What is the time complexity of Minimum Time to Break Locks I?

The optimal solution has a time complexity of O(n * 2^n).

What is the space complexity of Minimum Time to Break Locks I?

The space complexity is O(2^n).

What data structures are used to solve Minimum Time to Break Locks I?

This problem uses Array, Dynamic Programming, Backtracking, Bit Manipulation, Depth-First Search, Bitmask. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Time to Break Locks I?

Insight 1: The problem can be solved using dynamic programming because the optimal solution for breaking a subset of locks can be used to find the optimal solution for breaking a larger subset. Insight 2: Bitmasking is an effective way to represent subsets of locks. Each bit in the mask corresponds to a lock; if the bit is set, the lock is included in the subset.

Which companies ask LeetCode 3376. Minimum Time to Break Locks I in interviews?

This problem is frequently asked at IVP.

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Minimum Time to Break Locks I - LeetCode 3376 Solution

Minimum Time to Break Locks I - Complete Solution Guide

Minimum Time to Break Locks I is LeetCode problem 3376, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 i th 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

Detailed Explanation

The problem asks us to find the minimum time required to break 'n' locks, given an array 'strength' representing the energy needed for each lock, and an increment factor 'k'. We start with a sword having 0 energy and a factor 'x' initialized to 1. Each minute, the sword's energy increases by the current factor 'x'. To break a lock, the sword's energy must be at least equal to the lock's strength. After breaking a lock, the sword's energy resets to 0, and the factor 'x' increases by 'k'. The goal is to determine the optimal order to break the locks in order to minimize the total time taken.

Solution Approach

The provided code uses dynamic programming with bitmasking to solve the problem. The `dp` array stores the minimum time required to break a particular subset of locks represented by a bitmask. The algorithm iterates through all possible subsets of locks. For each subset, it considers each lock in the subset as the last lock to be broken. It calculates the time required to break that lock, given the current factor 'x', and adds it to the minimum time required to break the remaining locks in the subset (obtained from the `dp` array). The algorithm updates the `dp` array with the minimum time found so far.

Step-by-Step Algorithm

Step 1: Initialize a `dp` array of size 2^n with infinity, where dp[mask] represents the minimum time to break the locks corresponding to the bits set in the mask.
Step 2: Set `dp[0]` to 0, representing the initial state where no locks are broken.
Step 3: Iterate through all possible masks from 1 to 2^n - 1. Each mask represents a subset of locks.
Step 4: For each mask, calculate the number of broken locks (number of set bits in the mask) to determine the current factor 'x' as 1 + (num_broken - 1) * k.
Step 5: Iterate through each lock `j` from 0 to n-1. If the j-th bit is set in the current mask, it means that lock `j` is part of the subset represented by the mask.
Step 6: Calculate the `prev_mask` by unsetting the j-th bit in the current mask. `prev_mask` represents the subset of locks without the j-th lock.
Step 7: Calculate the time required to break the j-th lock: `time_for_this_lock = (strength[j] + current_factor - 1) // current_factor`.
Step 8: Update `dp[mask]` with the minimum of its current value and `dp[prev_mask] + time_for_this_lock`.
Step 9: After iterating through all masks, `dp[(1 << n) - 1]` will contain the minimum time required to break all locks. Return this value.

Key Insights

Insight 1: The problem can be solved using dynamic programming because the optimal solution for breaking a subset of locks can be used to find the optimal solution for breaking a larger subset.
Insight 2: Bitmasking is an effective way to represent subsets of locks. Each bit in the mask corresponds to a lock; if the bit is set, the lock is included in the subset.
Insight 3: The order in which locks are broken matters. Breaking locks with lower strengths earlier might lead to a smaller overall time due to the factor 'x' increasing after each lock is broken.
Insight 4: The number of locks is constrained to be small (n <= 8), which makes a bitmasking and dynamic programming approach feasible with reasonable time and space complexity.

Complexity Analysis

Time Complexity: O(n * 2^n)

Space Complexity: O(2^n)

Topics

This problem involves: Array, Dynamic Programming, Backtracking, Bit Manipulation, Depth-First Search, Bitmask.

Companies

Asked at: IVP.

Minimum Time to Break Locks I | LeetCode 3376

Medium

Array Dynamic Programming Backtracking Bit Manipulation Depth-First Search Bitmask

Time: O(n * 2^n)Space: O(2^n)

Solution Code

Available in:

import java.util.List;
import java.util.Arrays;

class Solution {
    public int findMinimumTime(List<Integer> strength, int k) {
        int n = strength.size();
        
        long[] dp = new long[1 << n];
        Arrays.fill(dp, Integer.MAX_VALUE);
        dp[0] = 0;

        for (int mask = 1; mask < (1 << n); mask++) {
            int num_broken = Integer.bitCount(mask);
            long current_factor = 1 + (num_broken - 1) * (long)k;
            
            for (int j = 0; j < n; j++) {
                if ((mask >> j) % 2 == 1) {
                    int prev_mask = mask ^ (1 << j);
                    int s = strength.get(j);
                    
                    long time_for_this_lock = (s + current_factor - 1) / current_factor;
                    
                    dp[mask] = Math.min(dp[mask], dp[prev_mask] + time_for_this_lock);
                }
            }
        }
                        
        return (int)dp[(1 << n) - 1];
    }
}

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Problem Info

Description

Problem 3376: Minimum Time to Break Locks I - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Backtracking Bit Manipulation Depth-First Search Bitmask

Companies

IVP

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3376

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Initialize a `dp` array of size 2^n with infinity, where dp[mask] represents the minimum time to break the locks corresponding to the bits set in the mask.
2
Step 2: Set `dp[0]` to 0, representing the initial state where no locks are broken.
3
Step 3: Iterate through all possible masks from 1 to 2^n - 1. Each mask represents a subset of locks.
4
Step 4: For each mask, calculate the number of broken locks (number of set bits in the mask) to determine the current factor 'x' as 1 + (num_broken - 1) * k.
5
Step 5: Iterate through each lock `j` from 0 to n-1. If the j-th bit is set in the current mask, it means that lock `j` is part of the subset represented by the mask.
6
Step 6: Calculate the `prev_mask` by unsetting the j-th bit in the current mask. `prev_mask` represents the subset of locks without the j-th lock.
7
Step 7: Calculate the time required to break the j-th lock: `time_for_this_lock = (strength[j] + current_factor - 1) // current_factor`.
8
Step 8: Update `dp[mask]` with the minimum of its current value and `dp[prev_mask] + time_for_this_lock`.
9
Step 9: After iterating through all masks, `dp[(1 << n) - 1]` will contain the minimum time required to break all locks. Return this value.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n * 2^n)

💾

Space Complexity

O(2^n)

📝

Why?

The time complexity arises from the nested loops that iterate through all subsets and locks. The space complexity is due to the `dp` array, which stores intermediate results for each subset.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Backtracking

A backtracking approach could explore all possible orders of breaking the locks and find the minimum time. However, since n <= 8, this approach will also be quite fast but likely a bit slower than DP due to redundant computations

Time: O(n!) in the worst case, as it explores all possible permutations of breaking the locks.Space: O(n) due to the recursion depth.

Trade-offs: Backtracking is simpler to implement but less efficient than dynamic programming, especially as 'n' increases. The DP solution avoids redundant calculations by storing and reusing intermediate results.

✨

Key Insights

1Insight 1: The problem can be solved using dynamic programming because the optimal solution for breaking a subset of locks can be used to find the optimal solution for breaking a larger subset.
2Insight 2: Bitmasking is an effective way to represent subsets of locks. Each bit in the mask corresponds to a lock; if the bit is set, the lock is included in the subset.
3Insight 3: The order in which locks are broken matters. Breaking locks with lower strengths earlier might lead to a smaller overall time due to the factor 'x' increasing after each lock is broken.
4Insight 4: The number of locks is constrained to be small (n <= 8), which makes a bitmasking and dynamic programming approach feasible with reasonable time and space complexity.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: When k is very large, breaking a lock might take only 1 minute if strength[i] is small. The algorithm correctly handles this by calculating the time required using integer division.
Edge case 2: When strength is very large and K is 1, then you may require more time to break locks
Edge case 3: Empty strength array. The problem states 1 <= n <= 8 so the case where the strength array is empty is not a concern.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to consider the factor 'x' increasing after breaking each lock. Failing to update the factor correctly will lead to incorrect time calculations.

✗Common mistake 2: Incorrectly calculating the time required to break a lock, particularly when using integer division. Ensure that the result is rounded up correctly.

✗Common mistake 3: Not initializing the `dp` array with appropriate initial values (e.g., infinity) for subsets that are not yet visited.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Task scheduling with dependencies and increasing efficiency. Breaking locks can be seen as completing tasks, where each task requires a certain amount of 'resource' (energy). Completing a task increases overall 'efficiency' (factor k), so the order of completing tasks influences the overall time taken.

🏦Real-world scenario 2: Resource allocation in projects. Different tasks require different amounts of resources and unlocking those tasks might improve available resources in the future (modeled by the increasing factor). The challenge lies in determining an optimal sequence to minimize completion time.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 15, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Time to Break Locks I"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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