How do you solve Frog Jump on LeetCode?

Frog Jump (LeetCode #403) 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: The frog can only jump forward, which limits the problem's complexity.

What is the brute force solution for Frog Jump?

The brute force approach for Frog Jump is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach explores all possible jumps the frog can make recursively. It checks each stone to see if the frog can reach the last stone by trying every possible jump size. The optimal Optimal Solution approach improves this to O(n²).

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

Frog Jump uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Hash Map, Recursive Backtracking.

What are the interview tips for Frog Jump?

Always clarify the problem constraints and edge cases before diving into coding. Think about how you can optimize your solution after writing the brute force version. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Frog Jump?

Not considering the case where the frog cannot jump due to large gaps between stones. Failing to check if the jump size is valid (i.e., must be greater than 0).

#403

Frog Jump

Hard

ArrayDynamic ProgrammingDynamic ProgrammingHash MapRecursive Backtracking

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 with a hashmap to store reachable states. This avoids redundant calculations and allows us to efficiently check if the frog can reach the last stone.

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Algorithm

4 steps

1Step 1: Create a hashmap to store reachable positions and their corresponding jump sizes.
2Step 2: Iterate through each stone and for each stone, check if the frog can jump to the next stones using the allowed jump sizes.
3Step 3: Update the hashmap with new reachable positions and their jump sizes.
4Step 4: At the end, check if the last stone is reachable.

solution.py9 lines

1def canCross(stones):
2    dp = {stone: set() for stone in stones}
3    dp[0].add(0)
4    for stone in stones:
5        for jump in dp[stone]:
6            for next_jump in [jump - 1, jump, jump + 1]:
7                if next_jump > 0 and stone + next_jump in dp:
8                    dp[stone + next_jump].add(next_jump)
9    return len(dp[stones[-1]]) > 0

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loops, but we use a hashmap to store reachable states, which optimizes space usage and avoids recalculating paths.

1The frog can only jump forward, which limits the problem's complexity.
2Using a hashmap to track reachable positions helps avoid redundant calculations.

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