How do I handle the state transition in Jump Game VII?

In Jump Game VII, handle state transitions by tracking the valid indices you can jump to within the range [i + minJump, i + maxJump], ensuring that all positions fall within this interval.

What is the time complexity of the Jump Game VII problem?

The time complexity of Jump Game VII depends on the approach, but a well-optimized solution using dynamic programming or sliding window would typically run in O(n) time.

How does dynamic programming apply to Jump Game VII?

Dynamic programming helps track reachable indices and manage overlapping subproblems efficiently. Each state transition from index to index is stored, reducing redundant checks.

What are the challenges when solving Jump Game VII?

Challenges include managing the range of jumps efficiently, optimizing the process to avoid recalculating all potential jumps, and handling large input sizes without exceeding time and space limits.

Can I use greedy algorithms to solve Jump Game VII?

Greedy algorithms are not ideal for Jump Game VII, as they may overlook optimal paths. The problem requires dynamic programming to ensure that all possible jumps are considered and evaluated.

#1871

Medium

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

跳跃游戏 VII

给你一个下标从 0 开始的二进制字符串 s 和两个整数 minJump 和 maxJump 。一开始，你在下标 0 处，且该位置的值一定为 '0' 。当同时满足如下条件时，你可以从下标 i 移动到下标 j 处： i + minJump 且 s[j] == '0' . 如果你可以到达 s 的下标 s.…

字符串动态规划滑动窗口前缀和

题目描述

给你一个下标从 0 开始的二进制字符串 s 和两个整数 minJump 和 maxJump 。一开始，你在下标 0 处，且该位置的值一定为 '0' 。当同时满足如下条件时，你可以从下标 i 移动到下标 j 处：

i + minJump <= j <= min(i + maxJump, s.length - 1) 且
s[j] == '0'.

如果你可以到达 s 的下标 s.length - 1 处，请你返回 true ，否则返回 false 。

示例 1：

输入：s = "011010", minJump = 2, maxJump = 3
输出：true
解释：
第一步，从下标 0 移动到下标 3 。
第二步，从下标 3 移动到下标 5 。

示例 2：

输入：s = "01101110", minJump = 2, maxJump = 3
输出：false

提示：

2 <= s.length <= 10⁵
s[i] 要么是 '0' ，要么是 '1'
s[0] == '0'
1 <= minJump <= maxJump < s.length

lightbulb

解题思路

方法一：前缀和 + 动态规划

我们定义一个长度为 $n+1$ 的前缀和数组 $pre$ ，其中 $pre[i]$ 表示 $s$ 的前 $i$ 个位置中能够到达的个数。定义一个长度为 $n$ 的布尔数组 $f$ ，其中 $f[i]$ 表示 $s[i]$ 是否能够到达。初始时 $pre[1] = 1$ ，而 $f[0] = true$ 。

考虑 $i \in [1, n)$ ，如果 $s[i] = 0$ ，那么我们需要判断 $s$ 的前 $i$ 个位置中是否存在一个位置 $j$ ，满足 $j$ 能够到达且 $j$ 到 $i$ 的距离在 $[minJump, maxJump]$ 之间。如果存在这样的位置 $j$ ，那么我们就有 $f[i] = true$ ，否则 $f[i] = false$ 。在判断 $j$ 是否存在时，我们可以通过前缀和数组 $pre$ 在 $O(1)$ 的时间内判断是否存在这样的位置 $j$ 。

最终答案即为 $f[n-1]$ 。

时间复杂度 $O(n)$ ，空间复杂度 $O(n)$ 。其中 $n$ 为字符串 $s$ 的长度。

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class Solution:
    def canReach(self, s: str, minJump: int, maxJump: int) -> bool:
        n = len(s)
        pre = [0] * (n + 1)
        pre[1] = 1
        f = [True] + [False] * (n - 1)
        for i in range(1, n):
            if s[i] == "0":
                l, r = max(0, i - maxJump), i - minJump
                f[i] = l <= r and pre[r + 1] - pre[l] > 0
            pre[i + 1] = pre[i] + f[i]
        return f[-1]

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Look for a candidate's understanding of state transition DP.
question_mark
Evaluate their ability to optimize using sliding windows.
question_mark
See how they manage time and space complexity when implementing dynamic programming in a real-world scenario.

warning

常见陷阱

外企场景

error
Overcomplicating the solution by checking all possible combinations instead of limiting the range of valid moves.
error
Failure to optimize the solution by using a sliding window, leading to inefficient time complexity.
error
Not managing memory effectively when using dynamic programming to track reachable indices, potentially causing excessive space usage.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Consider cases where the string starts with a larger number of '1's, blocking early jumps.
arrow_right_alt
What if there were more than two jump ranges instead of just minJump and maxJump?
arrow_right_alt
What if the string had a large number of consecutive '1's, requiring careful interval handling?

help

常见问题

外企场景

继续练习

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