Minimum time to reach a point with +t and -t moves at time t

https://www.geeksforgeeks.org/minimum-time-to-reach-a-point-with-t-and-t-moves-at-time-t/

Given a positive coordinate ‘X’ and you are at coordinate ‘0’, the task is to find the minimum time required to get to coordinate ‘X’ with the following move :
At time ‘t’, you can either stay at the same position or take a jump of length exactly ‘t’ either to the left or to the right. In other words, you can be at coordinate ‘x – t’, ‘x’ or ‘x + t’ at time ‘t’ where ‘x’ is the current position.

Examples:

Input: 9
Output: 4
At time 1, do not jump i.e x = 0 
At time 2, jump from x = 0 to x = 2 (x = x + 2)
At time 3, jump from x = 2 to x = 5 (x = x + 3)
At time 4, jump from x = 5 to x = 9 (x = x + 4)
So, minimum required time is 4.

Approach: The following greedy strategy works:
We just find the minimum ‘t’ such that 1 + 2 + 3 + ... + t >= X.

• If (t * (t + 1)) / 2 = X then answer is ‘t’.
• Else if (t * (t + 1)) / 2 > X, then we find (t * (t + 1)) / 2 – X and remove this number from the sequence [1, 2, 3, ..., t]. The resulting sequence sums up to ‘X’.

    static long cal_minimum_time(long X)

    {

  

        // Stores the minimum time

        long t = 0;

        long sum = 0;

  

        while (sum < X) {

  

            // increment 't' by 1

            t++;

  

            // update the sum

            sum = sum + t;

        }

  

        return t;

    }