What is the time complexity of Minimum Time to Remove All Cars Containing Illegal Goods?

The optimal solution for Minimum Time to Remove All Cars Containing Illegal Goods runs in O(n) time and uses O(n) space. This complexity is due to the linear traversal of the string to fill the prefix and suffix arrays.

What is the brute force solution for Minimum Time to Remove All Cars Containing Illegal Goods?

The brute force approach for Minimum Time to Remove All Cars Containing Illegal Goods is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible way to remove the illegal cars by trying all combinations of operations. This is straightforward but inefficient, as it may involve many redundant calculations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Time to Remove All Cars Containing Illegal Goods?

Minimum Time to Remove All Cars Containing Illegal Goods uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Prefix Sum, Suffix Array.

What are the interview tips for Minimum Time to Remove All Cars Containing Illegal Goods?

Always think about edge cases, like all cars being '0' or '1'. Explain your thought process clearly while coding. Consider both time and space complexity in your solution.

What are common mistakes when solving Minimum Time to Remove All Cars Containing Illegal Goods?

Not considering all combinations of removals. Overlooking the cost difference between removal types.

#2167

Minimum Time to Remove All Cars Containing Illegal Goods

Hard

StringDynamic ProgrammingDynamic ProgrammingPrefix SumSuffix Array

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach involves calculating the minimum time to remove illegal cars using prefix and suffix arrays. This allows us to efficiently compute the time required without redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Create an array withoutFirst where withoutFirst[i] stores the minimum time to remove all illegal cars from the suffix starting at index i without using type 1 operations.
2Step 2: Create an array onlyFirst where onlyFirst[i] stores the minimum time to remove all illegal cars from the prefix ending at index i using only type 1 operations.
3Step 3: Combine results from both arrays to find the minimum time to remove all illegal cars.

solution.py16 lines

1def minTimeToRemoveCars(s):
2    n = len(s)
3    withoutFirst = [0] * n
4    onlyFirst = [0] * n
5    for i in range(n - 1, -1, -1):
6        if s[i] == '1':
7            withoutFirst[i] = 2 + (withoutFirst[i + 1] if i + 1 < n else 0)
8        else:
9            withoutFirst[i] = withoutFirst[i + 1] if i + 1 < n else 0
10    for i in range(n):
11        if s[i] == '1':
12            onlyFirst[i] = i + 1
13        else:
14            onlyFirst[i] = onlyFirst[i - 1] if i > 0 else 0
15    min_time = min(onlyFirst[i] + withoutFirst[i + 1] for i in range(n - 1))
16    return min_time

ℹ

Complexity note: This complexity is due to the linear traversal of the string to fill the prefix and suffix arrays.

1Removing from the ends is cheaper than from the middle.
2Using prefix and suffix calculations can reduce redundant operations.

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