] = i!
static void pre()
{
  
    // 0! = 1
    fact[0] = 1;
    for (int i = 1; i < N; i++)
    {
  
        // i! = i * (i - 1)!
        fact[i] = i * fact[i - 1];
    }
}
  
// Function to return the count
// of possible permutations
static int CountPermutation(int a[], int n)
{
  
    // To store the result
    int ways = 1;
  
    // Sort the array
    Arrays.sort(a);
  
    // Initial size of the block
    int size = 1;
    for (int i = 1; i < n; i++) 
    {
  
        // Increase the size of block
        if (a[i] == a[i - 1]) 
        {
            size++;
        }
        else
        {
  
            // Update the result for
            // the previous block
            ways *= fact[size];
  
            // Reset the size to 1
            size = 1;
        }
    }
  
    // Update the result for
    // the last block
    ways *= fact[size];
  
    return ways;
}
  
// Driver Code
public static void main (String[] args) 
{
    int a[] = { 1, 2, 4, 4, 2, 4 };
    int n = a.length;
      
    // Pre-calculating factorials
    pre();
      
    System.out.println (CountPermutation(a, n));
}
}
  
// This code is contributed by Sachin

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach 
N = 20
  
# To store the factorials 
fact = [0] * N; 
  
# Function to update fact[] array 
# such that fact[i] = i! 
def pre() :
  
    # 0! = 1 
    fact[0] = 1
    for i in range(1, N):
  
        # i! = i * (i - 1)! 
        fact[i] = i * fact[i - 1]; 
  
# Function to return the count 
# of possible permutations 
def CountPermutation(a, n): 
  
    # To store the result 
    ways = 1
  
    # Sort the array 
    a.sort();
  
    # Initial size of the block 
    size = 1
    for i in range(1, n):
  
        # Increase the size of block 
        if (a[i] == a[i - 1]):
            size += 1
          
        else :
  
            # Update the result for 
            # the previous block 
            ways *= fact[size]; 
  
            # Reset the size to 1 
            size = 1
  
    # Update the result for 
    # the last block 
    ways *= fact[size]; 
  
    return ways; 
  
# Driver code 
if __name__ == "__main__"
  
    a = [ 1, 2, 4, 4, 2, 4 ]; 
    n = len(a); 
  
    # Pre-calculating factorials 
    pre(); 
  
    print(CountPermutation(a, n)); 
      
# This code is contributed by AnkitRai01

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
  
class GFG
{
      
static int N = 20;
  
// To store the factorials
static int []fact = new int[N];
  
// Function to update fact[] array
// such that fact[i] = i!
static void pre()
{
  
    // 0! = 1
    fact[0] = 1;
    for (int i = 1; i < N; i++)
    {
  
        // i! = i * (i - 1)!
        fact[i] = i * fact[i - 1];
    }
}
  
// Function to return the count
// of possible permutations
static int CountPermutation(int []a, int n)
{
  
    // To store the result
    int ways = 1;
  
    // Sort the array
    Array.Sort(a);
  
    // Initial size of the block
    int size = 1;
    for (int i = 1; i < n; i++) 
    {
  
        // Increase the size of block
        if (a[i] == a[i - 1]) 
        {
            size++;
        }
        else
        {
  
            // Update the result for
            // the previous block
            ways *= fact[size];
  
            // Reset the size to 1
            size = 1;
        }
    }
  
    // Update the result for
    // the last block
    ways *= fact[size];
  
    return ways;
}
  
// Driver Code
static public void Main ()
{
    int []a = { 1, 2, 4, 4, 2, 4 };
    int n = a.Length;
      
    // Pre-calculating factorials
    pre();
      
    Console.Write(CountPermutation(a, n));
}
}
  
// This code is contributed by Sachin.

chevron_right


Output:

12

Time Complexity: O(N * logN)



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

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 Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : AnkitRai01, Sach_Code