Discover integers such that A accommodates precisely X distinct integers better than A{i]

Given an array A[] of N integers, the duty is to print the variety of integers i from 1 to N, for which A accommodates precisely X distinct integers (for X = 0 to N-1) better than Ai

Examples:

Enter: N = 6, A[] = {2, 7, 1, 8, 2, 8}
Output: {2, 1, 2, 1, 0, 0}
Clarification: Allow us to take into account X = 2.
A[1] = 2: A accommodates 2 distinct integers better than 2 (7 and eight)
A[2] = 7: A accommodates 1 distinct integer better than 7 (8)
A[3] = 1: A accommodates 3 distinct integers better than 1 (2, 7 and eight)
A[4] = 8: A doesn’t comprise any distinct integers better than 8
A[5] = 2: A accommodates 2 distinct integers better than 2 (7 and eight)
A[6] = 8: A doesn’t comprise any distinct integers better than 8
So, the given situation is happy for i=1 and 5 in case of X=2.

Enter: N = 1, A[] = {1}
Output: 1

Strategy: To unravel the issue observe the under observations:

Observations:

If we retailer frequency of every component of array in a hashmap and type it in reducing order by the keys, then:

Let the hashmap be – ((u₁, f₁), (u₂, f₂)….(u_n, f_n)) the place f_i is the frequency of component u_i and u1 > u2 > ……un.

• For every i = 1 to n, there are i – 1 distinct integers that’s better than u_i. So, for every X=0 to n-1 the reply for X could be f_X+1.
• Reply for X = n to N – 1 could be 0.

Primarily based on the above statement following strategy can be utilized to unravel the issue:

• Declare a map (say mp) and insert the weather of array A into it.
• Iterate the map from the top and print the frequencies of the weather.
• For the remaining parts (i.e. N-n), print N-n zeroes.

Following is the code based mostly on the above strategy :

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