Count permutation such that sequence is non decreasing

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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++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
#define MAX 20
 
// To store the factorials
int fact[MAX];
 
// Function to update fact[] array
// such that fact[i] = i!
void pre()
{
 
    // 0! = 1
    fact[0] = 1;
    for (int i = 1; i < MAX; 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;
}

Java

//Java implementation of the approach
import java.util.Arrays; 
import java.io.*;
 
class GFG 
{
static int MAX = 20;
 
// To store the factorials
static int []fact=new int[MAX];
 
// Function to update fact[] array
// such that fact[i] = i!
static void pre()
{
 
    // 0! = 1
    fact[0] = 1;
    for (int i = 1; i < MAX; 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

Python3

# Python3 implementation of the approach 
MAX = 20
 
# To store the factorials 
fact = [0] * MAX; 
 
# Function to update fact[] array 
# such that fact[i] = i! 
def pre() :
 
    # 0! = 1 
    fact[0] = 1; 
    for i in range(1, MAX):
 
        # 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

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
static int MAX = 20;
 
// To store the factorials
static int []fact = new int[MAX];
 
// Function to update fact[] array
// such that fact[i] = i!
static void pre()
{
 
    // 0! = 1
    fact[0] = 1;
    for (int i = 1; i < MAX; 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.

Javascript

<script>
 
// Javascript implementation of the approach
 
const MAX = 20;
 
// To store the factorials
let fact = new Array(MAX);
 
// Function to update fact[] array
// such that fact[i] = i!
function pre()
{
 
    // 0! = 1
    fact[0] = 1;
    for (let i = 1; i < MAX; i++) {
 
        // i! = i * (i - 1)!
        fact[i] = i * fact[i - 1];
    }
}
 
// Function to return the count
// of possible permutations
function CountPermutation(a, n)
{
 
    // To store the result
    let ways = 1;
 
    // Sort the array
    a.sort();
 
    // Initial size of the block
    let size = 1;
    for (let 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
 
    let a = [ 1, 2, 4, 4, 2, 4 ];
    let n = a.length;
 
    // Pre-calculating factorials
    pre();
 
    document.write(CountPermutation(a, n));
 
</script>

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
#define MAX 20
 
// To store the factorials
int fact[MAX];
 
// Function to update fact[] array
// such that fact[i] = i!
void pre()
{
 
    // 0! = 1
    fact[0] = 1;
    for (int i = 1; i < MAX; 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;
}

Java

//Java implementation of the approach
import java.util.Arrays; 
import java.io.*;
 
class GFG 
{
static int MAX = 20;
 
// To store the factorials
static int []fact=new int[MAX];
 
// Function to update fact[] array
// such that fact[i] = i!
static void pre()
{
 
    // 0! = 1
    fact[0] = 1;
    for (int i = 1; i < MAX; 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

Python3

# Python3 implementation of the approach 
MAX = 20
 
# To store the factorials 
fact = [0] * MAX; 
 
# Function to update fact[] array 
# such that fact[i] = i! 
def pre() :
 
    # 0! = 1 
    fact[0] = 1; 
    for i in range(1, MAX):
 
        # 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

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
static int MAX = 20;
 
// To store the factorials
static int []fact = new int[MAX];
 
// Function to update fact[] array
// such that fact[i] = i!
static void pre()
{
 
    // 0! = 1
    fact[0] = 1;
    for (int i = 1; i < MAX; 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.

Javascript

<script>
 
// Javascript implementation of the approach
 
const N = 20;
 
// To store the factorials
let fact = new Array(N);
 
// Function to update fact[] array
// such that fact[i] = i!
function pre()
{
 
    // 0! = 1
    fact[0] = 1;
    for (let i = 1; i < N; i++) {
 
        // i! = i * (i - 1)!
        fact[i] = i * fact[i - 1];
    }
}
 
// Function to return the count
// of possible permutations
function CountPermutation(a, n)
{
 
    // To store the result
    let ways = 1;
 
    // Sort the array
    a.sort();
 
    // Initial size of the block
    let size = 1;
    for (let 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
 
    let a = [ 1, 2, 4, 4, 2, 4 ];
    let n = a.length;
 
    // Pre-calculating factorials
    pre();
 
    document.write(CountPermutation(a, n));
 
</script>

Output:

Time Complexity: O(N * logN)
Auxiliary Space: O(MAX), where max is the size of the factorial array.

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