Maximizing the elements with a[i+1] > a[i]

Given an array of N integers, rearrange the array elements such that the next array element is greater than the previous element (A_{(i+1)} > A_i).

Examples:

Input : arr[] = {20, 30, 10, 50, 40}
Output : 4
We rearrange the array as 10, 20, 30, 40, 50. As 20 > 10, 30 > 20, 40 > 30, 50 > 40, so we get 4 indices i such that A_{(i+1)} > A_i.



Input : arr[] = {200, 100, 100, 200}
Output : 2
We get optimal arrangement as 100 200 100 200.

If all elements are distinct, then answer is simply n-1 where n is the number of elements in the array. If there are repeating elements, then answer is n – max_freq.

Java

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import java.util.*;
  
class GFG {
  
    // returns the number of positions where A(i + 1) is
    // greater than A(i) after rearrangement of the array
    static int countMaxPos(int[] arr)
    {
        int n = arr.length;
  
        // Creating a HashMap containing char
        // as a key and occurrences as  a value
        HashMap<Integer, Integer> map
            = new HashMap<Integer, Integer>();
        for (int x : arr) {
            if (map.containsKey(x))
                map.put(x, map.get(x) + 1);
            else
                map.put(x, 1);
        }
  
        // Find the maximum frequency
        int max_freq = 0;
        for (Map.Entry entry : map.entrySet())
            max_freq = Math.max(max_freq, (int)entry.getValue());
  
        return n - max_freq;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int[] arr = { 20, 30, 10, 50, 40 };
        System.out.println(countMaxPos(arr));
    }
}

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Python3





# Python3 implementation of the above approach


# Returns the number of positions where

# A(i + 1) is greater than A(i) after

# rearrangement of the array

def countMaxPos(arr): 


        
          
          
          
        

            

	n = len(arr) 


	# Creating a HashMap containing char

	# as a key and occurrences as a value

	Map = {}

	for x in arr:

		if x in Map:

			Map[x] += 1

		else:

			Map[x] = 1


	# Find the maximum frequency

	max_freq = 0

	for entry in Map:

		max_freq = max(max_freq, Map[entry]) 


	return n – max_freq 


# Driver code

if __name__ == “__main__”:


	arr = [20, 30, 10, 50, 40]

	print(countMaxPos(arr)) 


# This code is contributed by Rituraj Jain





Output:


4






          
          
          
          

            

    

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