How do you solve Minimum Skips to Arrive at Meeting On Time on LeetCode?

Minimum Skips to Arrive at Meeting On Time (LeetCode #1883) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Understanding how to manage time with skips is crucial for optimizing travel.

What is the brute force solution for Minimum Skips to Arrive at Meeting On Time?

The brute force approach for Minimum Skips to Arrive at Meeting On Time is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying every possible combination of skips for the roads and checking if we can arrive on time. This method is straightforward but inefficient, as it explores all possibilities. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Minimum Skips to Arrive at Meeting On Time?

Minimum Skips to Arrive at Meeting On Time uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Greedy Algorithms.

What are the interview tips for Minimum Skips to Arrive at Meeting On Time?

Explain your thought process clearly when discussing the DP approach. Be prepared to discuss edge cases, such as when it's impossible to arrive on time. Practice explaining how you would optimize a brute-force solution.

What are common mistakes when solving Minimum Skips to Arrive at Meeting On Time?

Not considering the effect of skipping rests on subsequent roads. Failing to initialize the DP table correctly, leading to incorrect results.

#1883

Minimum Skips to Arrive at Meeting On Time

Hard

ArrayDynamic ProgrammingDynamic ProgrammingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n²)Space O(n)

The optimal approach uses dynamic programming to keep track of the minimum time required to reach each road with a certain number of skips. This allows us to efficiently calculate the minimum skips needed to arrive on time.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP table where dp[i][j] represents the minimum time to reach the i-th road with j skips.
2Step 2: Iterate through each road and update the DP table based on whether we skip the rest or not.
3Step 3: After processing all roads, check the last row of the DP table to find the minimum skips needed to arrive on time.

solution.py17 lines

1# Full working Python code
2
3def minSkips(dist, speed, hoursBefore):
4    n = len(dist)
5    dp = [[float('inf')] * (n + 1) for _ in range(n + 1)]
6    dp[0][0] = 0
7
8    for i in range(n):
9        for j in range(i + 1):
10            dp[i + 1][j] = min(dp[i + 1][j], dp[i][j] + dist[i] / speed)
11            if j > 0:
12                dp[i + 1][j] = min(dp[i + 1][j], math.ceil(dp[i][j]) + dist[i] / speed)
13
14    for j in range(n + 1):
15        if dp[n][j] <= hoursBefore:
16            return j
17    return -1

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loops, but we use O(n) space for the DP table to store intermediate results, which helps avoid recalculating.

1Understanding how to manage time with skips is crucial for optimizing travel.
2Dynamic programming allows us to break down the problem into manageable states.

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