Maximum partitions possible of given Array with cost at most K and equal count of odd and even elements

Given two integers N, K, and an array, arr[] of size N, containing an equal number of even and odd elements, and also given that the cost of partitioning the array by making a cut between index i  and i+1 is equal to the abs(arr[i]-arr[i+1]), the task is to find the maximum partitions of the array, such that each partition has an equal number of odd and even elements and has a total cost less than or equal to K.

Examples:

Input: N = 6, K = 4, arr[] = {1, 2, 5, 10, 15, 20}
Output: 1
Explanation:
The only possible way to partition is by making a cut between index 1 and index 2. 
Cost of partition = |arr[1]( =2)- arr[2](= 5)| = 3, which is less than and equal to K 
Arrays after partition: {1, 2} and {5, 10, 15, 20}.

Input: N = 4, K = 10, arr[] = {1, 3, 2, 4}
Output: 0

Approach: The given problem can be solved based on observations that it is always possible to make a valid partition by making a cut between the index i and i+i, if the count of even and odd elements in the prefix of i is equal. Follow the steps to solve the problem.

• Initialize a vector V to store the costs of all possible cuts in the array.
• Also, initialize variables, say odd as 0 and even as 0 which store the count of even and odd elements.
• Traverse the array arr[], using the variable i and perform the following steps:
  • If the current element is odd, then increment odd by 1. Otherwise, increment even by 1.
  • If the value of odd is equal to the value of even, then append the value of |arr[i]–arr[i+1]| to V.
• Sort the vector V in ascending order.
• Initialize an integer variable ans as 1, to store the number of partitions of the array.
• Traverse the vector V, using the variable i and perform the following steps:
  • If the value of V[i] is less than or equal to K then update the value K as K – V[i] and increment ans by 1.
  • Otherwise, break out of the loop.
• Finally, after completing the above steps, print the value of ans as the answer obtained.

Below is the implementation of the above approach:

C++

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

// Function to find the maximum
// partitions of the array with
// equal even and odd elements
// with cost less than k
int maximumcut(int arr[], int N, int K)
{
    // Stores count of odd elements
    int odd = 0;
    // Stores count of even elements
    int even = 0;

    // Stores the cost of partitions
    vector<int> V;

    for (int i = 0; i < N - 1; i++) {
        // Odd element
        if (arr[i] % 2 == 1) {
            odd++;
        }
        // Even element
        else {
            even++;
        }

        // Partition is possible
        if (odd == even) {
            int cost = abs(arr[i] - arr[i + 1]);

            // Append the cost of partition
            V.push_back(cost);
        }
    }
    // Stores the maximum number of
    // partitions
    int ans = 0;

    // Sort the costs in ascending order
    sort(V.begin(), V.end());

    // Traverse the vector V
    for (int i = 0; i < V.size(); i++) {
        // Check if cost is less than K
        if (V[i] <= K) {
            // Update the value of K
            K = K - V[i];

            // Update the value of ans
            ans++;
        }
        else {
            break;
        }
    }
    // Return ans
    return ans;
}

// Driver code
int main()
{
    // Given Input
    int arr[] = {1, 2, 5, 10, 15, 20};
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 4;
    // Function call
    cout << maximumcut(arr, N, K);
    return 0;
}

Java

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

class GFG
{

// Function to find the maximum
// partitions of the array with
// equal even and odd elements
// with cost less than k
public static int maximumcut(int arr[], int N, int K)
{

    // Stores count of odd elements
    int odd = 0;

    // Stores count of even elements
    int even = 0;

    // Stores the cost of partitions
    ArrayList<Integer> V = new ArrayList<Integer>();

    for (int i = 0; i < N - 1; i++) {
        // Odd element
        if (arr[i] % 2 == 1) {
            odd++;
        }
        // Even element
        else {
            even++;
        }

        // Partition is possible
        if (odd == even) {
            int cost = Math.abs(arr[i] - arr[i + 1]);

            // Append the cost of partition
            V.add(cost);
        }
    }
    // Stores the maximum number of
    // partitions
    int ans = 0;

    // Sort the costs in ascending order
    V.sort(null);

    // Traverse the vector V
    for (int i = 0; i < V.size(); i++)
    {

        // Check if cost is less than K
        if (V.get(i) <= K)
        {

            // Update the value of K
            K = K - V.get(i);

            // Update the value of ans
            ans++;
        }
        else {
            break;
        }
    }
    // Return ans
    return ans;
}

// Driver code
public static void  main(String args[])
{
    // Given Input
    int arr[] = {1, 2, 5, 10, 15, 20};
    int N = arr.length;
    int K = 4;
    // Function call
    System.out.println(maximumcut(arr, N, K));
}

}

// This code is contributed by gfgking.

Python3

# Python3 code for the above approach

# Function to find the maximum
# partitions of the array with
# equal even and odd elements
# with cost less than k
def maximumcut(arr, N, K):

    # Stores count of odd elements
    odd = 0

    # Stores count of even elements
    even = 0

    # Stores the cost of partitions
    V = []

    for i in range(0, N - 1, 1):

        # Odd element
        if (arr[i] % 2 == 1):
            odd += 1

        # Even element
        else:
            even += 1

        # Partition is possible
        if (odd == even):
            cost = abs(arr[i] - arr[i + 1])

            # Append the cost of partition
            V.append(cost)

    # Stores the maximum number of
    # partitions
    ans = 0

    # Sort the costs in ascending order
    V.sort()

    # Traverse the vector V
    for i in range(len(V)):

        # Check if cost is less than K
        if (V[i] <= K):

            # Update the value of K
            K = K - V[i]

            # Update the value of ans
            ans += 1

        else:
            break

    # Return ans
    return ans

# Driver code
if __name__ == '__main__':

    # Given Input
    arr = [ 1, 2, 5, 10, 15, 20 ]
    N = len(arr)
    K = 4

    # Function call
    print(maximumcut(arr, N, K))

# This code is contributed by SURENDRA_GANGWAR

C#

// C# code for the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function to find the maximum
// partitions of the array with
// equal even and odd elements
// with cost less than k
static int maximumcut(int []arr, int N, int K)
{

    // Stores count of odd elements
    int odd = 0;

    // Stores count of even elements
    int even = 0;

    // Stores the cost of partitions
    List<int> V = new List<int>();

    for(int i = 0; i < N - 1; i++)
    {

        // Odd element
        if (arr[i] % 2 == 1)
        {
            odd++;
        }

        // Even element
        else
        {
            even++;
        }

        // Partition is possible
        if (odd == even)
        {
            int cost = Math.Abs(arr[i] - arr[i + 1]);

            // Append the cost of partition
            V.Add(cost);
        }
    }

    // Stores the maximum number of
    // partitions
    int ans = 0;

    // Sort the costs in ascending order
    V.Sort();

    // Traverse the vector V
    for(int i = 0; i < V.Count; i++)
    {

        // Check if cost is less than K
        if (V[i] <= K)
        {

            // Update the value of K
            K = K - V[i];

            // Update the value of ans
            ans++;
        }
        else
        {
            break;
        }
    }

    // Return ans
    return ans;
}

// Driver code
public static void Main()
{

    // Given Input
    int []arr = { 1, 2, 5, 10, 15, 20 };
    int N = arr.Length;
    int K = 4;

    // Function call
    Console.Write(maximumcut(arr, N, K));
}
}

// This code is contributed by ipg2016107

Javascript

<script>

        // JavaScript code for the above approach

        // Function to find the maximum
        // partitions of the array with
        // equal even and odd elements
        // with cost less than k
        function maximumcut(arr, N, K) {
            // Stores count of odd elements
            let odd = 0;
            // Stores count of even elements
            let even = 0;

            // Stores the cost of partitions
            var V = [];

            for (let i = 0; i < N - 1; i++) {
                // Odd element
                if (arr[i] % 2 == 1) {
                    odd++;
                }
                // Even element
                else {
                    even++;
                }

                // Partition is possible
                if (odd == even) {
                    let cost = Math.abs(arr[i] - arr[i + 1]);

                    // Append the cost of partition
                    V.push(cost);
                }
            }
            // Stores the maximum number of
            // partitions
            let ans = 0;

            // Sort the costs in ascending order
            V.sort();

            // Traverse the vector V
            for (let i = 0; i < V.length; i++) {
                // Check if cost is less than K
                if (V[i] <= K) {
                    // Update the value of K
                    K = K - V[i];

                    // Update the value of ans
                    ans++;
                }
                else {
                    break;
                }
            }
            // Return ans
            return ans;
        }

        // Driver code

        // Given Input
        let arr = [1, 2, 5, 10, 15, 20];
        let N = arr.length;
        let K = 4;
        // Function call
        document.write(maximumcut(arr, N, K));

// This code is contributed by Potta Lokesh

</script>

Output:

1

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