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++ 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 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 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# 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. |
<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++ 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 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 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# 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. |
<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> |
12
Time Complexity: O(N * logN)
Auxiliary Space: O(MAX), where max is the size of the factorial array.