LeetCode 2403: Minimum Time to Kill All Monsters
LeetCode 2403 Solution 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
dpof sizen x 2wherenis the number of monsters anddp[i][j]represents the minimum time to kill monsters at positioniwhile facing directionj(either left or right). - Initialize
dp[0][0] = dp[0][1] = 0and set all other values toINF. - 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])
- For each position
- 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.
:
LeetCode 2403 Solutions in Java, C++, Python
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]);
}
}Interactive Code Editor for LeetCode 2403
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