Minimum Time To Finish The Race - Leetcode Solution

Problem Statement:

You are given a sequence of integers representing heights of a racetrack. You are also given a constant speed that you can cover in one unit of time. You start at the beginning of the racetrack and finish the race when you reach the end of the racetrack. However, you can choose to take some jumps, but when you take a jump, you have to pay the cost equal to the difference between the height of the current position and the height of the position you jump to. You can only jump to heights that are equal to or greater than the current position. You can take jumps any number of times, but your time will increase depending on the cost of each jump. Your task is to find the minimum time required to finish the race.

Solution:

First, we can understand that we need to cover all the positions of the racetrack, so our distance to cover is equal to the length of the racetrack. Let's say that the length of the racetrack is 'n'. So, our target point is 'n-1'.

Now, in order to find the minimum time taken to finish the race, we can use dynamic programming.

For each position of the racetrack, we can maintain an array of minimum time taken to reach that position. Let's say the array is 'dp'.

Initially, set dp[0] = 0, as it takes 0 time to reach the first position.

Now, for each position i, we can move either by walking or by jumping to a position higher or equal to the current position. We can calculate the minimum time taken to reach position i from position j (j<i) in two ways:

1. By Walking: We can move to position i by walking. In this case, the minimum time taken to reach position i will be equal to the minimum time taken to reach position j plus the time taken to move from position j to i at the given speed. So, we can write:

dp[i] = min(dp[i], dp[j] + (i-j)/speed)

Here, the (i-j)/speed is the time taken to move from position j to i, as we are given the speed at which we can cover one unit distance in one unit of time.

2. By Jumping: We can move from position j to position i by taking a jump, but in this case, we have to pay a cost equal to the difference between the height of position i and position j. So, the time taken to move from position j to i will be (i-j)/speed plus the cost of the jump. So, we can write:

dp[i] = min(dp[i], dp[j]+(i-j)/speed+cost)

Here, 'cost' will be the difference in height between position i and position j.

Finally, the minimum time taken to reach the end of the racetrack will be stored in dp[n-1].

Code:

Here is the Python code for the above approach:

def minTime(height: List[int], speed: int) -> int: n = len(height) dp = [float('inf')]*n dp[0] = 0

for i in range(1,n):
    for j in range(i):
        cost = max(0, height[i] - height[j])
        dp[i] = min(dp[i], dp[j]+(i-j)/speed+cost)

return dp[n-1]

Time Complexity:

The time complexity of the above approach is O(n^2), as we are iterating over all the possible pairs of positions. However, we can optimize the solution by using a heap-based approach that will reduce the time complexity to O(n*log(n)).

1