Leetcode #2297: Jump Game VIII
In this guide, we solve Leetcode #2297 Jump Game VIII 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
You are given a 0-indexed integer array nums of length n. You are initially standing at index 0.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Stack, Graph, Array, Dynamic Programming, Shortest Path, Monotonic Stack
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: nums = [3,2,4,4,1], costs = [3,7,6,4,2]
Output: 8
Explanation: You start at index 0.
- Jump to index 2 with a cost of costs[2] = 6.
- Jump to index 4 with a cost of costs[4] = 2.
The total cost is 8. It can be proven that 8 is the minimum cost needed.
Two other possible paths are from index 0 -> 1 -> 4 and index 0 -> 2 -> 3 -> 4.
These have a total cost of 9 and 12, respectively.
Python Solution
class Solution:
def minCost(self, nums: List[int], costs: List[int]) -> int:
n = len(nums)
g = defaultdict(list)
stk = []
for i in range(n - 1, -1, -1):
while stk and nums[stk[-1]] < nums[i]:
stk.pop()
if stk:
g[i].append(stk[-1])
stk.append(i)
stk = []
for i in range(n - 1, -1, -1):
while stk and nums[stk[-1]] >= nums[i]:
stk.pop()
if stk:
g[i].append(stk[-1])
stk.append(i)
f = [inf] * n
f[0] = 0
for i in range(n):
for j in g[i]:
f[j] = min(f[j], f[i] + costs[j])
return f[n - 1]
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.