How do you solve Frog Jump II on LeetCode?

Frog Jump II (LeetCode #2498) 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 minimize the maximum jump by strategically choosing which stones to jump to.

What is the time complexity of Frog Jump II?

The optimal solution for Frog Jump II runs in O(n) time and uses O(1) space. This complexity is linear because we only traverse the stones array once to compute the maximum jump lengths.

What is the brute force solution for Frog Jump II?

The brute force approach for Frog Jump II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible paths the frog can take to jump from the first stone to the last and back. We calculate the maximum jump length for each path and return the minimum of these maximums. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Frog Jump II?

Frog Jump II uses the following concepts: Array, Binary Search, Greedy. The recommended patterns to study are: Greedy, Two Pointers.

What are the interview tips for Frog Jump II?

Always start with a brute-force approach to understand the problem better. Look for patterns or properties in the problem that can lead to optimizations. Be prepared to explain your thought process clearly and justify your choices.

What are common mistakes when solving Frog Jump II?

Not considering the optimal strategy of skipping stones. Overcomplicating the path calculations instead of focusing on maximum distances.

#2498

Frog Jump II

Medium

ArrayBinary SearchGreedyGreedyTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal solution leverages the observation that the frog can minimize the maximum jump by strategically skipping stones. By jumping to every second stone on the way there and back, we can effectively minimize the maximum jump length.

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Algorithm

3 steps

1Step 1: Calculate the maximum jump length by considering the distance between the first and last stone.
2Step 2: Check the maximum jump lengths between adjacent stones to find the minimum possible maximum jump.
3Step 3: Return the minimum of these maximum jumps.

solution.py6 lines

1def frogJump(stones):
2    n = len(stones)
3    max_jump = stones[-1] - stones[0]
4    for i in range(1, n):
5        max_jump = min(max_jump, max(stones[i] - stones[i-1], stones[-1] - stones[i]))
6    return max_jump

ℹ

Complexity note: This complexity is linear because we only traverse the stones array once to compute the maximum jump lengths.

1The frog can minimize the maximum jump by strategically choosing which stones to jump to.
2The maximum jump length is determined by the distance between the first and last stones.

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