Smallest occurring element in each subsequence

Given an array of N distinct integers. For each element the task is to find the count of subsequence from all the possible subsequence whose minimum element is the current element.
Examples: 

Input: arr[] = {1, 2} 
Output: 2 1 
Explanation: 
Subsequences are {1}, {2}, {1, 2}. 
The count of the smallest element in each subsequence is: 
1 = 2, 2 = 1
Input: arr[] = {1, 2, 3} 
Output: 4 2 1

Naive Approach: The idea is to generate all possible subsequences of the given array and count the smallest element in each subsequence and print its count for each element in the array. 
Time Complexity: O(2^N) 
Auxiliary Space: O(N)
 

Efficient Approach: The idea is to observe a pattern i.e., so observe that minimum element occurs 2^{n - 1} times, the second minimum occurs 2^{n - 2} times and so on …. For Example:

Let the array be arr[] = {1, 2, 3} 
Subsequences are {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3} 
Minimum of each subsequence: {1}, {2}, {3}, {1}, {1}, {2}, {1}. 
where 
1 occurs 4 times i.e. 2^{n - 1} where n = 3. 
2 occurs 2 times i.e. 2^{n - 2} where n = 3. 
3 occurs 1 times i.e. 2^{n - 3} where n = 3.

Below are the steps:  

1. Store the index of each element in a Map such that we can print the element in the order of the original array.
2. Sort the given array.
3. Now the elements are in increasing order and from the above observation traverse the given array and keep the count of subsequence such that each element is the smallest element is given by pow(2, N - 1 - i).
4. Now traverse the map and print count of subsequence according to the element in the original array.

Below is the implementation of the above approach:

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

// Function that count the subsequence
// such that each element as the
// minimum element in the subsequence
void solve(int arr[], int N)
{
    map<int, int> M;

    // Store index in a map
    for (int i = 0; i < N; i++) {
        M[i] = arr[i];
    }

    // Sort the array
    sort(arr, arr + N);

    // To store count of subsequence
    unordered_map<int, int> Count;

    // Traverse the array
    for (int i = 0; i < N; i++) {

        // Store count of subsequence
        Count[arr[i]] = pow(2, N - i - 1);
    }

    // Print the count of subsequence
    for (auto& it : M) {
        cout << Count[M[it.second]] << ' ';
    }
}

// Driver code
int main()
{
    int arr[] = { 5, 2, 1 };
    int N = sizeof arr / sizeof arr[0];

    // Function call
    solve(arr, N);
}

// Java program for the above approach
import java.util.*;
class GFG{

// Function that count the subsequence
// such that each element as the
// minimum element in the subsequence
static void solve(int arr[], int N)
{
    HashMap<Integer,
              Integer> M = new HashMap<>();

    // Store index in a map
    for (int i = 0; i < N; i++)
    {
        M.put(i, arr[i]);
    }

    // Sort the array
    Arrays.sort(arr);

    // To store count of subsequence
    HashMap<Integer,
              Integer> Count = new HashMap<>();

    // Traverse the array
    for (int i = 0; i < N; i++) 
    {

        // Store count of subsequence
        Count.put(arr[i], 
                 (int)Math.pow(2, N - i - 1));
    }

    // Print the count of subsequence
    for (Map.Entry<Integer,
                    Integer> m : M.entrySet())
    {
        System.out.print(Count.get(m.getValue()) + " ");
    }
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 5, 2, 1 };
    int N = arr.length;

    // Function call
    solve(arr, N);
}
}

// This code is contributed by Amit Katiyar

# Python3 program for the above approach

# Function that count the subsequence
# such that each element as the
# minimum element in the subsequence
def solve(arr, N):

    M = {}

    # Store index in a map
    for i in range(N):
        M[i] = arr[i]
    
    # Sort the array
    arr.sort()

    # To store count of subsequence
    Count = {}

    # Traverse the array
    for i in range(N):

        # Store count of subsequence
        Count[arr[i]] = pow(2, N - i - 1)

    # Print the count of subsequence
    for it in Count.values():
        print(it, end = " ")

# Driver code
if __name__ == "__main__":

    arr = [ 5, 2, 1 ]
    N = len(arr)

    # Function call
    solve(arr, N)

# This code is contributed by chitranayal

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function that count the subsequence
// such that each element as the
// minimum element in the subsequence
static void solve(int []arr, int N)
{
    Dictionary<int,
               int> M = new Dictionary<int,
                                         int>();

    // Store index in a map
    for (int i = 0; i < N; i++)
    {
        M.Add(i, arr[i]);
    }

    // Sort the array
    Array.Sort(arr);

    // To store count of subsequence
    Dictionary<int,
               int> Count = new Dictionary<int,
                                             int>();

    // Traverse the array
    for (int i = 0; i < N; i++) 
    {

        // Store count of subsequence
        Count.Add(arr[i], 
                 (int)Math.Pow(2, N - i - 1));
    }
  // Print the count of subsequence
   foreach(KeyValuePair<int, int> m in M)
{
    Console.Write(Count[m.Value]);   
}  
}

// Driver code
public static void Main(String[] args)
{
    int []arr = { 5, 2, 1 };
    int N = arr.Length;

    // Function call
    solve(arr, N);
}
}

// This code is contributed by PrinciRaj1992 

<script>

// Javascript program for the above approach

// Function that count the subsequence
// such that each element as the
// minimum element in the subsequence
function solve(arr, N)
{
    var M = new Map();

    // Store index in a map
    for (var i = 0; i < N; i++) {
        M.set(i, arr[i]);
    }

    // Sort the array
    arr.sort((a,b)=>a-b);

    // To store count of subsequence
    var Count = new Map();

    // Traverse the array
    for (var i = 0; i < N; i++) {

        // Store count of subsequence
        Count.set(arr[i], parseInt(Math.pow(2, N - i - 1)));
    }

    // Print the count of subsequence
    M.forEach((value, key) => {
        document.write(Count.get(value) + ' ');
    });
    
}

// Driver code
var arr = [5, 2, 1];
var N = arr.length;
// Function call
solve(arr, N);

// This code is contributed by rrrtnx.
</script> 

4 2 1

Time Complexity: O(N*log N) 
Auxiliary Space: O(N)