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; ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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 ` |

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**Output:**

4

**Time Complexity:** O(1)

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