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Generate Array with components in given vary and median as Ok


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Given two integers N and Ok and a variety [L, R], the duty is to construct an array whose components are distinctive and within the vary [L, R] and the median of the array is Ok.

Examples:

Enter: N = 6, Ok = -1, L = -5, R = 5
Output: -4 -3 -2 0 1 2
Rationalization: Median = (-2 + 0)/2 = -1 which is the same as Ok.

Enter: N = 5, Ok = 4, L = 3, R = 8
Output: -1

 

Strategy: The issue may be solved utilizing the next mathematical concept:

  • If the array is of wierd size the median is the (N/2 + 1)th aspect after sorting. 
  • In any other case, the median would be the common of (N/2) anf (N/2 + 1)th components. 

So the optimum approach is to make an array such that the minimal worth and the utmost worth are as near the median as attainable.

This may be accomplished by merely protecting the distinction between the weather as 1. 
Due to this fact the minimal aspect of the array wll be (Ok – N/2) and the utmost aspect might be (Ok + N/2).

Observe: In case of N being even Ok is the typical of two center components. So these two components have a distinction of two between them and Ok won’t be current within the array.

Comply with the steps talked about beneath to implement the commentary:

  • Discover the attainable minimal and most values of the array.
  • If both L is larger than the minimal or R is lower than the utmost then the array can’t be constructed.
  • In any other case, repair the minimal and generate all the weather by sustaining a niche of 1 between adjoining components.
    • Within the case of N being even, the hole between the center two components might be 2.
  • Return the generated array as the reply.

Under is the implementation of the above strategy:

C++

  

#embody <bits/stdc++.h>

utilizing namespace std;

  

vector<int> constructArray(int N, int Ok, int L, int R)

  

int most important()

{

    int N = 6, Ok = -1;

    int L = -5, R = 5;

  

    

    vector<int> ans

        = constructArray(N, Ok, L, R);

    for (int x : ans)

        cout << x << " ";

    return 0;

}

Time Complexity: O(N)
Auxiliary Area: O(N)

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