How do you solve Minimum Sideway Jumps on LeetCode?

Minimum Sideway Jumps (LeetCode #1824) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The frog can jump to any lane at the same point if there are no obstacles.

What is the brute force solution for Minimum Sideway Jumps?

The brute force approach for Minimum Sideway Jumps is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach explores all possible paths the frog can take to reach the end, checking each lane at every point. This method is simple but inefficient, as it can lead to many redundant calculations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Sideway Jumps?

Minimum Sideway Jumps uses the following concepts: Array, Dynamic Programming, Greedy. The recommended patterns to study are: Dynamic Programming, Greedy Algorithms.

What are the interview tips for Minimum Sideway Jumps?

Always think about edge cases, such as no obstacles or all lanes blocked. Explain your thought process clearly while coding; it shows your understanding. Practice similar problems to recognize patterns and improve your problem-solving speed.

What are common mistakes when solving Minimum Sideway Jumps?

Forgetting to check if the next position is blocked before moving. Not considering all possible lanes when an obstacle is encountered.

#1824

Minimum Sideway Jumps

Medium

ArrayDynamic ProgrammingGreedyDynamic ProgrammingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses a dynamic programming approach to track the minimum side jumps needed to reach each lane at each point. This avoids redundant calculations and efficiently finds the answer.

⚙️

Algorithm

3 steps

1Step 1: Initialize a dp array where dp[i] holds the minimum side jumps needed to reach point i in each lane.
2Step 2: Iterate through each point from 0 to n, updating the dp values based on obstacles and possible lane transitions.
3Step 3: Return the minimum value from the dp array at point n.

solution.py11 lines

1def minSideJumps(obstacles):
2    n = len(obstacles) - 1
3    dp = [float('inf')] * 3
4    dp[1] = 0
5    for i in range(n):
6        if obstacles[i] == 0:
7            for j in range(3):
8                dp[j] = min(dp[j], dp[j] + (1 if obstacles[i + 1] != 0 else 0))
9        else:
10            dp[obstacles[i] - 1] = float('inf')
11    return min(dp)

ℹ

Complexity note: The time complexity is O(n) because we iterate through the obstacles array once, updating the dp values in constant time for each point.

1The frog can jump to any lane at the same point if there are no obstacles.
2Dynamic programming is effective for minimizing transitions in state-based problems.

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