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# 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++

 `// C++ program for the above approach` `#include` `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`

## Java

 `// 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`

## Python3

 `# 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))`

## C#

 `// 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`

## Javascript

 ``

Output:

`4`

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

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