Frog Jump II - You are given a 0-indexed integer array stones sorted in strictly increasing order representing the positions of stones in a river. A frog, initially on the first stone, wants to travel to the last stone and then return to the first stone. However, it can jump to any stone at most once.

In-depth solution and explanation for LeetCode 2498. Frog Jump II in Python, Java, C++ and more. Intuitions, example walk through, and complexity analysis. Better than official and forum solutions.

2023年1月25日 · 2498. Frog Jump II. Difficulty: Medium. Related Topics: Array, Binary Search, Greedy. Similar Questions: Climbing Stairs, Koko Eating Bananas. Problem. You are given a 0-indexed integer array stones sorted in strictly increasing order …

Frog Jump - 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. The frog can jump on a stone, but it must not jump into the water.

2024年11月11日 · From any stair i, the frog has two options: it can either jump to the (i+1)-th stair or the (i+2)-th stair. The cost of a jump is the absolute difference in height between the two stairs. The task is to find the minimum total cost required for the frog to reach the top. Examples: This problem is mainly a variation of finding n-th Fibonacci Number.

You are given a 0-indexed integer array stones sorted in strictly increasing order representing the positions of stones in a river. A frog, initially on the first stone, wants to travel to the last stone and then return to the first stone. However, it can jump to any stone at most once.

class Solution: def maxJump (self, stones: list [int])-> int: # Let's denote the forwarding path as F and the backwarding path as B. # 'F1 B2 B1 F2' is no better than 'F1 B2 F2 B1' since the distance between # F1 and F2 increase, resulting a larger `ans`. if len (stones) == 2: return stones [1]-stones [0] return max (stones [i]-stones [i-2] for ...