Maximum Sum of Array with given MEX

Given 3 integers N, K, and X, the task is to construct an array arr[] with the below conditions:

• Size of the array = N
• MEX of the array = K
• All array elements should be at most X

Among all the array that follows the above condition print the one having the maximum sum of its elements or print -1 if no such array exists.

Constraints:

• 1<= N <= 100000
• 1<= K <= 100000
• 1<= X <= 100000

Examples:

Input: N=5, K=3, X=3
Output: [0, 1, 2, 2, 2]
Explanation: The MEX of the above array is 3, and sum of its element is 7, it is guaranteed that we cannot create an array having sum greater than 7

Input: N=4, K=7, X=5
Output: -1
Explanation: No valid array exists

Approach: The problem can be solved using the following approach:

Case 1: If min(N, X+1) < K, then answer does not exists i.e. answer = -1.
Reason:

• If N < K, then the maximum value of MEX that can be created using N numbers is equal to N. For example, if N=4, then arr[] = [0, 1, 2, 3] gives MEX = 4. Therefore, if N<K we cannot achieve MEX = K
• If X+1 < K, then also it would be impossible to achieve MEX = K due to the fact that for MEX = K, K - 1 should be present in the array, but X is the maximum number present in the array.

Case 2: If answer exists, and we want to maximize it.

Since we have to form MEX=K therefore it is necessary for all the elements from 0 to K-1 to appear at least once in the array. Now since our task is to maximize the sum, all the elements form 0 to K-1 should occur exactly once all the remaining elements will depend upon the value of X i.e.

• If X = K, then all the remaining values should be K-1 so that MEX = K is preserved
• If X ≠ K, then all the remaining values should be X.

Steps to solve the problem:

• If N < K or X + 1 < K, then the answer does not exist, so print -1.
• Else, since we want the MEX to be equal to K so push all the numbers from 0 to K-1 into our answer.
• After pushing number from 0 to K-1, check if we can push X as all the remaining elements.
  • If yes, then push all the remaining elements as X.
  • Else, push all the remaining elements as X - 1.
• Finally, print the answer array.

Below is the implementation of the above approach:

0 1 2 2 2 

Time Complexity: O(N)
Auxiliary Space: O(N), where N is the size of array.