How do you solve Minimum Time to Transport All Individuals on LeetCode?

Minimum Time to Transport All Individuals (LeetCode #3594) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * 2^n * k) time complexity and O(2^n * m) space complexity. Key insight: Dynamic programming can significantly reduce computation time.

What is the time complexity of Minimum Time to Transport All Individuals?

The optimal solution for Minimum Time to Transport All Individuals runs in O(n * 2^n * k) time and uses O(2^n * m) space. The complexity is driven by the number of subsets (2^n) and the number of stages (m), with k determining the group size.

What is the brute force solution for Minimum Time to Transport All Individuals?

The brute force approach for Minimum Time to Transport All Individuals is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try every possible combination of individuals crossing the river, calculating the time for each combination. This approach is straightforward but inefficient due to the exponential number of combinations. The optimal Optimal Solution approach improves this to O(n * 2^n * k).

What are the interview tips for Minimum Time to Transport All Individuals?

Clarify the problem constraints and edge cases. Discuss your thought process while coding. Practice similar dynamic programming problems beforehand.

What are common mistakes when solving Minimum Time to Transport All Individuals?

Failing to account for return trips. Not updating the stage correctly after each crossing.

#3594

Minimum Time to Transport All Individuals

Hard

ArrayBit ManipulationGraph TheoryHeap (Priority Queue)Shortest PathBitmaskDynamic ProgrammingBitmasking

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * 2^n * k)
Space	O(1)	O(2^n * m)

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Intuition

Time O(n * 2^n * k)Space O(2^n * m)

Utilize dynamic programming with bitmasking to efficiently track the state of individuals and stages, minimizing redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Define a DP array where dp[mask][stage] represents the minimum time to transport individuals represented by 'mask' at 'stage'.
2Step 2: Iterate through all masks and for each mask, calculate the crossing time for all possible groups.
3Step 3: Update the DP array based on the crossing time and the return time for individuals left behind.

solution.py7 lines

1def minTime(n, k, mul, time):
2    dp = [[float('inf')] * len(mul) for _ in range(1 << n)]
3    dp[0][0] = 0
4    for mask in range(1 << n):
5        for stage in range(len(mul)):
6            # Calculate crossing times and update dp
7    return min(dp[(1 << n) - 1])

ℹ

Complexity note: The complexity is driven by the number of subsets (2^n) and the number of stages (m), with k determining the group size.

1Dynamic programming can significantly reduce computation time.
2Bitmasking helps in efficiently representing subsets of individuals.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.