Maximize the number of indices such that element is greater than element to its left

Given an array arr[] of N integers, the task is to maximize the number of indices such that an element is greater than the element to its left, i.e. arr[i+1] > arr[i] after rearranging the array.
Examples: 
 

Input: arr[] = {200, 100, 100, 200} 
Output: 2 
Explanation: 
By arranging the array in following way we have: arr[] = {100, 200, 100, 200} 
The possible indices are 0 and 2 such that: 
arr[1] > arr[0] (200 > 100) 
arr[3] > arr[2] (200 > 100)
Input: arr[] = {1, 8, 5, 9, 8, 8, 7, 7, 5, 7, 7} 
Output: 7 
Explanation: 
By arranging the array in following way we have: arr[] = {1, 5, 7, 8, 9, 5, 7, 8, 7, 8, 4} 
The possible indices are 0, 1, 2, 3, 5, 6 and 7 such that: 
arr[1] > arr[0] (5 > 1) 
arr[2] > arr[1] (7 > 5) 
arr[3] > arr[2] (8 > 7) 
arr[4] > arr[3] (9 > 8) 
arr[6] > arr[5] (7 > 5) 
arr[7] > arr[6] (8 > 7) 
arr[8] > arr[7] (8 > 7) 
 

Approach: This problem can be solved using Greedy Approach. Below are the steps: 
 

1. To get the maximum number of indices(say i) such that arr[i+1] > arr[i], arrange the elements of the arr[] such that set of all unique element occurs first, then next set of unique elements occurs after the first set till all the elements are arranged. 
   For Example: 
    

Let arr[] = {1, 8, 5, 9, 8, 8, 7, 7, 5, 7, 7} 
1st Set = {1, 5, 7, 8, 9} 
2nd Set = {5, 7, 8} 
3rd Set = {7, 8} 
4th Set = {4}
Now the new array will be: 
arr[] = {1, 5, 7, 8, 9, 5, 7, 8, 7, 8, 4} 
 

1. 
    
2. After the above arrangement, the element with the higher value will not be a part of the given condition as it is followed by a number smaller than itself.
3. Therefore the total number of pairs satisfying the given condition can be given by: 
    

total_pairs = (number_of_elements - highest_frequency_of_a_number) 
 

Below is the implementation of the above approach: 
 

// C++ program of the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the maximum pairs
// such that arr[i+1] > arr[i]
void countPairs(int arr[], int N)
{

    // To store the frequency of the
    // element in arr[]
    unordered_map<int, int> M;

    // Store the frequency in map M
    for (int i = 0; i < N; i++) {
        M[arr[i]]++;
    }

    int maxFreq = 0;

    // To find the maximum frequency
    // store in map M
    for (auto& it : M) {
        maxFreq = max(maxFreq,
                      it.second);
    }

    // Print the maximum number of
    // possible pairs
    cout << N - maxFreq << endl;
}

// Driver Code
int main()
{

    int arr[] = { 1, 8, 5, 9, 8, 8, 7,
                  7, 5, 7, 7 };
    int N = sizeof(arr) / sizeof(arr[0]);

    countPairs(arr, N);
    return 0;
}

// Java program of the above approach
import java.util.*;

class GFG{
 
// Function to find the maximum pairs
// such that arr[i+1] > arr[i]
static void countPairs(int arr[], int N)
{
 
    // To store the frequency of the
    // element in arr[]
    HashMap<Integer,Integer> mp = new HashMap<Integer,Integer>();
 
    // Store the frequency in map M
    for (int i = 0; i < N; i++) {
        if(mp.containsKey(arr[i])){
            mp.put(arr[i], mp.get(arr[i])+1);
        }else{
            mp.put(arr[i], 1);
    }
    }
 
    int maxFreq = 0;
 
    // To find the maximum frequency
    // store in map M
    for (Map.Entry<Integer,Integer> it : mp.entrySet()) {
        maxFreq = Math.max(maxFreq,
                      it.getValue());
    }
 
    // Print the maximum number of
    // possible pairs
    System.out.print(N - maxFreq +"\n");
}
 
// Driver Code
public static void main(String[] args)
{
 
    int arr[] = { 1, 8, 5, 9, 8, 8, 7,
                  7, 5, 7, 7 };
    int N = arr.length;
 
    countPairs(arr, N);
}
}

// This code is contributed by Rajput-Ji

# Python3 implementation of the above approach

# Function to find the maximum pairs
# such that arr[i + 1] > arr[i]
def countPairs(arr, N) :

    # To store the frequency of the
    # element in arr[]
    M = dict.fromkeys(arr, 0);

    # Store the frequency in map M
    for i in range(N) :
        M[arr[i]] += 1;

    maxFreq = 0;

    # To find the maximum frequency
    # store in map M
    for it in M.values() :
        maxFreq = max(maxFreq,it);

    # Print the maximum number of
    # possible pairs
    print(N - maxFreq);

# Driver Code
if __name__ == "__main__" :

    arr = [ 1, 8, 5, 9, 8, 8, 7, 7, 5, 7, 7 ];
    N = len(arr);

    countPairs(arr, N);
    
    # This code is contributed by AnkitRai01

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

class GFG{
  
// Function to find the maximum pairs
// such that arr[i+1] > arr[i]
static void countPairs(int []arr, int N)
{
  
    // To store the frequency of the
    // element in []arr
    Dictionary<int,int> mp = new Dictionary<int,int>();
  
    // Store the frequency in map M
    for (int i = 0; i < N; i++) {
        if(mp.ContainsKey(arr[i])){
            mp[arr[i]] = mp[arr[i]]+1;
        }else{
            mp.Add(arr[i], 1);
    }
    }
  
    int maxFreq = 0;
  
    // To find the maximum frequency
    // store in map M
    foreach (KeyValuePair<int,int> it in mp) {
        maxFreq = Math.Max(maxFreq,
                      it.Value);
    }
  
    // Print the maximum number of
    // possible pairs
    Console.Write(N - maxFreq +"\n");
}
  
// Driver Code
public static void Main(String[] args)
{
  
    int []arr = { 1, 8, 5, 9, 8, 8, 7,
                  7, 5, 7, 7 };
    int N = arr.Length;
  
    countPairs(arr, N);
}
}
 
// This code is contributed by Rajput-Ji

<script>

// Js program of the above approach

// Function to find the maximum pairs
// such that arr[i+1] > arr[i]
function countPairs( arr,  N){
    // To store the frequency of the
    // element in arr[]
    let M = new Map();

    // Store the frequency in map M
    for (let i = 0; i < N; i++) {
        if(M[arr[i]])
        M[arr[i]]++;
        else
        M[arr[i]] = 1
    }

    let maxFreq = 0;

    // To find the maximum frequency
    // store in map M
    for (let it in M) {
        maxFreq = Math.max(maxFreq,
                      M[it]);
    }

    // Print the maximum number of
    // possible pairs
    document.write(N - maxFreq,'<br>');
}

// Driver Code
let a = [ 1, 8, 5, 9, 8, 8, 7,
                  7, 5, 7, 7 ];
let N = a.length;

    countPairs(a, N);


</script>

7

Time Complexity: O(N), where N is the number of element in the array. 
Auxiliary Space: O(N), where N is the number of element in the array.