LeetCode 818: Race Car Solution

Master LeetCode problem 818 (Race Car), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

818. Race Car

Hard

Dynamic Programming

Problem Explanation

Explanation:

To solve this problem, we can use a dynamic programming approach. We can create a dp array to store the minimum number of instructions needed to reach each position from the starting position 0. We iterate through each position up to the target and for each position, we check two scenarios:

Keep accelerating to reach the target from the current position in the forward direction.
Reverse at some point to reach the target from the current position in the reverse direction.

We update the dp array with the minimum of these two scenarios. Finally, we return the value at the target position in the dp array as the result.

Algorithm:

Initialize a dp array of size target + 1 to store the minimum instructions needed to reach each position from 0.
Iterate over positions from 1 to target:
- Calculate the total distance covered to reach the target from the current position by accelerating only.
- Calculate the minimum number of instructions needed to reach the target from the current position by reversing at some point.
- Update the dp array at the current position with the minimum of the above two scenarios.
Return the value at the target position in the dp array as the result.

Time Complexity: O(target * log(target))
Space Complexity: O(target)

Solution Code

class Solution {
    public int racecar(int target) {
        int[] dp = new int[target + 1];
        Arrays.fill(dp, Integer.MAX_VALUE);
        dp[0] = 0;

        for (int pos = 1; pos <= target; pos++) {
            int k = 32 - Integer.numberOfLeadingZeros(pos);
            if (pos == (1 << k) - 1) {
                dp[pos] = k;
                continue;
            }

            for (int i = 1; i < k; i++) {
                dp[pos] = Math.min(dp[pos], dp[pos - (1 << (k - 1)) + (1 << i)] + k - 1 + i + 1);
            }

            dp[pos] = Math.min(dp[pos], k + 1 + dp[(1 << k) - 1 - pos]);
        }

        return dp[target];
    }
}

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