LeetCode #3376 — MEDIUM

Minimum Time to Break Locks I

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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 the i^th 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 a given value k.

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 3^rd Lock	2
2	2	2	Nothing	2
3	4	2	Break 2^nd Lock	3
4	3	3	Break 1^st 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 1^st Lock	3
3	3	3	Nothing	3
4	6	3	Break 2ⁿ^d Lock	5
5	5	5	Break 3^r^d Lock	7

The locks cannot be broken in less than 5 minutes; thus, the answer is 5.

Constraints:

n == strength.length
1 <= n <= 8
1 <= K <= 10
1 <= strength[i] <= 10⁶

Step 01

Brute Force Baseline

Problem summary: 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 a given value k. 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Backtracking · Bit Manipulation

Example 1

[3,4,1]
1

Example 2

[2,5,4]
2

Step 02

Core Insight

What unlocks the optimal approach

Try all <code>n!</code> permutation ways of breaking the locks.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3376: Minimum Time to Break Locks I
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];
    }
}

// Accepted solution for LeetCode #3376: Minimum Time to Break Locks I
func findMinimumTime(strength []int, K int) int {
	n := len(strength)
	f := make([]int, 1<<n)
	for i := range f {
		f[i] = -1
	}
	var dfs func(int) int
	dfs = func(i int) int {
		if i == 1<<n-1 {
			return 0
		}
		if f[i] != -1 {
			return f[i]
		}
		x := 1 + K*bits.OnesCount(uint(i))
		f[i] = 1 << 30
		for j, s := range strength {
			if i>>j&1 == 0 {
				f[i] = min(f[i], dfs(i|1<<j)+(s+x-1)/x)
			}
		}
		return f[i]
	}
	return dfs(0)
}

# Accepted solution for LeetCode #3376: Minimum Time to Break Locks I
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)

// Accepted solution for LeetCode #3376: Minimum Time to Break Locks I
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3376: Minimum Time to Break Locks I
// 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];
//     }
// }

// Accepted solution for LeetCode #3376: Minimum Time to Break Locks I
function findMinimumTime(strength: number[], K: number): number {
    const n = strength.length;
    const f: number[] = Array(1 << n).fill(-1);
    const dfs = (i: number): number => {
        if (i === (1 << n) - 1) {
            return 0;
        }
        if (f[i] !== -1) {
            return f[i];
        }
        f[i] = Infinity;
        const x = 1 + K * bitCount(i);
        for (let j = 0; j < n; ++j) {
            if (((i >> j) & 1) == 0) {
                f[i] = Math.min(f[i], dfs(i | (1 << j)) + Math.ceil(strength[j] / x));
            }
        }
        return f[i];
    };
    return dfs(0);
}

function bitCount(i: number): number {
    i = i - ((i >>> 1) & 0x55555555);
    i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
    i = (i + (i >>> 4)) & 0x0f0f0f0f;
    i = i + (i >>> 8);
    i = i + (i >>> 16);
    return i & 0x3f;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.