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

0 0 2 minutes read

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:

Time Complexity: O(1)

Auxiliary Space: O(1)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!