Count subsequences which contains both the maximum and minimum array element
Given an array arr[] consisting of N integers, the task is to find the number of subsequences which contain the maximum as well as the minimum element present in the given array.
Example :
Input: arr[] = {1, 2, 3, 4}
Output: 4
Explanation:
There are 4 subsequence {1, 4}, {1, 2, 4}, {1, 3, 4}, {1, 2, 3, 4} which contains the maximum array element(= 4) and the minimum array element(= 1).
Input: arr[] = {4, 4, 4, 4}
Output: 15
Naive Approach: The simplest approach is to first, traverse the array and find the maximum and minimum of the array and then generate all possible subsequences of the given array. For each subsequence, check if it contains both the maximum and the minimum array element. For all such subsequences, increase the count by 1. Finally, print the count of such subsequences.
Time Complexity: O(2N)
Auxiliary Space: O(N)
Efficient Approach: Follow the steps below to optimize the above approach:
- Find the count of occurrences of the maximum element and the minimum element. Let i and j be the respective count.
- Check if the maximum and the minimum element are the same or not. If found to be true, then the possible subsequences are all non-empty subsequences of the array.
- Otherwise, for satisfying the condition of the subsequence, it should contain at least 1 element from i and at least 1 element from j. Therefore, the required count of subsequences is given by the following equation:
(pow(2, i) -1 ) * ( pow(2, j) -1 ) * pow(2, n-i-j)
Below is the implementation of the above approach:
C++
#include<bits/stdc++.h>
using namespace std;
int count( int arr[], int n, int value);
double countSubSequence( int arr[], int n)
{
int maximum = *max_element(arr, arr + n);
int minimum = *min_element(arr, arr + n);
if (maximum == minimum)
return pow (2, n) - 1;
int i = count(arr, n, maximum);
int j = count(arr, n, minimum);
double res = ( pow (2, i) - 1) *
( pow (2, j) - 1) *
pow (2, n - i - j);
return res;
}
int count( int arr[], int n, int value)
{
int sum = 0;
for ( int i = 0; i < n; i++)
if (arr[i] == value)
sum++;
return sum;
}
int main()
{
int arr[] = { 1, 2, 3, 4 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << countSubSequence(arr, n) << endl;
}
|
Java
import java.util.Arrays;
class GFG{
static double countSubSequence( int [] arr, int n)
{
int maximum = Arrays.stream(arr).max().getAsInt();
int minimum = Arrays.stream(arr).min().getAsInt();
if (maximum == minimum)
return Math.pow( 2 , n) - 1 ;
int i = count(arr, maximum);
int j = count(arr, minimum);
double res = (Math.pow( 2 , i) - 1 ) *
(Math.pow( 2 , j) - 1 ) *
Math.pow( 2 , n - i - j);
return res;
}
static int count( int [] arr, int value)
{
int sum = 0 ;
for ( int i = 0 ; i < arr.length; i++)
if (arr[i] == value)
sum++;
return sum;
}
public static void main(String[] args)
{
int [] arr = { 1 , 2 , 3 , 4 };
int n = arr.length;
System.out.println(countSubSequence(arr, n));
}
}
|
Python3
def countSubSequence(arr, n):
maximum = max (arr)
minimum = min (arr)
if maximum = = minimum:
return pow ( 2 , n) - 1
i = arr.count(maximum)
j = arr.count(minimum)
res = ( pow ( 2 , i) - 1 ) * ( pow ( 2 , j) - 1 ) * pow ( 2 , n - i - j)
return res
arr = [ 1 , 2 , 3 , 4 ]
n = len (arr)
print (countSubSequence(arr, n))
|
C#
using System;
using System.Linq;
class GFG{
static double countSubSequence( int [] arr,
int n)
{
int maximum = arr.Max();
int minimum = arr.Min();
if (maximum == minimum)
return Math.Pow(2, n) - 1;
int i = count(arr, maximum);
int j = count(arr, minimum);
double res = (Math.Pow(2, i) - 1) *
(Math.Pow(2, j) - 1) *
Math.Pow(2, n - i - j);
return res;
}
static int count( int [] arr, int value)
{
int sum = 0;
for ( int i = 0; i < arr.Length; i++)
if (arr[i] == value)
sum++;
return sum;
}
public static void Main(String[] args)
{
int [] arr = {1, 2, 3, 4};
int n = arr.Length;
Console.WriteLine(countSubSequence(arr, n));
}
}
|
Javascript
<script>
function countSubSequence(arr, n)
{
let maximum = Math.max(...arr);
let minimum = Math.min(...arr);
if (maximum == minimum)
return Math.pow(2, n) - 1;
let i = count(arr, maximum);
let j = count(arr, minimum);
let res = (Math.pow(2, i) - 1) *
(Math.pow(2, j) - 1) *
Math.pow(2, n - i - j);
return res;
}
function count(arr, value)
{
let sum = 0;
for (let i = 0; i < arr.length; i++)
if (arr[i] == value)
sum++;
return sum;
}
let arr = [ 1, 2, 3, 4 ];
let n = arr.length;
document.write(countSubSequence(arr, n));
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1)
Last Updated :
07 May, 2021
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