LeetCode 3385: Minimum Time to Break Locks II Solution
Master LeetCode problem 3385 (Minimum Time to Break Locks II), 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.
3385. Minimum Time to Break Locks II
Problem Explanation
Explanation
To solve this problem, we can use dynamic programming. The idea is to find the minimum time needed to break locks at each position by considering all possible previous positions. We will maintain a 2D dp array where dp[i][j] represents the minimum time to break locks from position 0 to i using j turns.
At each position i, we iterate over all possible previous positions j and calculate the cost of breaking locks from position j to i using one more turn. We update dp[i][j] with the minimum cost among all these possibilities.
Finally, the answer will be the minimum time to break locks at the last position using all available turns.
Time Complexity
The time complexity of this approach is O(n^2 * m), where n is the number of locks and m is the number of turns.
Space Complexity
The space complexity of this approach is O(n * m).
Solution Code
class Solution {
public int minTimeToOpenLocks(int[] locks, int turns) {
int n = locks.length;
int[][] dp = new int[n][turns + 1];
for (int i = 0; i < n; i++) {
dp[i][0] = Integer.MAX_VALUE;
}
for (int j = 1; j <= turns; j++) {
for (int i = 0; i < n; i++) {
dp[i][j] = Integer.MAX_VALUE;
for (int k = 0; k < n; k++) {
int cost = Math.abs(locks[i] - locks[k]) + 1;
if (i >= k) {
cost = Math.min(cost, Math.abs(locks[i] - locks[k] + n) + 1);
} else {
cost = Math.min(cost, Math.abs(locks[k] - locks[i] + n) + 1);
}
dp[i][j] = Math.min(dp[i][j], cost + dp[k][j - 1]);
}
}
}
int result = Integer.MAX_VALUE;
for (int i = 0; i < n; i++) {
result = Math.min(result, dp[i][turns]);
}
return result;
}
}Try It Yourself
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