Maximum number of distinct positive integers that can be used to represent N

Given an integer N, the task is to find the maximum number of distinct positive integers that can be used to represent N.

Examples:

Input: N = 5 
Output: 2 
5 can be represented as 1 + 4, 2 + 3, 3 + 2, 4 + 1 and 5. 
So maximum integers that can be used in the representation are 2.

Input: N = 10 
Output: 4 

Approach: We can always greedily choose distinct integers to be as small as possible to maximize the number of distinct integers that can be used. If we are using the first x natural numbers, let their sum be f(x).
So we need to find a maximum x such that f(x) < = n.

1 + 2 + 3 + … n < = n 
x*(x+1)/2 < = n 
x^2+x-2n < = 0 
We can solve the above equation by using quadratic formula X = (-1 + sqrt(1+8*n))/2. 

Below is the implementation of the above approach:  

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the required count
int count(int n)
{
    return int((-1 + sqrt(1 + 8 * n)) / 2);
}

// Driver code
int main()
{
    int n = 10;

    cout << count(n);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{
    // Function to return the required count
    static int count(int n)
    {
        return (int)(-1 + Math.sqrt(1 + 8 * n)) / 2;

    }

    // Driver code
    public static void main (String[] args)
    {
        int n = 10;

        System.out.println(count(n));
    }
}

// This code is contributed by ihritik

Python3

# Python3 implementation of the approach
from math import sqrt

# Function to return the required count
def count(n) :

    return (-1 + sqrt(1 + 8 * n)) // 2;

# Driver code
if __name__ == "__main__" :

    n = 10;

    print(count(n));

# This code is contributed by AnkitRai01

C#

// C# implementation of approach
using System;

class GFG
{

    // Function to return the required count
    public static int count(int n)
    {
        return (-1 + (int)Math.Sqrt(1 + 8 * n)) / 2;
    }

    // Driver Code
    public static void Main()
    {
        int n = 10;

        Console.Write(count(n));
    }
}

// This code is contributed by Mohit Kumar

Javascript

<script>

// Javascript implementation of the approach

// Function to return the required count
function count(n)
{
    return parseInt((-1 + Math.sqrt(1 + 8 * n)) / 2);
}

// Driver code
var n = 10;

document.write(count(n));

// This code is contributed by rutvik_56

</script>

Output:

4

Time Complexity: O(1)

Auxiliary Space: O(1)