Minimum Time to Break Locks II - Practice Coding Problems

Minimum Time to Break Locks II - Problem

Bob finds himself trapped in a mysterious dungeon with n magical locks blocking his escape route. Each lock requires a specific amount of energy to break, stored in an array strength[] where strength[i] represents the energy needed to break the i-th lock.

Bob wields a special sword with unique properties:

🗡️ Initial energy: 0
⚡ Growth factor X: starts at 1
📈 Energy increase: Every minute, sword energy increases by current factor X
🔓 Lock breaking: Sword energy must reach at least strength[i] to break lock i
🔄 Reset mechanism: After breaking any lock, energy resets to 0 and factor X increases by 1

Your mission is to determine the minimum time in minutes required for Bob to break all locks in the optimal order and escape the dungeon!

Example: With locks [3, 4, 1], Bob could break them as [1, 3, 4] taking 1 + 2 + 2 = 5 minutes total.

Input & Output

example_1.py — Basic Case

$ Input: [3, 4, 1]

› Output: 5

💡 Note: Optimal order is [2, 1, 0] (break locks with strength [1, 4, 3]). Lock 2: time = ceil(1/1) = 1, Lock 1: time = ceil(4/2) = 2, Lock 0: time = ceil(3/3) = 1. Total: 1 + 2 + 1 = 4. Wait, let me recalculate... Lock 2: 1 minute (factor=1), Lock 0: 2 minutes (factor=2, ceil(3/2)=2), Lock 1: 2 minutes (factor=3, ceil(4/3)=2). Total: 1 + 2 + 2 = 5.

example_2.py — Single Lock

$ Input: [5]

› Output: 5

💡 Note: Only one lock with strength 5. With initial factor 1, it takes exactly 5 minutes to accumulate enough energy.

example_3.py — Equal Strength

$ Input: [2, 2, 2]

› Output: 5

💡 Note: All locks have equal strength, so any order gives the same result. With factors 1, 2, 3: time = ceil(2/1) + ceil(2/2) + ceil(2/3) = 2 + 1 + 1 = 4. Actually, let me recalculate: 2 + 1 + 1 = 4, but we need to check if there's a better arrangement. All arrangements give the same result due to symmetry: 4 minutes total.

Visualization

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Understanding the Visualization

Initial State

Sword starts with 0 energy and growth factor 1

Energy Accumulation

Each minute, energy increases by current factor

Lock Breaking

When energy ≥ lock strength, break lock and reset

Factor Upgrade

After breaking lock, factor increases by 1

Optimal Strategy

Break weaker locks first to maximize factor for stronger locks

Key Takeaway

🎯 Key Insight: The optimal strategy is to break weaker locks first while the factor is low, saving the stronger locks for when the sword has a higher growth factor, thus minimizing total time.

Time & Space Complexity

Time Complexity

⏱️

O(n! × n)

n! permutations, each taking O(n) time to simulate

⚠ Quadratic Growth

Space Complexity

O(n)

Space for recursion stack and current permutation

⚡ Linearithmic Space

Constraints

1 ≤ n ≤ 8
1 ≤ strength[i] ≤ 10⁶
Time limit: 2 seconds per test case

Asked in

G Google 25 a Amazon 18 f Meta 12 ⊞ Microsoft 8

The optimal approach uses dynamic programming with bitmasks to efficiently explore all possible lock-breaking orders. The key insight is that locks with higher strength should generally be broken later when the sword's factor is higher, but the exact optimal order requires examining all possibilities. The bitmask DP solution runs in O(n² × 2ⁿ) time, making it practical for the given constraints (n ≤ 8).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Permutations)	O(n! × n)	O(n)	Generate all possible lock-breaking orders and find the minimum time
Dynamic Programming with Bitmask	O(n² × 2ⁿ)	O(2ⁿ)	Use bitmask DP to efficiently track which locks are broken and find optimal order

Brute Force (Try All Permutations) — Algorithm Steps

Generate all n! permutations of lock indices
For each permutation, simulate the lock-breaking process
Track current energy, factor X, and total time
For each lock, wait until energy >= strength[lock], then reset
Return the minimum time across all permutations

Visualization

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Step-by-Step Walkthrough

Generate Permutations

Create all possible orderings of locks

Simulate Each Order

For each order, calculate total time needed

Track Minimum

Keep track of the best time found so far

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

int minTime;

void swap(int* a, int* b) {
    int temp = *a;
    *a = *b;
    *b = temp;
}

void generatePermutations(int* strength, int* indices, int n, int start) {
    if (start == n) {
        int totalTime = 0;
        int factor = 1;
        
        for (int i = 0; i < n; i++) {
            int requiredEnergy = strength[indices[i]];
            int timeNeeded = (requiredEnergy + factor - 1) / factor;
            totalTime += timeNeeded;
            factor++;
        }
        
        if (totalTime < minTime) {
            minTime = totalTime;
        }
        return;
    }
    
    for (int i = start; i < n; i++) {
        swap(&indices[start], &indices[i]);
        generatePermutations(strength, indices, n, start + 1);
        swap(&indices[start], &indices[i]);
    }
}

int minimumTimeToBreakLocks(int* strength, int n) {
    minTime = INT_MAX;
    int* indices = malloc(n * sizeof(int));
    for (int i = 0; i < n; i++) {
        indices[i] = i;
    }
    
    generatePermutations(strength, indices, n, 0);
    free(indices);
    return minTime;
}

int main() {
    int n = 3;
    int strength[3];
    scanf("%d %d %d", &strength[0], &strength[1], &strength[2]);
    printf("%d\n", minimumTimeToBreakLocks(strength, n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n! × n)

n! permutations, each taking O(n) time to simulate

⚠ Quadratic Growth

Space Complexity

O(n)

Space for recursion stack and current permutation

⚡ Linearithmic Space

Constraints

1 ≤ n ≤ 8
1 ≤ strength[i] ≤ 10⁶
Time limit: 2 seconds per test case

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Try All Permutations) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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