Maximum Sum of Array with given MEX

Last Updated : 12 Jan, 2024

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:

C++

#include <bits/stdc++.h> #define ll long long using namespace std; void resultArray(ll N, ll K, ll X) {     // case when answer does not exist     if (N < K || X + 1 < K) {         cout << -1 << endl;         return;     }      // vector to store the answer     vector<ll> ans;     // Push all the numbers less than K into     // the answer, to get MEX = K     for (ll i = 0; i < K; i++) {         ans.push_back(i);     }      // If MEX is equal to maximum element allowed, then fill     // the remaining positions with second maximum element     if (X == K) {         for (int i = K; i < N; i++) {             ans.push_back(K - 1);         }     }     // Else fill the remaining positions with the maximum     // element allowed     else {         for (int i = K; i < N; i++) {             ans.push_back(X);         }     }     for (int i = 0; i < N; i++)         cout << ans[i] << " ";     cout << "\n"; } int main() {     // taking input     ll N = 5, K = 3, X = 3;      resultArray(N, K, X); }

Java

import java.util.*;  class GFG {     // Function to generate the result array     static void resultArray(long N, long K, long X) {         // Case when the answer does not exist         if (N < K || X + 1 < K) {             System.out.println("-1");             return;         }         // List to store the answer         List<Long> ans = new ArrayList<>();         // Push all the numbers less than K into          // the answer to get MEX = K         for (long i = 0; i < K; i++) {             ans.add(i);         }         if (X == K) {             for (long i = K; i < N; i++) {                 ans.add(K - 1);             }         }         // Else fill the remaining positions with          // the maximum element allowed         else {             for (long i = K; i < N; i++) {                 ans.add(X);             }         }         // Print the result array         for (Long num : ans) {             System.out.print(num + " ");         }         System.out.println();     }      public static void main(String[] args) {         // Example input values         long N = 5, K = 3, X = 3;         resultArray(N, K, X);     } }

Python3

def GFG(N, K, X):     # Case when the answer does not exist     if N < K or X + 1 < K:         print("-1")         return          # Array to store the result     ans = []          # Push all the numbers less than K into answer to get MEX = K     for i in range(K):         ans.append(i)          if X == K:         for i in range(K, N):             ans.append(K - 1)     else:         # Else fill the remaining positions with the maximum element allowed         for i in range(K, N):             ans.append(X)          # Print the result array     print(' '.join(map(str, ans)))  # input values N, K, X = 5, 3, 3 GFG(N, K, X)

using System; using System.Collections.Generic;  class GFG {     // Function to generate the result array     static void ResultArray(long N, long K, long X)     {         // Case when the answer does not exist         if (N < K || X + 1 < K)         {             Console.WriteLine("-1");             return;         }         // List to store the answer         List<long> ans = new List<long>();         // Push all the numbers less than K into        // the answer to get MEX = K         for (long i = 0; i < K; i++)         {             ans.Add(i);         }         if (X == K)         {             for (long i = K; i < N; i++)             {                 ans.Add(K - 1);             }         }         // Else fill the remaining positions with         // the maximum element allowed         else         {             for (long i = K; i < N; i++)             {                 ans.Add(X);             }         }         // Print the result array         foreach (var num in ans)         {             Console.Write(num + " ");         }         Console.WriteLine();     }     static void Main()     {         // Example input values         long N = 5, K = 3, X = 3;         ResultArray(N, K, X);     } }

JavaScript

function GFG(N, K, X) {     // Case when the answer does not exist     if (N < K || X + 1 < K) {         console.log("-1");         return;     }     // Array to store the result     const ans = [];     // Push all the numbers less than K into answer to get MEX = K     for (let i = 0; i < K; i++) {         ans.push(i);     }     if (X === K) {         for (let i = K; i < N; i++) {             ans.push(K - 1);         }     }     // Else fill the remaining positions with maximum element allowed     else {         for (let i = K; i < N; i++) {             ans.push(X);         }     }     // Print the result array     console.log(ans.join(' ')); } // input values const N = 5, K = 3, X = 3; GFG(N, K, X);

Output

0 1 2 2 2

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

Maximum XOR of Two Numbers in an Array | Set 2

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Maximum Sum of Array with given MEX

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