Baozi Leetcode Solution 290: Word Pattern
Problem Statement
pattern
and a string str
, find if str
follows the same pattern.pattern
and a non-empty word in str
.Input: pattern ="abba"
, str ="dog cat cat dog"
Output: true
Input:pattern ="abba"
, str ="dog cat cat fish"
Output: false
Input: pattern ="aaaa"
, str ="dog cat cat dog"
Output: false
Input: pattern ="abba"
, str ="dog dog dog dog"
Output: false
You may assume
pattern
contains only lowercase letters, and str
contains lowercase letters that may be separated by a single space.Problem link
Video Tutorial
You can find the detailed video tutorial here
Thought Process
A very simple problem thus normally can solve it in multiple ways.
Encode string and patterns then compare
Since we are comparing "patterns" here, one straightforward way is encode the pattern into a string, then use the same encoding algorithm to encode the str array into a string, then compare the string.
What encoding should we choose? Well it's not really an encoding per se. What I did is just convert any word or character to a character staring with ('a' + an index). If we see this character before, we just directly return from the hash map lookup. For example, "duck dog dog" would be encoded as "abb" while "bcc" would also be encoded as "abb".
Use bijection mapping
Note in the problem description it mentions it is a bijection mapping (i.e., a one to one mapping).
As shown in the graph below, you see the differences between injection, surjection and bijection. That said, bijection does not allow duplicates. We can build a one to one mapping between the pattern and string, since it's bijection, if two characters in the pattern map to the same string, then it's not a valid bijection, therefore return false.
Ref: https://en.wikipedia.org/wiki/Injective_function |
Solutions
Encode string and patterns then compare
// This way will also work, just a little bit more work by encoding each string into the same one
// Kinda similar to the isomorphic string
public boolean wordPatternEncoding(String pattern, String str) {
if (str == null || str.isEmpty() || pattern == null || pattern.isEmpty()) {
return false;
} String[] s = str.split(" ");
if (pattern.length() != s.length) {
return false;
} // encode pattern
String patternEncoded = this.encodeString(pattern);
// encode the string array
String strEncoded = this.encodeArray(s); // compare
return patternEncoded.equals(strEncoded);
} private String encodeArray(String[] s) {
Map<String, Character> lookup = new HashMap<>(); int index = 0; // starting from 'a'
StringBuilder sb = new StringBuilder();
for (String ss : s) {
if (lookup.containsKey(ss)) {
sb.append(lookup.get(ss));
} else {
char c = (char)('a' + index);
sb.append(c);
index++;
lookup.put(ss, c);
}
}
return sb.toString();
} // encode it to base to a, this is not really encoding, but mapping a char to a completely different one using
// the same order as encodeArray
private String encodeString(String s) {
Map<Character, Character> lookup = new HashMap<>();
int index = 0; // starting from 'a'
StringBuilder sb = new StringBuilder(); for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i); if (lookup.containsKey(c)) {
sb.append(lookup.get(c));
} else {
char t = (char)('a' + index);
sb.append(t);
index++;
lookup.put(c, t);
}
} return sb.toString();
}
Time Complexity: O(N), N is the length of pattern or string array, we loop it 3 times, but still O(N)
Space Complexity: O(N), N is the length of pattern or string array, we need the extra map and string to store the results
Use bijection mapping (Recommended)
// I recommend this solution: just need map to keep the mapping relationship
public boolean wordPattern(String pattern, String str) {
if (pattern == null || pattern.length() == 0 || str == null || str.length() == 0) {
return false;
} String[] strs = str.trim().split(" "); if (pattern.length() != strs.length) {
return false;
} Map<Character, String> lookup = new HashMap<>();
// As it says, it is a bijection, so it needs to be 1 to 1 mapping, cannot exist a case one key maps to different value case
// E.g., need this set for abba, dog dog dog dog -> false case
Set<String> mapped = new HashSet<>(); for (int i = 0; i < pattern.length(); i++) {
char c = pattern.charAt(i); if (lookup.containsKey(c)) {
if (!lookup.get(c).equals(strs[i])) {
return false;
}
} else {
// shit, just know put actually returns a V, which is the previous value, or null if not exist (or an associated null value)
lookup.put(c, strs[i]);
if (mapped.contains(strs[i])) {
return false;
}
mapped.add(strs[i]);
}
} return true;
}
There is also a clever implementation like below. The key point is use the index to compare, if there is duplicate index, meaning there are two keys already mapped to the same value. Also, remember java put() returns a valid, not a void :)
// Reference: https://leetcode.com/problems/word-pattern/discuss/73402/8-lines-simple-Java
public boolean wordPattern(String pattern, String str) {
String[] words = str.split(" ");
if (words.length != pattern.length())
return false;
Map index = new HashMap();
for (Integer i=0; i<words.length; ++i)
if (index.put(pattern.charAt(i), i) != index.put(words[i], i))
return false;
return true;
}
Time Complexity: N is the length of pattern or string array
Space Complexity: O(N), N is the length of pattern or string array, we need a map regardless
References
Baozi Leetcode Solution 290: Word Pattern的更多相关文章
- 【leetcode】290. Word Pattern
problem 290. Word Pattern 多理解理解题意!!! 不过博主还是不理解,应该比较的是单词的首字母和pattern的顺序是否一致.疑惑!知道的可以分享一下下哈- 之前理解有误,应该 ...
- Leetcode solution 291: Word Pattern II
Problem Statement Given a pattern and a string str, find if str follows the same pattern. Here follo ...
- 【一天一道LeetCode】#290. Word Pattern
一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 Given a ...
- 【LeetCode】290. Word Pattern 解题报告(Python)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.c ...
- leetcode 290. Word Pattern 、lintcode 829. Word Pattern II
290. Word Pattern istringstream 是将字符串变成字符串迭代器一样,将字符串流在依次拿出,比较好的是,它不会将空格作为流,这样就实现了字符串的空格切割. C++引入了ost ...
- leetcode面试准备: Word Pattern
leetcode面试准备: Word Pattern 1 题目 Given a pattern and a string str, find if str follows the same patte ...
- [LeetCode] 290. Word Pattern 词语模式
Given a pattern and a string str, find if str follows the same pattern. Here follow means a full mat ...
- LeetCode 290 Word Pattern(单词模式)(istringstream、vector、map)(*)
翻译 给定一个模式,和一个字符串str.返回str是否符合同样的模式. 这里的符合意味着全然的匹配,所以这是一个一对多的映射,在pattern中是一个字母.在str中是一个为空的单词. 比如: pat ...
- [LeetCode] 290. Word Pattern 单词模式
Given a pattern and a string str, find if str follows the same pattern. Here follow means a full mat ...
随机推荐
- ansible(二)
一.软件相关模块 1.yum(下载包) 正常操作 yum 与rpm的区别 yum可以解决依赖关系rpm 全称readhat package manager(红帽包管理工具),需要自己解决依赖 yum源 ...
- python的实用函数
>>> file=open('txt.txt','a') >>> print >> file,'hello,world' >>> fi ...
- excel表格处理
xlrd模块 是python中一个第三方的用于读取excle表格的模块,很多企业在没有使用计算机管理前大多使用表格来管理数据,所以导入表格还是非常常用的! exlce结构分析 一个excle表 ...
- wed前端html/css简单理解
开发工具: txt文本 / dreamwave:DW(cs6/cc) / Hbuilder / webstorm / sublime / vscode 前端: 知识架构: 3层: 结构 / 表现 / ...
- SpringBoot从入门到精通二(SpringBoot整合myBatis的两种方式)
前言 通过上一章的学习,我们已经对SpringBoot有简单的入门,接下来我们深入学习一下SpringBoot,我们知道任何一个网站的数据大多数都是动态的,也就是说数据是从数据库提取出来的,而非静态数 ...
- Python自学day-1
一.Python介绍 1.python擅长领域: WEB开发:Django. pyramid. Tornado. Bottle. Flask. WebPy 网络编程:Twisted(牛 ...
- Angular4.0从入门到实战打造在线竞拍网站学习笔记之一--组件
Angular4.0基础知识之组件 Angular4.0基础知识之路由 Angular4.0依赖注入 Angular4.0数据绑定&管道 最近搞到手了一部Angular4的视频教程,这几天正好 ...
- Appcan 自定义数字加减控件
DIV部分: *这里的三个ID:as_sub_3.as_now_3.as_add_3里面的“3”可以自定义,这个对于生成任意个数的列表形式很有帮助 *cb 为执行成功后可进行回调 <div cl ...
- SSM(五)Mybatis配置缓存
1.在没有配置的情况下,mybatis默认开启一级缓存. Object object=mapper.getXxx(object); Object object2=mapper.getXxx(objec ...
- Xmanager 5远程连接CentOS7图形化界面
1.安装Xmanager 5下载链接:https://pan.baidu.com/s/1JwBk3UB4ErIDheivKv4-NA提取码:cw04 双击xmgr5_wm.exe进行安装 点击‘下一步 ...