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)
{
if (N < K || X + 1 < K) {
cout << -1 << endl;
return;
}
vector<ll> ans;
for (ll i = 0; i < K; i++) {
ans.push_back(i);
}
if (X == K) {
for (int i = K; i < N; i++) {
ans.push_back(K - 1);
}
}
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()
{
ll N = 5, K = 3, X = 3;
resultArray(N, K, X);
}
|
Java
import java.util.*;
class GFG {
static void resultArray(long N, long K, long X) {
if (N < K || X + 1 < K) {
System.out.println("-1");
return;
}
List<Long> ans = new ArrayList<>();
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 {
for (long i = K; i < N; i++) {
ans.add(X);
}
}
for (Long num : ans) {
System.out.print(num + " ");
}
System.out.println();
}
public static void main(String[] args) {
long N = 5, K = 3, X = 3;
resultArray(N, K, X);
}
}
|
Python3
def GFG(N, K, X):
if N < K or X + 1 < K:
print("-1")
return
ans = []
for i in range(K):
ans.append(i)
if X == K:
for i in range(K, N):
ans.append(K - 1)
else:
for i in range(K, N):
ans.append(X)
print(' '.join(map(str, ans)))
N, K, X = 5, 3, 3
GFG(N, K, X)
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static void ResultArray(long N, long K, long X)
{
if (N < K || X + 1 < K)
{
Console.WriteLine("-1");
return;
}
List<long> ans = new List<long>();
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
{
for (long i = K; i < N; i++)
{
ans.Add(X);
}
}
foreach (var num in ans)
{
Console.Write(num + " ");
}
Console.WriteLine();
}
static void Main()
{
long N = 5, K = 3, X = 3;
ResultArray(N, K, X);
}
}
|
Javascript
function GFG(N, K, X) {
if (N < K || X + 1 < K) {
console.log("-1");
return;
}
const ans = [];
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 {
for (let i = K; i < N; i++) {
ans.push(X);
}
}
console.log(ans.join(' '));
}
const N = 5, K = 3, X = 3;
GFG(N, K, X);
|
Time Complexity: O(N)
Auxiliary Space: O(N), where N is the size of array.
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