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Maximum number of distinct points on a 2D plane

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  • Last Updated : 23 Feb, 2023
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Given an array X[] of length N where X[i]2 (1 ≤ i ≤ N). Divide each element of X[] into two integers such that the sum of those two integers is equal to the divided element of X[], Formally X[i] = (a + b) for all (1 ≤ i ≤ N), Where (a, b > 0), a and b can be duplicate as well. Mark (a, b) as a point in the 2D plane, the task is to find the maximum number of distinct points on a 2D plane that can be created.

Examples:

Input: N = 4, X[] = {2, 3, 4, 5}
Output: 4
Explanation: Optimal Division of elements takes place as:

  • X[1] = 2, divided into (1, 1)
  • X[2] = 3, divided into (1, 2)
  • X[3] = 4, divided into (1, 3)
  • X[4] = 5, divided into (1, 4)

Total 4 distinct points (1, 1), (1, 2), (1, 3) and (1, 4) are on a 2D plane. Which is the maximum possible for this case.

Input: N = 6, X[] = {2, 2, 4, 4, 2, 6}
Output: 4
Explanation: Optimal Division of elements takes place as:

  • X[1] = 2, divided into (1, 1)
  • X[2] = 2, divided into (1, 1)
  • X[3] = 4, divided into (3, 1)
  • X[4] = 4, divided into (1, 3)
  • X[5] = 2, divided into (1, 1)
  • X[6] = 6, divided into (3, 2)

Total 4 distinct points (1, 1), (1, 3), (3, 1), and (3, 2) are on a 2D plane. Which is the maximum possible for this case. 

Approach: Implement the idea below to solve the problem:

The problem is observation based and can be solved by using HashMap and counting the frequency of elements present in X[]. The concept is first to initialize HashMap with the frequency of elements present in X[]. Then make a Count variable and initialize it to 0, Traverse over HashMap and then increment the count by min(element – 1, frequency) in each iteration.

Steps were taken to solve the problem:

  • Declare the Hashmap of <Integer, Integer> type.
  • Initialize the map with the frequency of elements in X[].
  • Create a Count variable and initialize it to 0.
  • Traverse on HashMap using loop and increment count by min(element-1, frequency) in each operation.

Below is the code to implement the approach:

C++




// C++ code to implement the approach
 
#include <bits/stdc++.h>
using namespace std;
 
 // Function for returning maximum number
 // of distinct points
 long long Max_Points(int n, int x[])
 {
    // Map declared for counting frequency
    map <int, int> mp;
     
        // Loop for initializing Map with frequency
        for (int i = 0; i < n; i++) {
            mp[x[i]]++;
        }
 
        // Variable to store maximum number
        // of distinct points
        long long count = 0;
 
        // Loop for traversing over Map
        for(auto i : mp){
          int element = i.first;
          int value = i.second;
          count += min(element - 1, value);
        }
 
        // Returning the value of count
        return count;
    }
 
int main() {
 
   // Input value of N
   int N = 6;
   int X[] = { 2, 2, 4, 4, 2, 6 };
 
   // Function call
    cout << Max_Points(N, X) << endl;
   
  return 0;
}


Java




// Java code to implement the approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
    // Driver function
    public static void main(String[] args)
        throws java.lang.Exception
    {
        // Input value of N
        int N = 6;
        int[] X = { 2, 2, 4, 4, 2, 6 };
 
        // Function call
        System.out.println(Max_Points(N, X));
    }
 
    // Function for returning maximum number
    // of distinct points
    static long Max_Points(int n, int[] X)
    {
 
        // HashMap declared for counting
        // frequency
        HashMap<Integer, Integer> map = new HashMap<>();
 
        // Loop for initializing HashMap with
        // frequency
        for (int i = 0; i < n; i++) {
            int d = X[i];
            map.put(d, map.get(d) == null ? 1
                                          : map.get(d) + 1);
        }
 
        // Variable to store maximum number
        // of distinct points
        long count = 0;
 
        // Loop for traversing over HashMap
        for (Map.Entry<Integer, Integer> set :
             map.entrySet()) {
            count += Math.min(set.getKey() - 1,
                              set.getValue());
        }
 
        // Returning the value of count
        return count;
    }
}


C#




// C# code to implement the approach
using System;
using System.Collections.Generic; 
 
class GFG {
 
    // Driver function
    static public void Main ()
    {
        // Input value of N
        int N = 6;
        int[] X = { 2, 2, 4, 4, 2, 6 };
 
        // Function call
        Console.WriteLine(Max_Points(N, X));
    }
 
    // Function for returning maximum number
    // of distinct points
    static long Max_Points(int n, int[] X)
    {
 
        // HashMap declared for counting
        // frequency
        Dictionary<int,int> map=new Dictionary<int,int>();
 
        // Loop for initializing HashMap with
        // frequency
        for (int i = 0; i < n; i++) {
            int d = X[i];
            if(!map.ContainsKey(d))
            {
                map.Add(d,1);
            }
            else
               map[d]+=1;
        }
 
        // Variable to store maximum number
        // of distinct points
        long count = 0;
 
        // Loop for traversing over HashMap
        foreach(KeyValuePair<int,int> set in map)
        {
            count += Math.Min(set.Key - 1,set.Value);
        }
         
        // Returning the value of count
        return count;
    }
}
 
// This code is contributed by Pushpesh Raj.


Python




def Max_Points(n, x):
    # Map declared for counting frequency
    mp = {}
      
    # Loop for initializing Map with frequency
    for i in range(n):
        if x[i] in mp:
            mp[x[i]] += 1
        else:
            mp[x[i]] = 1
  
    # Variable to store maximum number
    # of distinct points
    count = 0
  
    # Loop for traversing over Map
    for element, value in mp.items():
        count += min(element - 1, value)
  
    # Returning the value of count
    return count
 
N = 6
X = [2, 2, 4, 4, 2, 6]
print(Max_Points(N,X))


Javascript




// JS code to implement the approach
 
 // Function for returning maximum number
 // of distinct points
 function Max_Points(n, x)
 {
    // Map declared for counting frequency
    let mp = new Map();
     
    // Loop for initializing Map with frequency
    for (let i=0;i<n;i++) {
        if (mp.has(x[i])) {
 
            // If number is present in mp,
            // incrementing it's count by 1
            mp.set(x[i], mp.get(x[i]) + 1);
        }
        else {
 
            // If integer is not present in mp,
            // putting this integer to mp with 1 as it's value
            mp.set(x[i], 1);
        }
    }
 
    // Variable to store maximum number
    // of distinct points
    let count = 0;
 
    // Loop for traversing over Map
    for (let [key, value] of mp.entries())
    {
        count += Math.min(key - 1, value);
    }
 
    // Returning the value of count
    return count;
}
 
// Input value of N
let N = 6;
let X = [ 2, 2, 4, 4, 2, 6 ];
 
// Function call
document.write(Max_Points(N, X));


Output

4

Time Complexity: O(N)
Auxiliary Space: O(N), as HashMap is used for counting frequencies.

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