LeetCode 2403: Minimum Time to Kill All Monsters Solution

Master LeetCode problem 2403 (Minimum Time to Kill All Monsters), 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.

2403. Minimum Time to Kill All Monsters

Hard

Array Dynamic Programming Bit Manipulation Bitmask

Problem Explanation

Explanation:

To solve this problem, we can use a dynamic programming approach. We can maintain a 2D array to represent the minimum time to kill monsters at each position while considering the left and right direction separately. We initialize this array with a value of INF (infinity) for each position except the first position which is initialized with 0. Then, we iterate through each position and calculate the minimum time to kill all monsters by considering the time taken to kill monsters in the left and right directions. The final answer is the minimum time to kill all monsters at the last position.

Algorithm:

Initialize a 2D array dp of size n x 2 where n is the number of monsters and dp[i][j] represents the minimum time to kill monsters at position i while facing direction j (either left or right).
Initialize dp[0][0] = dp[0][1] = 0 and set all other values to INF.
Iterate through each position from 1 to n-1:
- For each position i, calculate the minimum time to kill monsters facing left and right directions separately:
  - dp[i][0] = min(dp[i-1][0] + A[i], dp[i-1][1] + 2*A[i])
  - dp[i][1] = min(dp[i-1][1] + A[i], dp[i-1][0] + 2*A[i])
The final answer is min(dp[n-1][0], dp[n-1][1]).

Time Complexity:

The time complexity of this approach is O(n) where n is the number of monsters.

Space Complexity:

The space complexity is O(n) for the 2D array dp.

Solution Code

class Solution {
    public int minTimeToKillAllMonsters(int[] A) {
        int n = A.length;
        int[][] dp = new int[n][2];
        dp[0][0] = dp[0][1] = 0;
        
        for (int i = 1; i < n; i++) {
            dp[i][0] = Math.min(dp[i-1][0] + A[i], dp[i-1][1] + 2*A[i]);
            dp[i][1] = Math.min(dp[i-1][1] + A[i], dp[i-1][0] + 2*A[i]);
        }
        
        return Math.min(dp[n-1][0], dp[n-1][1]);
    }
}

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