Delete odd and even numbers at alternate step such that sum of remaining elements is minimized

Given an array arr[] of N elements. At any step, we can delete a number of a different parity from the just previous step, i.e., if, at the previous step, an odd number was deleted then delete an even number in the current step or vice versa. 
It is allowed to start by deleting any number. Deletion is possible till we can delete numbers of different parity at every step. The task is to find the minimum possible sum of the elements left at the end.

Examples: 

Input: arr[] = {1, 5, 7, 8, 2} 
Output: 0 
Delete elements in the order 1, 2, 5, 8 and finally 7. 
There are multiple ways of deletion, 
resulting in the same minimized sum. 

Input: arr[] = {2, 2, 2, 2} 
Output: 6 
Delete 2 in first step. 
Cannot delete any number, since there are no odd numbers left. 
Hence, the leftover elements sum is 6. 

Approach: The following ways can be followed to solve the above problem: 

• Count the number of odd and even elements and store in vectors v1 and v2.
• Check if the number of odd and even elements are the same or differ by 1, then we can perform N steps, resulting in leftover sum as 0.
• If the size differs by more than 1, then only there will be leftover elements.
• In order to minimize the leftover sum of elements, we select the larger elements first.
• Hence, the sum of the X smaller elements will be the answer, where X is v2.size() - v1.size() - 1 or v1.size() - v2.size() - 1 depending on the count of even and odd elements.

Below is the implementation of the above approach: 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the minimized sum
int MinimizeleftOverSum(int a[], int n)
{
    vector<int> v1, v2;
    for (int i = 0; i < n; i++) {

        if (a[i] % 2)
            v1.push_back(a[i]);
        else
            v2.push_back(a[i]);
    }

    // If more odd elements
    if (v1.size() > v2.size()) {

        // Sort the elements
        sort(v1.begin(), v1.end());
        sort(v2.begin(), v2.end());

        // Left-over elements
        int x = v1.size() - v2.size() - 1;

        int sum = 0;
        int i = 0;

        // Find the sum of leftover elements
        while (i < x) {
            sum += v1[i++];
        }

        // Return the sum
        return sum;
    }

    // If more even elements
    else if (v2.size() > v1.size()) {

        // Sort the elements
        sort(v1.begin(), v1.end());
        sort(v2.begin(), v2.end());

        // Left-over elements
        int x = v2.size() - v1.size() - 1;

        int sum = 0;
        int i = 0;

        // Find the sum of leftover elements
        while (i < x) {
            sum += v2[i++];
        }

        // Return the sum
        return sum;
    }

    // If same elements
    else
        return 0;
}

// Driver code
int main()
{

    int a[] = { 2, 2, 2, 2 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << MinimizeleftOverSum(a, n);

    return 0;
}

// Java implementation of the approach
import java.util.*;

class GFG
{

// Function to find the minimized sum
static int MinimizeleftOverSum(int a[], int n)
{
    Vector<Integer> v1 = new Vector<Integer>(), 
                    v2 = new Vector<Integer>();
    for (int i = 0; i < n; i++)
    {

        if (a[i] % 2 == 1)
            v1.add(a[i]);
        else
            v2.add(a[i]);
    }

    // If more odd elements
    if (v1.size() > v2.size())
    {

        // Sort the elements
        Collections.sort(v1);
        Collections.sort(v2);

        // Left-over elements
        int x = v1.size() - v2.size() - 1;

        int sum = 0;
        int i = 0;

        // Find the sum of leftover elements
        while (i < x)
        {
            sum += v1.get(i++);
        }

        // Return the sum
        return sum;
    }

    // If more even elements
    else if (v2.size() > v1.size())
    {

        // Sort the elements
        Collections.sort(v1);
        Collections.sort(v2);

        // Left-over elements
        int x = v2.size() - v1.size() - 1;

        int sum = 0;
        int i = 0;

        // Find the sum of leftover elements
        while (i < x)
        {
            sum += v2.get(i++);
        }

        // Return the sum
        return sum;
    }

    // If same elements
    else
        return 0;
}

// Driver code
public static void main(String[] args) 
{
    int a[] = { 2, 2, 2, 2 };
    int n = a.length;
    System.out.println(MinimizeleftOverSum(a, n));
}
}

// This code is contributed by Princi Singh

# Python3 implementation of the approach 

# Function to find the minimized sum 
def MinimizeleftOverSum(a, n) :
    
    v1, v2 = [], []; 
    for i in range(n) : 
        
        if (a[i] % 2) :
            v1.append(a[i]); 
        else :
            v2.append(a[i]); 
    
    # If more odd elements 
    if (len(v1) > len(v2)) : 

        # Sort the elements 
        v1.sort(); 
        v2.sort(); 

        # Left-over elements 
        x = len(v1) - len(v2) - 1; 

        sum = 0; 
        i = 0; 

        # Find the sum of leftover elements 
        while (i < x) : 
            sum += v1[i];
            i += 1

        # Return the sum 
        return sum; 
    
    # If more even elements 
    elif (len(v2) > len(v1)) :

        # Sort the elements 
        v1.sort(); 
        v2.sort(); 

        # Left-over elements 
        x = len(v2) - len(v1) - 1; 

        sum = 0; 
        i = 0; 

        # Find the sum of leftover elements 
        while (i < x) :
            sum += v2[i];
            i += 1
        
        # Return the sum 
        return sum; 
    
    # If same elements 
    else :
        return 0; 

# Driver code 
if __name__ == "__main__" : 

    a = [ 2, 2, 2, 2 ]; 
    n = len(a); 
    
    print(MinimizeleftOverSum(a, n)); 

# This code is contributed by Ryuga

// C# implementation of the approach
using System;
using System.Collections.Generic;

class GFG
{

// Function to find the minimized sum
static int MinimizeleftOverSum(int []a, 
                               int n)
{
    List<int> v1 = new List<int>(), 
              v2 = new List<int>();
    for (int i = 0; i < n; i++)
    {

        if (a[i] % 2 == 1)
            v1.Add(a[i]);
        else
            v2.Add(a[i]);
    }

    // If more odd elements
    if (v1.Count > v2.Count)
    {

        // Sort the elements
        v1.Sort();
        v2.Sort();

        // Left-over elements
        int x = v1.Count - v2.Count - 1;

        int sum = 0;
        int i = 0;

        // Find the sum of leftover elements
        while (i < x)
        {
            sum += v1[i++];
        }

        // Return the sum
        return sum;
    }

    // If more even elements
    else if (v2.Count > v1.Count)
    {

        // Sort the elements
        v1.Sort();
        v2.Sort();

        // Left-over elements
        int x = v2.Count - v1.Count - 1;

        int sum = 0;
        int i = 0;

        // Find the sum of leftover elements
        while (i < x)
        {
            sum += v2[i++];
        }

        // Return the sum
        return sum;
    }

    // If same elements
    else
        return 0;
}

// Driver code
public static void Main(String[] args) 
{
    int []a = { 2, 2, 2, 2 };
    int n = a.Length;
    Console.WriteLine(MinimizeleftOverSum(a, n));
}
}

// This code is contributed by PrinciRaj1992 

<script>
// javascript implementation of the approach

// Function to find the minimized sum
function MinimizeleftOverSum(a , n)
{
   var v1 = [], v2 =[];
    for (i = 0; i < n; i++)
    {

        if (a[i] % 2 == 1)
            v1.push(a[i]);
        else
            v2.push(a[i]);
    }

    // If more odd elements
    if (v1.length > v2.length)
    {

        // Sort the elements
        v1.sort();
        v2.sort();

        // Left-over elements
        var x = v1.length - v2.length - 1;

        var sum = 0;
        var i = 0;

        // Find the sum of leftover elements
        while (i < x)
        {
            sum += v1[i++];
        }

        // Return the sum
        return sum;
    }

    // If more even elements
    else if (v2.length > v1.length)
    {

        // Sort the elements
        v1.sort();
        v2.sort();

        // Left-over elements
        var x = v2.length - v1.length - 1;

        var sum = 0;
        var i = 0;

        // Find the sum of leftover elements
        while (i < x)
        {
            sum += v2[i++];
        }

        // Return the sum
        return sum;
    }

    // If same elements
    else
        return 0;
}

// Driver code
    var a = [ 2, 2, 2, 2 ];
    var n = a.length;
    document.write(MinimizeleftOverSum(a, n));

// This code is contributed by Rajput-Ji 
</script>

6

Time Complexity: O(n * log n), as in the worst case we will be using an inbuilt sort function to sort an array of size n. Where N is the number of elements in the array.
Auxiliary Space: O(n), as we are using extra space for the array v1 and v2. Where N is the number of elements in the array.