514. Freedom Trail
In the video game Fallout 4, the quest "Road to Freedom" requires players to reach a metal dial called the "Freedom Trail Ring", and use the dial to spell a specific keyword in order to open the door.
Given a string ring, which represents the code engraved on the outer ring and another string key, which represents the keyword needs to be spelled. You need to find the minimum number of steps in order to spell all the characters in the keyword.
Initially, the first character of the ring is aligned at 12:00 direction. You need to spell all the characters in the string key one by one by rotating the ring clockwise or anticlockwise to make each character of the string key aligned at 12:00 direction and then by pressing the center button.
At the stage of rotating the ring to spell the key character key[i]:
- You can rotate the ring clockwise or anticlockwise one place, which counts as 1 step. The final purpose of the rotation is to align one of the string ring's characters at the 12:00 direction, where this character must equal to the character key[i].
- If the character key[i] has been aligned at the 12:00 direction, you need to press the center button to spell, which also counts as 1 step. After the pressing, you could begin to spell the next character in the key (next stage), otherwise, you've finished all the spelling.
Example:

Input: ring = "godding", key = "gd"
Output: 4
Explanation:
For the first key character 'g', since it is already in place, we just need 1 step to spell this character.
For the second key character 'd', we need to rotate the ring "godding" anticlockwise by two steps to make it become "ddinggo".
Also, we need 1 more step for spelling.
So the final output is 4.
Note:
- Length of both ring and key will be in range 1 to 100.
- There are only lowercase letters in both strings and might be some duplcate characters in both strings.
- It's guaranteed that string key could always be spelled by rotating the string ring.
Approach #1: DP. [C++]
class Solution {
public:
int findRotateSteps(string ring, string key) {
int N = ring.length();
int M = key.length();
int ans = INT_MAX;
// dp[i][j] denote means the minimum of steps to spell the key[0:i] which key[i] at the index of j in ring.
vector<vector<int>> dp(M, vector<int>(N, INT_MAX));
vector<vector<int>> char_idx(26, vector<int>());
for (int i = 0; i < N; ++i) {
int c = ring[i] - 'a';
char_idx[c].push_back(i);
}
for (int i = 0; i < M; ++i) {
int cur_char = key[i] - 'a';
for (int cur_pos : char_idx[cur_char]) {
if (i == 0) {
dp[i][cur_pos] = min(cur_pos, N - cur_pos) + 1;
} else {
int pre_char = key[i-1] - 'a';
for (int pre_pos : char_idx[pre_char]) {
int diff = min(abs(cur_pos-pre_pos), N - abs(cur_pos-pre_pos));
dp[i][cur_pos] = min(dp[i][cur_pos], dp[i-1][pre_pos]+diff+1);
//cout << dp[i][cur_pos] << endl;
}
}
if (i == M-1)
ans = min(ans, dp[M-1][cur_pos]);
}
}
return ans;
}
};
Analysis:
https://leetcode.com/problems/freedom-trail/discuss/194601/DP-solution-with-detailed-text-and-video-explanation
514. Freedom Trail的更多相关文章
- 514 Freedom Trail 自由之路
详见:https://leetcode.com/problems/freedom-trail/description/ C++: class Solution { public: int findRo ...
- [LeetCode] Freedom Trail 自由之路
In the video game Fallout 4, the quest "Road to Freedom" requires players to reach a metal ...
- [Swift]LeetCode514. 自由之路 | Freedom Trail
In the video game Fallout 4, the quest "Road to Freedom" requires players to reach a metal ...
- 动态规划——Freedom Trail
题目:https://leetcode.com/problems/freedom-trail/ 额...不解释大意了,题目我也不想写过程了有点繁琐,直接给出代码: public int findRot ...
- LeetCode All in One题解汇总(持续更新中...)
突然很想刷刷题,LeetCode是一个不错的选择,忽略了输入输出,更好的突出了算法,省去了不少时间. dalao们发现了任何错误,或是代码无法通过,或是有更好的解法,或是有任何疑问和建议的话,可以在对 ...
- LeetCode Weekly Contest 22
1. 532. K-diff Pairs in an Array 分析:由于要唯一,所以要去重,考虑k=0,时候,相同的数字需要个数大于1,所以,先用map统计个数,对于k=0,特判,对于其他的,遍历 ...
- 算法与数据结构基础 - 深度优先搜索(DFS)
DFS基础 深度优先搜索(Depth First Search)是一种搜索思路,相比广度优先搜索(BFS),DFS对每一个分枝路径深入到不能再深入为止,其应用于树/图的遍历.嵌套关系处理.回溯等,可以 ...
- All LeetCode Questions List 题目汇总
All LeetCode Questions List(Part of Answers, still updating) 题目汇总及部分答案(持续更新中) Leetcode problems clas ...
- Leetcode problems classified by company 题目按公司分类(Last updated: October 2, 2017)
All LeetCode Questions List 题目汇总 Sorted by frequency of problems that appear in real interviews. Las ...
随机推荐
- android学习-Eclipse中修改Android项目图标
参考原文:http://blog.csdn.net/wpwbb510582246/article/details/52556753 方法一:替换res文件夹下的ic_launcher-web.png图 ...
- Django之ModelForm篇
ModelForm a. class Meta: model, # 对应Model的 fields=None, # 字段 exclude=None, # 排除字段 labels=None, # 提示信 ...
- 前端开发之JavaScript HTML DOM理论篇一
主要内容: 1.DOM简介 2.DOM 节点 3.DOM 方法和属性 4.DOM 访问和修改 一.DOM简介 1.什么是 DOM? DOM 是 W3C(万维网联盟)的标准. DOM 定义了访问 HTM ...
- 用Pylint规范化Python代码,附PyCharm配置
Pylint一个可以检查Python代码错误,执行代码规范的工具.它还可以对代码风格提出建议. 官网:https://pylint.readthedocs.io pip install pylint ...
- 阿里巴巴Java开发规约扫描插件-Alibaba Java Coding Guidelines 在idea上安装使用教程
经过247天的持续研发,阿里巴巴于10月14日在杭州云栖大会上,正式发布众所期待的<阿里巴巴Java开发规约>扫描插件!该插件由阿里巴巴P3C项目组研发.P3C是世界知名的反潜机,专门对付 ...
- RimLight(轮廓光) - Shader
[RimLight(轮廓光) - Shader] RimLight指的是物体的轮廓光.效果如下: 轮廓光的强度通过 1.0 - dot(normal, eye_vector)来计算.使用这个公式,则指 ...
- JAVA中的数组对象
代码:Student [] sd=new Student[5];//新建一个学生类的数组对象sd. sd[0]=new Student("kj",13);//为数组对 ...
- 563. Binary Tree Tilt 子节点差的绝对值之和
[抄题]: Given a binary tree, return the tilt of the whole tree. The tilt of a tree node is defined as ...
- GitHub 上的十一款热门开源安全工具
作为开源开发领域的基石,“所有漏洞皆属浅表”已经成为一条著名的原则甚至是信条.作为广为人知的Linus定律,当讨论开源模式在安全方面的优势时,开放代码能够提高项目漏洞检测效率的理论也被IT专业人士们所 ...
- "软掩膜"和“硬掩膜”-智能IC卡
目录 一.“软掩膜”和“硬掩膜”... 2 二.EMV迁移进程... 3 三.PBOC规范和EMV规范对比... 3 四.总结... 5 五.关于SDA和DDA. 6 一.“软掩膜”和“硬掩膜” “软 ...