Rearrange array elements to maximize the sum of MEX of all prefix arrays

Given an array arr[] of size N, the task is to rearrange the array elements such that the sum of MEX of all prefix arrays is the maximum possible.

Note: MEX of a sequence is the minimum non-negative number not present in the sequence. 

Examples:

Input: arr[] = {2, 0, 1}
Output: 0, 1, 2
Explanation:
Sum of MEX of all prefix arrays of all possible permutations of the given array:
arr[] = {0, 1, 2}, Mex(0) + Mex(0, 1) + Mex(0, 1, 2) = 1 + 2 + 3 = 6
arr[] = {1, 0, 2}, Mex(1) + Mex(1, 0) + Mex(1, 0, 2) = 0 + 2 + 3 = 5
arr[] = {2, 0, 1}, Mex(2) + Mex(2, 0) + Mex(2, 0, 1) = 0 + 1 + 3 = 4
arr[] = {0, 2, 1}, Mex(0) + Mex(0, 2) + Mex(0, 2, 1) = 1 + 1 + 3 = 5
arr[] = {1, 2, 0}, Mex(1) + Mex(1, 2) + Mex(1, 2, 0) = 0 + 0 + 3 = 3
arr[] = {2, 1, 0}, Mex(2) + Mex(2, 1) + Mex(2, 1, 0) = 0 + 0 + 3 = 3
Hence, the maximum sum possible is 6.

Input: arr[] = {1, 0, 0}
Output: 0, 1, 0
Explanation:
Sum of MEX of all prefix arrays of all possible permutations of the given array:
arr[]={1, 0, 0}, Mex(1) + Mex(1, 0) + Mex(1, 0, 0) = 0 + 2 + 2 = 4
arr[]={0, 1, 0}, Mex(0) + Mex(0, 1) + Mex(0, 1, 0) = 1 + 2 + 2 = 5
arr[]={0, 0, 1}, Mex(0) + Mex(0, 0) + Mex(0, 0, 1) = 1 + 1 + 2 = 4
Hence, the maximum value is 5 for the arrangement, arr[]={0, 1, 0}.

Naive Approach: The simplest approach is to generate all possible permutations of the given array arr[] and then for each permutation, find the value of MEX of all the prefix arrays, while keeping track of the overall maximum value. After iterating over all possible permutations, print the permutation having the largest value.

Time Complexity: O(N² * N!)
Auxiliary Space: O(N)

Efficient Approach: The optimal idea is based on the observation that the sum of MEX of prefix arrays will be maximum when all the distinct elements are arranged in increasing order and the duplicates are present at the end of the array. 
Follow the steps below to solve the problem:

• Initialize a vector of integers, say ans, to store the required arrangement.
• Sort the array arr[] in increasing order.
• Traverse the array arr[] and push all distinct elements into the array ans[].
• Again traverse the array arr[] and push all the remaining duplicate elements into the array ans[].
• After completing the above steps, print the array ans[].

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the maximum sum of
// MEX of prefix arrays for any
// arrangement of the given array
void maximumMex(int arr[], int N)
{

    // Stores the final arrangement
    vector<int> ans;

    // Sort the array in increasing order
    sort(arr, arr + N);

    // Iterate over the array arr[]
    for (int i = 0; i < N; i++) {
        if (i == 0 || arr[i] != arr[i - 1])
            ans.push_back(arr[i]);
    }

    // Iterate over the array, arr[]
    // and push the remaining occurrences
    // of the elements into ans[]
    for (int i = 0; i < N; i++) {
        if (i > 0 && arr[i] == arr[i - 1])
            ans.push_back(arr[i]);
    }

    // Print the array, ans[]
    for (int i = 0; i < N; i++)
        cout << ans[i] << " ";
}

// Driver Code
int main()
{

    // Given array
    int arr[] = { 1, 0, 0 };

    // Store the size of the array
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    maximumMex(arr, N);

    return 0;
}

Java

// Java program for the above approach
import java.util.Arrays;

class GFG{

// Function to find the maximum sum of
// MEX of prefix arrays for any
// arrangement of the given array
static void maximumMex(int arr[], int N)
{

    // Stores the final arrangement
    int ans[] = new int[2 * N];

    // Sort the array in increasing order
    Arrays.sort(ans);
    int j = 0;

    // Iterate over the array arr[]
    for(int i = 0; i < N; i++)
    {
        if (i == 0 || arr[i] != arr[i - 1])
        {
            j += 1;
            ans[j] = arr[i];
        }
    }

    // Iterate over the array, arr[]
    // and push the remaining occurrences
    // of the elements into ans[]
    for(int i = 0; i < N; i++)
    {
        if (i > 0 && arr[i] == arr[i - 1])
        {
            j += 1;
            ans[j] = arr[i];
        }
    }

    // Print the array, ans[]
    for(int i = 0; i < N; i++)
        System.out.print(ans[i] + " ");
}

// Driver Code
public static void main (String[] args)
{

    // Given array
    int arr[] = { 1, 0, 0 };

    // Store the size of the array
    int N = arr.length;

    // Function Call
    maximumMex(arr, N);
}
}

// This code is contributed by AnkThon

Python3

# Python3 program for the above approach

# Function to find the maximum sum of
# MEX of prefix arrays for any
# arrangement of the given array
def maximumMex(arr, N):

    # Stores the final arrangement
    ans = []

    # Sort the array in increasing order
    arr = sorted(arr)

    # Iterate over the array arr[]
    for i in range(N):
        if (i == 0 or arr[i] != arr[i - 1]):
            ans.append(arr[i])

    # Iterate over the array, arr[]
    # and push the remaining occurrences
    # of the elements into ans[]
    for i in range(N):
        if (i > 0 and arr[i] == arr[i - 1]):
            ans.append(arr[i])

    # Print the array, ans[]
    for i in range(N):
        print(ans[i], end = " ")

# Driver Code
if __name__ == '__main__':

    # Given array
    arr = [1, 0, 0 ]

    # Store the size of the array
    N = len(arr)

    # Function Call
    maximumMex(arr, N)

# This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;

class GFG{

// Function to find the maximum sum of
// MEX of prefix arrays for any
// arrangement of the given array
static void maximumMex(int []arr, int N)
{

    // Stores the final arrangement
    int []ans = new int[2 * N];

    // Sort the array in increasing order
    Array.Sort(ans);
    int j = 0;

    // Iterate over the array arr[]
    for(int i = 0; i < N; i++)
    {
        if (i == 0 || arr[i] != arr[i - 1])
        {
            j += 1;
            ans[j] = arr[i];
        }
    }

    // Iterate over the array, arr[]
    // and push the remaining occurrences
    // of the elements into ans[]
    for(int i = 0; i < N; i++)
    {
        if (i > 0 && arr[i] == arr[i - 1])
        {
            j += 1;
            ans[j] = arr[i];
        }
    }

    // Print the array, ans[]
    for(int i = 0; i < N; i++)
        Console.Write(ans[i] + " ");
}

// Driver Code
public static void Main (string[] args)
{

    // Given array
    int []arr = { 1, 0, 0 };

    // Store the size of the array
    int N = arr.Length;

    // Function Call
    maximumMex(arr, N);
}
}

// This code is contributed by AnkThon

Javascript

<script>

// Javascript program for the above approach

// Function to find the maximum sum of
// MEX of prefix arrays for any
// arrangement of the given array
function maximumMex(arr, N)
{

    // Stores the final arrangement
    var ans = [];

    // Sort the array in increasing order
    arr.sort();
    var i;
    // Iterate over the array arr[]
    for (i = 0; i < N; i++) {
        if (i == 0 || arr[i] != arr[i - 1])
            ans.push(arr[i]);
    }

    // Iterate over the array, arr[]
    // and push the remaining occurrences
    // of the elements into ans[]
    for (i = 0; i < N; i++) {
        if (i > 0 && arr[i] == arr[i - 1])
            ans.push(arr[i]);
    }

    // Print the array, ans[]
    for (i = 0; i < N; i++)
        document.write(ans[i] + " ");
}

// Driver Code
    // Given array
    var arr = [1, 0, 0];

    // Store the size of the array
    var N = arr.length;

    // Function Call
    maximumMex(arr, N);

</script>

Output

0 1 0

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