Leetcode #1824: Minimum Sideway Jumps
In this guide, we solve Leetcode #1824 Minimum Sideway Jumps 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
There is a 3 lane road of length n that consists of n + 1 points labeled from 0 to n. A frog starts at point 0 in the second lane and wants to jump to point n.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Greedy, 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: obstacles = [0,1,2,3,0]
Output: 2
Explanation: The optimal solution is shown by the arrows above. There are 2 side jumps (red arrows).
Note that the frog can jump over obstacles only when making side jumps (as shown at point 2).
Python Solution
class Solution:
def minSideJumps(self, obstacles: List[int]) -> int:
f = [1, 0, 1]
for v in obstacles[1:]:
for j in range(3):
if v == j + 1:
f[j] = inf
break
x = min(f) + 1
for j in range(3):
if v != j + 1:
f[j] = min(f[j], x)
return min(f)
Complexity
The time complexity is , where is the length of the array . 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.