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.
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.
- 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’.
Below is the implementation of the above approach:
The minimum time required is : 3
Time Complexity: O(n), since there is a while loop that runs for n times.
Auxiliary Space: O(1), since no extra space has been taken.
Approach#2: Using math
The problem can be solved using dynamic programming. We can start from the base case where n=0, and build the solution recursively by considering all possible jumps that can be taken at each step. We can use a memoization table to store the minimum time required to reach a point, so that we can avoid recomputing the same subproblems multiple times.
1. Create a memoization table dp of size n+1 and initialize all values to infinity.
2. Set dp to 0.
For i from 1 to n, compute dp[i] as follows:
If i is a triangular number, set dp[i] to the square root of 2*i.
Otherwise, set dp[i] to dp[i-1] + 1.
For each jump size j from 1 to i, compute the time required to reach i-j*(j+1)/2 and add j to it.
If this time is less than dp[i], update dp[i] with this new minimum time.
3. Return dp[n].
Time complexity: Computing the time required to reach a point takes O(1) time.
The outer loop runs for n iterations, and the inner loop runs for a maximum of sqrt(2*n) iterations.
Therefore, the time complexity of the algorithm is O(n*sqrt(n)).
Space complexity: We need an array of size n+1 to store the memoization table.
Therefore, the space complexity of the algorithm is O(n).
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