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(2N)
Auxiliary Space: O(N)
Efficient Approach: The idea is to observe a pattern i.e., so observe that minimum element occurs 2n – 1 times, the second minimum occurs 2n – 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. 2n – 1 where n = 3.
2 occurs 2 times i.e. 2n – 2 where n = 3.
3 occurs 1 times i.e. 2n – 3 where n = 3.
Below are the steps:
- Store the index of each element in a Map such that we can print the element in the order of the original array.
- Sort the given array.
- 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).
- 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)