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++ 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 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 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# 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 |
<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> |
4
Time Complexity: O(1)
Auxiliary Space: O(1)