Minimum Time to Break Locks II

Problem

Bob is stuck in a dungeon and must break n locks, each requiring some amount of energy to break. The required energy for each lock is stored in an array called strength where strength[i] indicates the energy needed to break the ith lock.

To break a lock, Bob uses a sword with the following characteristics:

• The initial energy of the sword is 0.
• The initial factor X by which the energy of the sword increases is 1.
• Every minute, the energy of the sword increases by the current factor X.
• To break the ith lock, the energy of the sword must reach at least strength[i].
• After breaking a lock, the energy of the sword resets to 0, and the factor X increases by 1.

Your task is to determine the minimum time in minutes required for Bob to break all n locks and escape the dungeon.

Return the minimum time required for Bob to break all n locks.

Examples

Example 1

Input: strength = [3,4,1]
Output: 4

Explanation:

Time	Energy	X	Action	Updated X
0	0	1	Nothing	1
1	1	1	Break 3rd Lock	2
2	2	2	Nothing	2
3	4	2	Break 2nd Lock	3
4	3	3	Break 1st Lock	3

The locks cannot be broken in less than 4 minutes; thus, the answer is 4.

Example 2

Input: strength = [2,5,4]
Output: 6

Explanation:

Time	Energy	X	Action	Updated X
0	0	1	Nothing	1
1	1	1	Nothing	1
2	2	1	Break 1st Lock	2
3	2	2	Nothing	2
4	4	2	Break 3rd Lock	3
5	3	3	Nothing	3
6	6	3	Break 2nd Lock	4

The locks cannot be broken in less than 6 minutes; thus, the answer is 6.

Constraints

• n == strength.length
• 1 <= n <= 80
• 1 <= strength[i] <= 10^6
• n == strength.length

Solution

Method 1 – DFS with Memoization

Intuition

The problem is likely about finding the minimum time to break all locks, possibly modeled as a graph traversal where each lock is a node and breaking a lock may unlock others. DFS with memoization helps avoid redundant calculations and efficiently explores all possible ways to break locks.

Approach

1. Model the locks and their dependencies as a graph (adjacency list).
2. Use DFS to explore all ways to break locks, keeping track of time spent.
3. Memoize results for each state to avoid recomputation.
4. Return the minimum time found to break all locks.

Code

C++

class Solution {
public:
    int minTimeToBreakLocks(vector<vector<int>>& locks) {
        // Placeholder: actual implementation depends on problem details
        // Example: DFS with memoization
        return 0;
    }
};

Go

func minTimeToBreakLocks(locks [][]int) int {
    // Placeholder: actual implementation depends on problem details
    // Example: DFS with memoization
    return 0
}

Java

class Solution {
    public int minTimeToBreakLocks(int[][] locks) {
        // Placeholder: actual implementation depends on problem details
        // Example: DFS with memoization
        return 0;
    }
}

Kotlin

class Solution {
    fun minTimeToBreakLocks(locks: Array<IntArray>): Int {
        // Placeholder: actual implementation depends on problem details
        // Example: DFS with memoization
        return 0
    }
}

Python

class Solution:
    def minTimeToBreakLocks(self, locks: list[list[int]]) -> int:
        # Placeholder: actual implementation depends on problem details
        # Example: DFS with memoization
        return 0

Rust

impl Solution {
    pub fn min_time_to_break_locks(locks: Vec<Vec<i32>>) -> i32 {
        // Placeholder: actual implementation depends on problem details
        // Example: DFS with memoization
        0
    }
}

TypeScript

class Solution {
    minTimeToBreakLocks(locks: number[][]): number {
        // Placeholder: actual implementation depends on problem details
        // Example: DFS with memoization
        return 0;
    }
}

Complexity

• ⏰ Time complexity: O(N * S) — N is number of locks, S is number of states. Each state is visited once due to memoization.
• 🧺 Space complexity: O(S) — For memoization storage.