Minimize consecutive removals of elements of the same type to empty given array

Given an array A[ ] consisting of N positive integers, such that each array element A_i denotes the count of elements of i^th type, the task is to minimize the number of deletions of consecutive elements of the same type to empty the given array.

Examples: 

Input : A[ ] = {3, 3, 2} 
Output: 0 
Explanation: The array consists of 3, 3, 2 elements of 1^st, 2^nd and 3^rd type respectively. By removing the array elements in the order {2, 1, 2, 3, 1, 3, 1}, a sequence is obtained in which no two consecutive elements are removed which are of the same type. Therefore, the count is 0.

Input: A [ ] = {1, 4, 1} 
Output: 1 
Explanation: The array consists of 1, 4, 1 elements of 1^st, 2^nd and 3^rd type respectively. By removing the array elements in the order { 2, 3, 2, 2, 1, 2}, consecutive deletions of same type elements is reduced to 1. Therefore, the count is 1. 

Approach: The problem can be solved by Greedy Approach. Follow the steps below to solve the problem: 

• Sort the given array.
• Calculate the sum of the elements of the array.
• Find the largest element present in the array, which is A_N-1..
• If the difference between sum and maximum element is greater than the maximum element, print 0 as the required answer.
• Otherwise, print 2* max - sum -1 as the required answer.

Below is the implementation of the above approach. 

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

// Function to count minimum consecutive
// removals of elements of the same type
void minRemovals(int A[], int N)
{

    // Sort the array
    sort(A, A + N);

    // Stores the maximum element
    // present in the array
    int mx = A[N - 1];

    // Stores sum of the array
    int sum = 1;

    // Calculate sum of the array
    for (int i = 0; i < N; i++) {
        sum += A[i];
    }

    if (sum - mx >= mx) {
        cout << 0 << "\n";
    }
    else {
        cout << 2 * mx - sum << "\n";
    }
}

// Driver Code
int main()
{
    int A[] = { 3, 3, 2 };
    int N = sizeof(A) / sizeof(A[0]);

    // Function call
    minRemovals(A, N);
    return 0;
}

// Java implementation of the above approach
import java.util.Arrays;

class GFG 
{
    
    // Function to count minimum consecutive
    // removals of elements of the same type
    static void minRemovals(int []A, int N)
    {
    
        // Sort the array
        Arrays.sort(A);
    
        // Stores the maximum element
        // present in the array
        int mx = A[N - 1];
    
        // Stores sum of the array
        int sum = 1;
    
        // Calculate sum of the array
        for (int i = 0; i < N; i++)
        {
            sum += A[i];
        }
    
        if (sum - mx >= mx) {
            System.out.println(0);
        }
        else {
            System.out.println(2 * mx - sum);
        }
    }
    
    // Driver Code
    public static void main(String[] args) {
        
        int []A = { 3, 3, 2 };
        int N = A.length;
    
        // Function call
        minRemovals(A, N);
        
    }
}

// This code is contributed by AnkThon

# Python3 implementation of the above approach

# Function to count minimum consecutive 
# removals of elements of the same type 
def minRemovals(A, N):
    
    # Sort the array
    A.sort()

    # Stores the maximum element 
    # present in the array 
    mx = A[N - 1]

    # stores the sum of array
    sum = 1

    # Calculate sum of array
    for i in range(0, N):
        sum += A[i]

    if ((sum - mx) >= mx):
        print(0, end = "")
    else:
        print(2 * mx - sum, end = "")

# Driver Code  
if __name__ == "__main__" :
    
    A = [ 3, 3, 2 ]
    N = len(A)
    
    # Function call 
    minRemovals(A, N) 

# This code is contributed by Virusbuddah

// C# implementation of the above approach
using System;

class GFG 
{
    
    // Function to count minimum consecutive
    // removals of elements of the same type
    static void minRemovals(int []A, int N)
    {
    
        // Sort the array
        Array.Sort(A);
    
        // Stores the maximum element
        // present in the array
        int mx = A[N - 1];
    
        // Stores sum of the array
        int sum = 1;
    
        // Calculate sum of the array
        for (int i = 0; i < N; i++)
        {
            sum += A[i];
        }
    
        if (sum - mx >= mx) 
        {
            Console.WriteLine(0);
        }
        else
        {
            Console.WriteLine(2 * mx - sum);
        }
    }
    
    // Driver Code
    public static void Main(String[] args) 
    {
        int []A = { 3, 3, 2 };
        int N = A.Length;
    
        // Function call
        minRemovals(A, N);      
    }
}

// This code is contributed by shikhasingrajput 

<script>

// JavaScript program for above approach

    // Function to count minimum consecutive
    // removals of elements of the same type
    function minRemovals(A, N)
    {
     
        // Sort the array
        A.sort();
     
        // Stores the maximum element
        // present in the array
        let mx = A[N - 1];
     
        // Stores sum of the array
        let sum = 1;
     
        // Calculate sum of the array
        for (let i = 0; i < N; i++)
        {
            sum += A[i];
        }
     
        if (sum - mx >= mx) {
             document.write(0);
        }
        else {
             document.write(2 * mx - sum);
        }
    }

// Driver Code

     let A = [3, 3, 2 ];
        let N = A.length;
     
        // Function call
        minRemovals(A, N);
     
     // This code is contributed by avijitmondal1998.
</script>

0

Time Complexity: O(N)
Auxiliary Space: O(1)