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.
Input: 6 Output: 3 At time 1, jump from x = 0 to x = 1 (x = x + 1) At time 2, jump from x = 1 to x = 3 (x = x + 2) At time 3, jump from x = 3 to x = 6 (x = x + 3) So, minimum required time is 3. 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.
(t * (t + 1)) / 2 = Xthen answer is ‘t’.
- Else if
(t * (t + 1)) / 2 > X, then we find
(t * (t + 1)) / 2 – Xand remove this number from the sequence
[1, 2, 3, ..., t]. The resulting sequence sums up to ‘X’.
Below is the implementation of the above approach:
The minimum time required is : 3
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