How do you solve Merge Operations for Minimum Travel Time on LeetCode?

Merge Operations for Minimum Travel Time (LeetCode #3538) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * k) time complexity and O(n * k) space complexity. Key insight: Merging reduces the number of signs, affecting total travel time.

What is the time complexity of Merge Operations for Minimum Travel Time?

The optimal solution for Merge Operations for Minimum Travel Time runs in O(n * k) time and uses O(n * k) space. The complexity arises from filling a DP table of size (k+1) x (n-k+1).

What is the brute force solution for Merge Operations for Minimum Travel Time?

The brute force approach for Merge Operations for Minimum Travel Time is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try all possible merges of adjacent signs and calculate the total travel time after each merge. This approach is straightforward but inefficient as it checks every possible combination. The optimal Optimal Solution approach improves this to O(n * k).

What algorithm or data structure is used to solve Merge Operations for Minimum Travel Time?

Merge Operations for Minimum Travel Time uses the following concepts: Array, Dynamic Programming, Prefix Sum. The recommended patterns to study are: Dynamic Programming, Greedy Algorithms.

What are the interview tips for Merge Operations for Minimum Travel Time?

Explain your thought process clearly. Discuss edge cases, like k = 0 or k = n-1. Practice similar DP problems to build confidence.

What are common mistakes when solving Merge Operations for Minimum Travel Time?

Not considering the impact of merges on total time. Failing to initialize the DP table correctly.

#3538

Merge Operations for Minimum Travel Time

Hard

ArrayDynamic ProgrammingPrefix SumDynamic ProgrammingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n * k)Space O(n * k)

Using dynamic programming, we can keep track of the minimum travel time after each merge operation, reducing unnecessary recalculations.

⚙️

Algorithm

3 steps

1Step 1: Create a DP array where dp[i][j] represents the minimum time with i merges and j signs.
2Step 2: Initialize the DP array with base cases for 0 merges.
3Step 3: Fill the DP table iteratively, considering merging adjacent signs and updating the time accordingly.

solution.py8 lines

1def minTravelTime(l, n, k, position, time):
2    dp = [[float('inf')] * (n - k + 1) for _ in range(k + 1)]
3    for j in range(n - k + 1):
4        dp[0][j] = sum(time)
5    for i in range(1, k + 1):
6        for j in range(n - i):
7            dp[i][j] = min(dp[i][j], dp[i - 1][j] - time[j])
8    return dp[k][0]

ℹ

Complexity note: The complexity arises from filling a DP table of size (k+1) x (n-k+1).

1Merging reduces the number of signs, affecting total travel time.
2Dynamic programming helps manage overlapping subproblems efficiently.

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