Count number of Subsequences in Array in which X and Y are min and max elements

Last Updated : 23 Jul, 2025

Given an array arr[] consisting of N unique elements, the task is to return the count of the subsequences in an array in which the minimum element is X and the maximum element is Y.

Examples:

Input: arr[] = {2, 4, 6, 7, 5}, X = 2, Y = 5
Output: 2
Explanation: Subsequences in which the minimum element is X and the maximum element is Y are {2, 5}, {2, 4, 5}.

Input: arr[] = {2, 4, 6, 7, 5, 1, 9, 10, 11}, X = 2, Y = 7
Output: 8
Explanation: subsequences in which the minimum element is X and the maximum element is Y are {2, 4, 6, 7, 5}, {2, 4, 7, 5}, {2, 4, 6, 7}, {2, 4, 7}, {2, 6, 7, 5}, {2, 7, 5}, {2, 6, 7}, {2, 7}

Naive Approach: The basic way to solve the problem is as follows:

The given problem can be solved by generating and counting all possible subsequences of the given array using recursion and checking if each subsequence has the minimum element as X and the maximum element as Y.

Below is the code for the above approach:

C++

// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;

int Subsequences(vector<int>& arr, int index,
                 vector<int>& subarr, int n, int x, int y)
{

    if (index == n) {

        // If the subsequence is valid
        // return 1;
        if (subarr.size() != 0 && subarr[0] == x
            && subarr[subarr.size() - 1] == y) {

            // Increment count of valid
            // subsequences
            return 1;
        }
        else
            return 0;
    }
    else {

        // Pick the current index into
        // the subsequence.
        subarr.push_back(arr[index]);

        // Recursive call to consider the
        // remaining elements in
        // the subsequence
        int pic
            = Subsequences(arr, index + 1, subarr, n, x, y);

        // Backtrack to remove the current
        // index from the subsequence
        subarr.pop_back();

        // Recursive call to not consider
        // the current element in
        // the subsequence
        int notpic
            = Subsequences(arr, index + 1, subarr, n, x, y);

        // Sum the counts of valid
        // subsequences
        return pic + notpic;
    }
}

// Drivers code
int main()
{
    vector<int> arr = { 2, 4, 6, 7, 5, 1, 9, 10, 11 };
    int x = 2, y = 7;
    vector<int> subarr;

    // Sort the array to simplify the search
    sort(arr.begin(), arr.end());

    // Call the recursive function to find
    // all valid subsequences
    cout << Subsequences(arr, 0, subarr, arr.size(), x, y);

    return 0;
}

Java

// / Java code for the above approach:
import java.util.*;

public class Main {
    public static int
    Subsequences(ArrayList<Integer> arr, int index,
                 ArrayList<Integer> subarr, int n, int x,
                 int y)
    {
        // If the subsequence is valid
        // return 1;
        if (index == n) {
            if (subarr.size() != 0 && subarr.get(0) == x
                && subarr.get(subarr.size() - 1) == y) {
                // Increment count of valid
                // subsequences
                return 1;
            }
            else {
                return 0;
            }
        }
        else {
            // Pick the current index into
            // the subsequence.
            subarr.add(arr.get(index));
            // Recursive call to consider the
            // remaining elements in
            // the subsequence
            int pic = Subsequences(arr, index + 1, subarr,
                                   n, x, y);
            // Backtrack to remove the current
            // index from the subsequence
            subarr.remove(subarr.size() - 1);
            // Recursive call to not consider
            // the current element in
            // the subsequence
            int notpic = Subsequences(arr, index + 1,
                                      subarr, n, x, y);
            // Sum the counts of valid
            // subsequences
            return pic + notpic;
        }
    }
    // Drivers code
    public static void main(String[] args)
    {
        ArrayList<Integer> arr = new ArrayList<Integer>(
            Arrays.asList(2, 4, 6, 7, 5, 1, 9, 10, 11));
        int x = 2, y = 7;
        ArrayList<Integer> subarr
            = new ArrayList<Integer>();
        // Sort the array to simplify the search
        Collections.sort(arr);
        // Call the recursive function to find
        // all valid subsequences
        System.out.println(
            Subsequences(arr, 0, subarr, arr.size(), x, y));
    }
}

Python3

# Python3 code for the above approach:

def Subsequences(arr, index, subarr, n, x, y):
    if index == n:
        # If the subsequence is valid, return 1
        if subarr and subarr[0] == x and subarr[-1] == y:
            # Increment count of valid subsequences
            return 1
        else:
            return 0
    else:
        # Pick the current index into the subsequence
        subarr.append(arr[index])

        # Recursive call to consider the remaining elements in the subsequence
        pic = Subsequences(arr, index + 1, subarr, n, x, y)

        # Backtrack to remove the current index from the subsequence
        subarr.pop()

        # Recursive call to not consider the current element in the subsequence
        notpic = Subsequences(arr, index + 1, subarr, n, x, y)

        # Sum the counts of valid subsequences
        return pic + notpic

# Driver code
if __name__ == "__main__":
  arr = [2, 4, 6, 7, 5, 1, 9, 10, 11]
  x, y = 2, 7
  subarr = []

  # Sort the array to simplify the search
  arr.sort()

  # Call the recursive function to find all valid subsequences
  print(Subsequences(arr, 0, subarr, len(arr), x, y))

using System;
using System.Collections.Generic;
using System.Linq;

class Program {
    static int Subsequences(List<int> arr, int index,
                            List<int> subarr, int n, int x,
                            int y)
    {

        if (index == n) {

            // If the subsequence is valid
            // return 1;
            if (subarr.Count != 0 && subarr[0] == x
                && subarr[subarr.Count - 1] == y) {

                // Increment count of valid
                // subsequences
                return 1;
            }
            else
                return 0;
        }
        else {

            // Pick the current index into
            // the subsequence.
            subarr.Add(arr[index]);

            // Recursive call to consider the
            // remaining elements in
            // the subsequence
            int pic = Subsequences(arr, index + 1, subarr,
                                   n, x, y);

            // Backtrack to remove the current
            // index from the subsequence
            subarr.RemoveAt(subarr.Count - 1);

            // Recursive call to not consider
            // the current element in
            // the subsequence
            int notpic = Subsequences(arr, index + 1,
                                      subarr, n, x, y);

            // Sum the counts of valid
            // subsequences
            return pic + notpic;
        }
    }

    static void Main(string[] args)
    {
        List<int> arr = new List<int>() {
            2, 4, 6, 7, 5, 1, 9, 10, 11
        };
        int x = 2;
        int y = 7;
        List<int> subarr = new List<int>();

        // Sort the array to simplify the search
        arr.Sort();

        // Call the recursive function to find
        // all valid subsequences
        Console.WriteLine(Subsequences(arr, 0, subarr,
                                       arr.Count(), x, y));
    }
}

JavaScript

// javascript code for the above approach:
function Subsequences(arr, index, subarr, n, x, y) {
  if (index == n) {
    // If the subsequence is valid, return 1
    if (subarr.length != 0 && subarr[0] == x && subarr[subarr.length - 1] == y) {
      // Increment count of valid subsequences
      return 1;
    } else {
      return 0;
    }
  } else {
    // Pick the current index into the subsequence.
    subarr.push(arr[index]);

    // Recursive call to consider the remaining elements in the subsequence.
    let pic = Subsequences(arr, index + 1, subarr, n, x, y);

    // Backtrack to remove the current index from the subsequence.
    subarr.pop();

    // Recursive call to not consider the current element in the subsequence.
    let notpic = Subsequences(arr, index + 1, subarr, n, x, y);

    // Sum the counts of valid subsequences.
    return pic + notpic;
  }
}

// Drivers code
let arr = [2, 4, 6, 7, 5, 1, 9, 10, 11];
let x = 2, y = 7;
let subarr = [];

// Sort the array to simplify the search.
arr.sort();

// Call the recursive function to find all valid subsequences.
console.log(Subsequences(arr, 0, subarr, arr.length, x, y));

Output

Time Complexity: O(2ⁿ)
Auxiliary Space: O(n)

Efficient Approach: To solve the problem follow the below idea:

We know that the subsequence will be formed will be in the range (X, Y), and the formula that will come for all the subsequence in the range (X, Y) excluding X and Y will be 2ⁿ where n is the number of elements in the range X and Y. Therefore will return the 2ⁿ as the answer.

Below are the steps for the above approach:

Initialize a counter variable say, cnt = 0 to keep track of the number of elements in the range [X, Y].
Run a loop from i = 0 to i < n and check if the element lies within the range (x, y),
- if (arr[i] > X && arr[i] < Y), increment the cnt variable by 1.
Return 2^cnt, which gives us the number of subsequences, and return 1 << cnt.

Below is the code for the above approach:

C++

// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;

int countSubsequences(int arr[], int n, int x, int y)
{

    // Initialize a counter to keep track
    // of the number of elements in
    // the range [x, y]
    int cnt = 0;
    for (int i = 0; i < n; i++) {

        // Check if the element lies
        // within the range (x, y)
        if (arr[i] > x && arr[i] < y) {

            // If so, increment the counter
            cnt++;
        }
    }

    // Return 2 raised to the power of cnt,
    // which gives us the number
    // of subsequences
    return (1 << cnt);
}

// Drivers code
int main()
{
    int arr[] = { 2, 4, 6, 7, 5 };

    // Calculate the length of the array
    int n = sizeof(arr) / sizeof(arr[0]);
    int x = 2, y = 5;

    // Call the countSubsequences function
    // and print the result
    cout << countSubsequences(arr, n, x, y) << endl;
    return 0;
}

Java

import java.util.*;

public class Main {
    public static int countSubsequences(int[] arr, int n, int x, int y) 
    {
      
        // Initialize a counter to keep track
        // of the number of elements in
        // the range [x, y]
        int cnt = 0;
        for (int i = 0; i < n; i++)
        {
          
            // Check if the element lies
            // within the range (x, y)
            if (arr[i] > x && arr[i] < y) 
            {
              
                // If so, increment the counter
                cnt++;
            }
        }

        // Return 2 raised to the power of cnt,
        // which gives us the number
        // of subsequences
        return (1 << cnt);
    }

    // Driver code
    public static void main(String[] args) {
        int[] arr = {2, 4, 6, 7, 5};

        // Calculate the length of the array
        int n = arr.length;
        int x = 2, y = 5;

        // Call the countSubsequences function
        // and print the result
        System.out.println(countSubsequences(arr, n, x, y));
    }
}

Python

def countSubsequences(arr, n, x, y):
    # Initialize a counter to keep track
    # of the number of elements in
    # the range [x, y]
    cnt = 0
    for i in range(n):
        # Check if the element lies
        # within the range (x, y)
        if arr[i] > x and arr[i] < y:
            # If so, increment the counter
            cnt += 1
    # Return 2 raised to the power of cnt,
    # which gives us the number
    # of subsequences
    return 1 << cnt

# Drivers code
arr = [2, 4, 6, 7, 5]

# Calculate the length of the array
n = len(arr)
x, y = 2, 5

# Call the countSubsequences function
# and print the result
print(countSubsequences(arr, n, x, y))

// C# program to implement the above approach
using System;

public class GFG {
    public static int countSubsequences(int[] arr, int n, int x, int y)
    {
    
        // Initialize a counter to keep track
        // of the number of elements in
        // the range [x, y]
        int cnt = 0;
        for (int i = 0; i < n; i++)
        {
        
            // Check if the element lies
            // within the range (x, y)
            if (arr[i] > x && arr[i] < y)
            {
            
                // If so, increment the counter
                cnt++;
            }
        }

        // Return 2 raised to the power of cnt,
        // which gives us the number
        // of subsequences
        return (1 << cnt);
    }

    // Driver code
    public static void Main() {
        int[] arr = {2, 4, 6, 7, 5};

        // Calculate the length of the array
        int n = arr.Length;
        int x = 2, y = 5;

        // Call the countSubsequences function
        // and print the result
        Console.WriteLine(countSubsequences(arr, n, x, y));
    }
}

// This code is contributed by Vaibhav Nandan

JavaScript

function countSubsequences(arr, x, y) {

    // Initialize a counter to keep track
    // of the number of elements in
    // the range [x, y]
    let cnt = 0;
    for (let i = 0; i < arr.length; i++) {
    
        // Check if the element lies
        // within the range (x, y)
        if (arr[i] > x && arr[i] < y) {
        
            // If so, increment the counter
            cnt++;
        }
    }
    
    // Return 2 raised to the power of cnt,
    // which gives us the number
    // of subsequences
    return Math.pow(2, cnt);
}

// Drivers code
const arr = [2, 4, 6, 7, 5];
const x = 2, y = 5;

console.log(countSubsequences(arr, x, y));

Output

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

Analysis of Algorithms

nk20102001

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Recursion

Count number of Subsequences in Array in which X and Y are min and max elements

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