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<bits/stdc++.h>using namespace std;int count(int arr[], int n, int value);// Function to calculate the// count of subsequencesdouble 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 Codeint 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 approachimport java.util.Arrays;class GFG{ // Function to calculate the// count of subsequencesstatic 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 subsequencesdef 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 Codearr = [1, 2, 3, 4]n = len(arr)# Function callprint(countSubSequence(arr, n)) |
C#
// C# program for // the above approachusing System;using System.Linq;class GFG{ // Function to calculate the// count of subsequencesstatic 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
<script>// JavaScript program for the above approach// Function to calculate the// count of subsequencesfunction countSubSequence(arr, n){ // Find the maximum // from the array let maximum = Math.max(...arr); // Find the minimum // from the array let minimum = Math.min(...arr); // If array contains only // one distinct element if (maximum == minimum) return Math.pow(2, n) - 1; // Find the count of maximum let i = count(arr, maximum); // Find the count of minimum let j = count(arr, minimum); // Finding the result // with given condition 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;} // Driver Code let arr = [ 1, 2, 3, 4 ]; let n = arr.length; // Function call document.write(countSubSequence(arr, n));// This code is contributed by souravghosh0416.</script> |
Output:
4
Time Complexity: O(N)
Auxiliary Space: O(1)
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