Count of subsequences having maximum distinct elements
Given an arr of size n. The problem is to count all the subsequences having maximum number of distinct elements.
Examples:
Input : arr[] = {4, 7, 6, 7}
Output : 2
The indexes for the subsequences are:
{0, 1, 2} - Subsequence is {4, 7, 6} and
{0, 2, 3} - Subsequence is {4, 6, 7}
Input : arr[] = {9, 6, 4, 4, 5, 9, 6, 1, 2}
Output : 8
Naive Approach: Consider all the subsequences having distinct elements and count the one’s having maximum distinct elements.
Efficient Approach: Create a hash table to store the frequency of each element of the array. Take product of all the frequencies.
The solution is based on the fact that there is always 1 subsequence possible when all elements are distinct. If elements repeat, every occurrence of repeating element makes a mew subsequence of distinct elements.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long int ull;
ull countSubseq( int arr[], int n)
{
unordered_map< int , int > um;
ull count = 1;
for ( int i = 0; i < n; i++)
um[arr[i]]++;
for ( auto itr = um.begin(); itr != um.end(); itr++)
count *= (itr->second);
return count;
}
int main()
{
int arr[] = { 4, 7, 6, 7 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << "Count = "
<< countSubseq(arr, n);
return 0;
}
|
Java
import java.util.HashMap;
class geeks
{
public static long countSubseq( int [] arr, int n)
{
HashMap<Integer, Integer> um = new HashMap<>();
long count = 1 ;
for ( int i = 0 ; i < n; i++)
{
if (um.get(arr[i]) != null )
{
int a = um.get(arr[i]);
um.put(arr[i], ++a);
}
else
um.put(arr[i], 1 );
}
for (HashMap.Entry<Integer, Integer> entry : um.entrySet())
{
count *= entry.getValue();
}
return count;
}
public static void main(String[] args)
{
int [] arr = { 4 , 7 , 6 , 7 };
int n = arr.length;
System.out.println( "Count = " + countSubseq(arr, n));
}
}
|
Python3
def countSubseq(arr, n):
um = {i: 0 for i in range ( 8 )}
count = 1
for i in range (n):
um[arr[i]] + = 1
for key, values in um.items():
if (values > 0 ):
count * = values
return count
if __name__ = = '__main__' :
arr = [ 4 , 7 , 6 , 7 ]
n = len (arr)
print ( "Count =" , countSubseq(arr, n))
|
C#
using System;
using System.Collections.Generic;
class GFG
{
public static long countSubseq( int [] arr,
int n)
{
Dictionary< int ,
int > um = new Dictionary< int ,
int >();
long count = 1;
for ( int i = 0; i < n; i++)
{
if (um.ContainsKey(arr[i]))
{
int a = um[arr[i]];
um.Remove(arr[i]);
um.Add(arr[i], ++a);
}
else
um.Add(arr[i], 1);
}
foreach (KeyValuePair< int , int > entry in um)
{
count *= entry.Value;
}
return count;
}
public static void Main(String[] args)
{
int [] arr = { 4, 7, 6, 7 };
int n = arr.Length;
Console.WriteLine( "Count = " +
countSubseq(arr, n));
}
}
|
Javascript
<script>
function countSubseq(arr, n)
{
var um = new Map();
var count = 1;
for ( var i = 0; i < n; i++)
{
if (um.has(arr[i]))
um.set(arr[i], um.get(arr[i])+1)
else
um.set(arr[i], 1);
}
um.forEach((value, key) => {
count *= value;
});
return count;
}
var arr = [4, 7, 6, 7];
var n = arr.length;
document.write( "Count = "
+ countSubseq(arr, n));
</script>
|
Output:
Count = 2
Time Complexity: O(n).
Auxiliary Space: O(n).
Last Updated :
28 Jul, 2022
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