Minimum Time to Break Locks I
MedPrompt
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
xby 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
ithlock, the energy of the sword must reach at leaststrength[i]. - After breaking a lock, the energy of the sword resets to 0, and the factor
xincreases by a given valuek.
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.
Example 1:
Input: strength = [3,4,1], k = 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], k = 2
Output: 5
Explanation:
| Time | Energy | x | Action | Updated x |
|---|---|---|---|---|
| 0 | 0 | 1 | Nothing | 1 |
| 1 | 1 | 1 | Nothing | 1 |
| 2 | 2 | 1 | Break 1st Lock | 3 |
| 3 | 3 | 3 | Nothing | 3 |
| 4 | 6 | 3 | Break 2nd Lock | 5 |
| 5 | 5 | 5 | Break 3rd Lock | 7 |
The locks cannot be broken in less than 5 minutes; thus, the answer is 5.
Constraints:
n == strength.length1 <= n <= 81 <= K <= 101 <= strength[i] <= 106
Approaches
2 approaches with complexity analysis and trade-offs.
This approach explores every possible order of breaking the locks. Since the total time depends on the sequence in which the locks are broken, we can generate all permutations of the locks, calculate the total time for each permutation, and find the minimum among them. Given the small constraint on n (up to 8), this method is feasible, although inefficient.
Algorithm
- Initialize
minTimeto a very large value. - Create a list of lock strengths from the input array.
- Implement a recursive function, say
generatePermutations(index, currentPermutation), to explore all possible orderings of locks. - The base case for the recursion is when
indexreaches the number of locksn. This signifies that a complete permutation has been formed. - In the base case, calculate the total time for the
currentPermutation: a. InitializetotalTime = 0andfactor = 1. b. Iterate through the locks in thecurrentPermutation. For each lock strengths: i. Calculate the time required:time = (s + factor - 1) / factor(this is equivalent toceil(s / factor)using integer arithmetic). ii. AddtimetototalTime. iii. Update the factor for the next lock:factor += k. c. Update the global minimum time:minTime = min(minTime, totalTime). - In the recursive step (for
ifromindexton-1): a. Swap the elements atindexandito create a new ordering to explore. b. Make a recursive call:generatePermutations(index + 1, ...). c. Swap the elements back to their original positions to backtrack and explore other possibilities. - Initiate the process by calling
generatePermutations(0, initialListOfStrengths). - The final value of
minTimeis the answer.
Walkthrough
The core idea is to recognize that any sequence of breaking n locks corresponds to a permutation of the lock indices {0, 1, ..., n-1}. We can use a recursive backtracking algorithm to generate all n! permutations of the lock strengths. For each permutation, we simulate the process of breaking the locks in that specific order. The factor x starts at 1. For the i-th lock in the sequence (1-indexed), the factor will be 1 + (i-1) * k. The time to break a lock with strength s and current factor x is ceil(s / x). We sum up the times for all n locks for a given permutation and keep track of the minimum total time found so far. After checking all n! permutations, this minimum value is the solution.
class Solution { long minTime = Long.MAX_VALUE; int k; int n; public int minimumTime(int[] strength, int k) { this.k = k; this.n = strength.length; java.util.List<Integer> strengthList = new java.util.ArrayList<>(); for (int s : strength) { strengthList.add(s); } generatePermutations(0, strengthList); return (int) minTime; } private void generatePermutations(int index, java.util.List<Integer> currentPermutation) { if (index == n) { calculateTime(currentPermutation); return; } for (int i = index; i < n; i++) { java.util.Collections.swap(currentPermutation, index, i); generatePermutations(index + 1, currentPermutation); java.util.Collections.swap(currentPermutation, index, i); // backtrack } } private void calculateTime(java.util.List<Integer> permutation) { long currentTime = 0; long factor = 1; for (int s : permutation) { // time = ceil(s / factor) long time = (s + factor - 1) / factor; currentTime += time; factor += k; } minTime = Math.min(minTime, currentTime); }}Complexity
Time
O(n * n!) - There are `n!` possible permutations of the locks. Generating all permutations using the standard recursive algorithm takes `O(n * n!)` time. For each complete permutation, we spend `O(n)` time to calculate the total time.
Space
O(n) - This space is used for the recursion stack, which can go up to `n` levels deep, and for storing the list representing the current permutation.
Trade-offs
Pros
Conceptually simple and directly models the problem statement.
Easy to implement using standard recursion and backtracking.
Cons
Highly inefficient due to its factorial time complexity.
Becomes infeasible for values of
nlarger than 10-12.
A more efficient approach uses dynamic programming with bitmasking. This technique is well-suited for problems with small constraints on n where we need to consider subsets of items. We can represent the set of broken locks using a bitmask. The state of our DP will be defined by this mask, and we'll compute the minimum time to break the corresponding set of locks by building up from smaller subsets to larger ones.
Algorithm
- Define a DP array,
dp, of size2^n.dp[mask]will store the minimum time to break the set of locks represented by themask. - Initialize all entries of
dpto a large value (infinity), except for the base case. - Set the base case:
dp[0] = 0, as it takes 0 time to break 0 locks. - Iterate through each
maskfrom1to(1<<n) - 1. - For each
mask, calculate the number of locks already broken,c = Integer.bitCount(mask). - The factor for breaking the
c-th lock in any sequence leading to this state isfactor = 1 + (long)(c - 1) * k. - Iterate through each lock
jfrom0ton-1. - Check if lock
jis part of the current set (i.e., if thej-th bit is set inmask). - If it is, this lock
jcould have been the last one broken to reach the statemask. The previous state would beprevMask = mask ^ (1 << j). - Calculate the time to break lock
j:timeForJ = (strength[j] + factor - 1) / factor. - Update the
dpvalue for the current mask:dp[mask] = min(dp[mask], dp[prevMask] + timeForJ). - After iterating through all masks, the final answer is
dp[(1<<n) - 1], which is the minimum time to break allnlocks.
Walkthrough
We use a DP array, dp, of size 2^n. dp[mask] stores the minimum time required to break the locks represented by the mask. A 1 at the i-th bit of the mask means the i-th lock (from the original input) has been broken. The state transition relies on identifying which lock was the last one broken to reach the current state mask. The number of set bits in a mask (popcount(mask)) tells us how many locks have been broken. If c = popcount(mask), we are about to break the c-th lock in a sequence. The factor for this c-th break will always be 1 + (c-1) * k, regardless of which specific c locks were broken or in what order. This allows us to build the solution iteratively.
class Solution { public int minimumTime(int[] strength, int k) { int n = strength.length; int numMasks = 1 << n; long[] dp = new long[numMasks]; java.util.Arrays.fill(dp, Long.MAX_VALUE); dp[0] = 0; for (int mask = 1; mask < numMasks; mask++) { int brokenLocksCount = Integer.bitCount(mask); long factor = 1 + (long)(brokenLocksCount - 1) * k; for (int j = 0; j < n; j++) { // Check if lock j is in the current set (mask) if ((mask & (1 << j)) != 0) { int prevMask = mask ^ (1 << j); if (dp[prevMask] != Long.MAX_VALUE) { long timeForJ = (strength[j] + factor - 1) / factor; dp[mask] = Math.min(dp[mask], dp[prevMask] + timeForJ); } } } } return (int) dp[numMasks - 1]; }}Complexity
Time
O(n * 2^n) - We iterate through `2^n` masks. For each mask, we perform an inner loop that runs `n` times to check each bit. The operations inside the loop are constant time.
Space
O(2^n) - We use a DP array of size `2^n` to store the minimum time for each subset of locks.
Trade-offs
Pros
Significantly more efficient than the brute-force permutation approach.
Guarantees finding the optimal solution by systematically exploring all valid state transitions.
Cons
The exponential time and space complexity limit its use to problems with small
n.Can be less intuitive to formulate compared to the direct brute-force approach.
Solutions
Solution
class Solution {private List<Integer> strength;private Integer[] f;private int k;private int n;public int findMinimumTime(List<Integer> strength, int K) { n = strength.size(); f = new Integer[1 << n]; k = K; this.strength = strength; return dfs(0); }private int dfs(int i) { if (i == (1 << n) - 1) { return 0; } if (f[i] != null) { return f[i]; } int cnt = Integer.bitCount(i); int x = 1 + cnt * k; f[i] = 1 << 30; for (int j = 0; j < n; ++j) { if ((i >> j & 1) == 0) { f[i] = Math.min(f[i], dfs(i | 1 << j) + (strength.get(j) + x - 1) / x); } } return f[i]; }}class Solution {public: int findMinimumTime(vector<int> &strength, int K) { int n = strength.size(); int f[1 << n]; memset(f, -1, sizeof(f)); int k = K; auto dfs = [&](auto &&dfs, int i) -> int { if (i == (1 << n) - 1) { return 0; } if (f[i] != -1) { return f[i]; } int cnt = __builtin_popcount(i); int x = 1 + k * cnt; f[i] = INT_MAX; for (int j = 0; j < n; ++j) { if (i >> j & 1 ^ 1) { f[i] = min(f[i], dfs(dfs, i | 1 << j) + (strength[j] + x - 1) / x); } } return f[i]; }; return dfs(dfs, 0); }};class Solution: def findMinimumTime(self, strength: List[int], K: int) -> int: @ cache def dfs(i: int) -> int: if i == (1 << len(strength)) - 1: return 0 cnt = i . bit_count() x = 1 + cnt * K ans = inf for j, s in enumerate(strength): if i >> j & 1 ^ 1: ans = min(ans, dfs(i | 1 << j) + (s + x - 1) // x) return ans return dfs(0)Video walkthrough
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