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














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# 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 



















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C#














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// C# implementation of the approach 


using System; 


using System.Collections.Generic;              


  


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 


        Dictionary<int,  


                   int> map = new Dictionary<int, 


                                             int>(); 


        foreach (int x in arr)  


        { 


            if (map.ContainsKey(x)) 


                map[x] = map[x] + 1; 


            else


                map.Add(x, 1); 


        } 


  


        // Find the maximum frequency 


        int max_freq = 0; 


        foreach(KeyValuePair<int, int> entry in map) 


            max_freq = Math.Max(max_freq, entry.Value); 


  


        return n - max_freq; 


    } 


  


    // Driver code 


    public static void Main(String[] args) 


    { 


        int[] arr = { 20, 30, 10, 50, 40 }; 


        Console.WriteLine(countMaxPos(arr)); 


    } 


} 


  


// This code is contributed by 29AjayKumar 



















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Output:


4






          
          
          
          

            

    

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