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++

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// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
 
int count(int arr[], int n, int value);
 
// Function to calculate the
// count of subsequences
double countSubSequence(int arr[], int n)
{
   
    // Find the maximum
    // from the array
    int maximum = *max_element(arr, arr + n);
  
    // Find the minimum
    // from the array
    int minimum = *min_element(arr, arr + n);
  
    // If array contains only
    // one distinct element
    if (maximum == minimum)
        return pow(2, n) - 1;
  
    // Find the count of maximum
    int i = count(arr, n, maximum);
  
    // Find the count of minimum
    int j = count(arr, n, minimum);
  
    // Finding the result
    // with given condition
    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;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Function call
    cout << countSubSequence(arr, n) << endl;
}
 
// This code is contributed by rutvik_56

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Java

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// Java program for the above approach
import java.util.Arrays;
 
class GFG{
     
// Function to calculate the
// count of subsequences
static double countSubSequence(int[] arr, int n)
{
     
    // Find the maximum
    // from the array
    int maximum = Arrays.stream(arr).max().getAsInt();
 
    // Find the minimum
    // from the array
    int minimum = Arrays.stream(arr).min().getAsInt();
 
    // If array contains only
    // one distinct element
    if (maximum == minimum)
        return Math.pow(2, n) - 1;
 
    // Find the count of maximum
    int i = count(arr, maximum);
 
    // Find the count of minimum
    int j = count(arr, minimum);
 
    // Finding the result
    // with given condition
    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;
}
 
// Driver Code
public static void main(String[] args)
{
    int[] arr = { 1, 2, 3, 4 };
    int n = arr.length;
 
    // Function call
    System.out.println(countSubSequence(arr, n));
}
}
 
// This code is contributed by Amit Katiyar

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Python3

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# Python3 program for the above approach
 
# Function to calculate the
# count of subsequences
def countSubSequence(arr, n):
 
    # Find the maximum
    # from the array
    maximum = max(arr)
 
    # Find the minimum
    # from the array
    minimum = min(arr)
 
    # If array contains only
    # one distinct element
    if maximum == minimum:
        return pow(2, n)-1
 
    # Find the count of maximum
    i = arr.count(maximum)
 
    # Find the count of minimum
    j = arr.count(minimum)
 
    # Finding the result
    # with given condition
    res = (pow(2, i) - 1) * (pow(2, j) - 1) * pow(2, n-i-j)
 
    return res
 
 
# Driver Code
arr = [1, 2, 3, 4]
n = len(arr)
 
# Function call
print(countSubSequence(arr, n))

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C#

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// C# program for
// the above approach
using System;
using System.Linq;
class GFG{
     
// Function to calculate the
// count of subsequences
static double countSubSequence(int[] arr,
                               int n)
{   
  // Find the maximum
  // from the array
  int maximum = arr.Max();
 
  // Find the minimum
  // from the array
  int minimum = arr.Min();
 
  // If array contains only
  // one distinct element
  if (maximum == minimum)
    return Math.Pow(2, n) - 1;
 
  // Find the count of maximum
  int i = count(arr, maximum);
 
  // Find the count of minimum
  int j = count(arr, minimum);
 
  // Finding the result
  // with given condition
  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;
}
 
// Driver Code
public static void Main(String[] args)
{
  int[] arr = {1, 2, 3, 4};
  int n = arr.Length;
 
  // Function call
  Console.WriteLine(countSubSequence(arr, n));
}
}
  
// This code is contributed by shikhasingrajput

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Output: 

4




Time Complexity: O(N)
Auxiliary Space: O(1)

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