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Maximize the number of indices such that element is greater than element to its left
  • Last Updated : 30 Jul, 2020

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}

  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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

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.

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