Leetcode #403: Frog Jump
In this guide, we solve Leetcode #403 Frog Jump in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
A frog is crossing a river. The river is divided into some number of units, and at each unit, there may or may not exist a stone.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Array, Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: stones = [0,1,3,5,6,8,12,17]
Output: true
Explanation: The frog can jump to the last stone by jumping 1 unit to the 2nd stone, then 2 units to the 3rd stone, then 2 units to the 4th stone, then 3 units to the 6th stone, 4 units to the 7th stone, and 5 units to the 8th stone.
Python Solution
class Solution:
def canCross(self, stones: List[int]) -> bool:
def dfs(i, k):
if i == n - 1:
return True
for j in range(k - 1, k + 2):
if j > 0 and stones[i] + j in pos and dfs(pos[stones[i] + j], j):
return True
return False
n = len(stones)
pos = {s: i for i, s in enumerate(stones)}
return dfs(0, 0)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.