opening

how are you doing?
I’m in San Diego right now.
- I’m currently based in San Diego.
I’m totally comfortable with relocation.
- Yes, I’m definitely open to relocating to Florida. I’ve always liked the vibe there — the beaches, the palm trees, the warm weather — it’s a place I’d genuinely enjoy living in.”
  - I’m definitely open to the West Coast. I’ve always liked the mix of tech and nature — like being close to the ocean or mountains is pretty awesome.
  - Yeah, I’d be cool with moving to the East Coast. I kind of enjoy having all four seasons.
  - Yeah, I’m open to the Midwest. I’ve heard Austin and Denver have a great food and music scene — I think I’d really enjoy living there.
when interviewer is late:
- They might say: Sorry, I’m late
  - u can say: no worries.
- That’s totally fine. I was a little anxious when I first joined and waited for a while, but now I’m feeling much more relaxed now It’s a good thing.
- No problem at all, I completely understand. I really appreciate you taking the time to speak with me today. I’m looking forward to our conversation.

when their audio was breaking up or not coming through clearly

I think u were cutting out.

Is this a very common saying in america?

Yes, “cutting out” is a very common and natural expression in American English, especially during phone or online voice/video calls.

You’ll often hear people say:

“You’re cutting out.”
“I think you cut out for a second.”
“Can you hear me? You were cutting in and out.”

These phrases are used to describe intermittent audio loss or unstable connection. They’re informal but widely used in both casual and professional settings.

coding

explain code

692. Top K Frequent Words

Given an array of strings words and an integer k, return the k most frequent strings.

Return the answer sorted by the frequency from highest to lowest. Sort the words with the same frequency by their lexicographical(/leksɪkəˈɡræfɪkl/ adj. 辞典编纂的) order.

Example 1:

Input: words = ["i","love","leetcode","i","love","coding"], k = 2
Output: ["i","love"]
Explanation: “i” and “love” are the two most frequent words. Note that “i” comes before “love” due to a lower alphabetical order.

Example 2:

Input: words = ["the","day","is","sunny","the","the","the","sunny","is","is"], k = 4
Output: ["the","is","sunny","day"]
Explanation: “the”, “is”, “sunny” and “day” are the four most frequent words, with the number of occurrence being 4, 3, 2 and 1 respectively.

Summary:

ok, let me read the problem statement …
to solve this problem,
First, I use a HashMap to count the frequency of each word.
Then, I use a min-heap with a fixed size of k to find the top k frequent words. the heap will store these words in an ascending frequency order, and if two words’ frequency are the same, the heap will sort them in reverse alphabetical order.
Next, I’m gonna iterate through the map, and In each iteration if the current word’s frequency is larger than the top element of the heap, or if their frequency is equal, I will compare these two words with their alphabetical order, if the current word’s alphabetical order is smaller, I insert it into the heap.
after processing, I convert this heap into a list, and because it’s a min-heap, I will reverse the list to get the right order.


import java.util.*;

class Solution {
    public List<String> topKFrequent(String[] ws, int k) {
        // first, we should add a null check
        if (words == null || words.length == 0 || k <= 0) {
            return new ArrayList<>();
        }

        // first, let me Create a HashMap with string keys and integer values, I use it to count how many times each word appears for later calculations.
        Map<String, Integer> map = new HashMap<>();
        // let's iterate through the given string array
        for (String w : ws) {
            // and store each string as a key and its frequency as the value in the map.
            map.put(w, map.getOrDefault(w, 0) + 1);
            // This map now contains each word and its frequency
        }

        // Does that make sense so far?

        // Create a priority queue, which is a min-heap, 
        // ordered first by frequency(sort it based on how many times each word appears),
        // and for those words with the same frequency, sort them in reverse alphabetical order.
        PriorityQueue<Object[]> q = new PriorityQueue<>(k, (a, b) -> {
            int c1 = (Integer) a[0], c2 = (Integer) b[0];
            // If their frequencies is not the same, sort them in an increasing/ascending frequency order
            if (c1 != c2) return c1 - c2;
            String s1 = (String) a[1], s2 = (String) b[1];
            // If frequencies are equal, sort by reverse lexicographical/alphabetical order
            return s2.compareTo(s1);
        });

        // so far so good ? 到目前为止还顺利吗？

        // then let's Go through each word in the map
        // then let's Go through the map
        for (String s : map.keySet()) {
            int cnt = map.get(s); // Get the frequency of the current word

            if (q.size() < k) {
                // If the heap hasn't reached size k, just add the element to the PriorityQueue
                q.add(new Object[]{cnt, s});
            } else {
                // let's Look at the top element of the heap (the "smallest" one)
                Object[] peek = q.peek();
                // because "the peek at index zero"("peek zero") is a object, we have to cast it to an integer
                if (cnt > (Integer) peek[0]) {
                    // If current word's frequency is higher than the smallest in heap, replace it
                    q.poll(); // Remove the smallest element
                    q.add(new Object[]{cnt, s}); // Add the new word to the PriorityQueue
                } else if (cnt == (Integer) peek[0]) {
                    // If frequencies are equal, compare the words in their lexicographical order
                    // let's cast "the peek at index one"("peek one") to a String,
                    // then assign it to the variable top.
                    String top = (String) peek[1];
                    if (s.compareTo(top) < 0) {
                        // If current word comes before the top word in dictionary order, replace the top word with current word
                        q.poll();
                        q.add(new Object[]{cnt, s});
                    }
                }
            }
        }

        // maybe we can `factor out` a method for the above block of code? **提取**出一个方法

        // let's define a array list with string values called "answer"
        List<String> ans = new ArrayList<>();
        // go through the priority queue and add each string to the array list
        while (!q.isEmpty()) {
            // Since it's a min-heap, we need to reverse the result later
            ans.add((String) q.poll()[1]);
        }
        // finally, let's reverse the array list to get descending frequency order, in this way, we can make the highest frequency words come first
        Collections.reverse(ans);
        return ans;
    }
}

public class Main {
    public static void main(String[] args) {
        // let's create a Solution object "sol"
        Solution sol = new Solution();
        // next, define a List with string values called "result"
        // then we use the object 'sol' to call its topKFrequent method , and then pass a String array and a integer 4 to this method
        List<String> res = sol.topKFrequent(new String[]{"the","day","is","sunny","the","the","the","sunny","is","is"}, 4);
        // finally, let me print out the result array list
        System.out.println(res);

        // I should comment out this line, it's useless now.
        Solution sol2 = new Solution();

        // finally, let's analyze the Time Complexity: O(nlogk)
        // I think the time complexity is big O of n log k, the n is number of unique words, and heap operations are big O of log k

        // Space Complexity: O(n)
        // its space complexity is big O of n, because we define a hash map and a priority queue.
    }
}

// class Pair {
//     int frequency = 0;
//     String word;

//     public Pair(int freq, String w) {
//         frequency = freq;
//         word = w;
//     }
// }

Digits

0.5 → zero point five
-0.5 → “negative zero point five” or “minus zero point five”
½ → “one half”
⅓ → “one third”
¼ → “one quarter” or “one fourth”
¾ → “three quarters”
⅔ → “two thirds”
1/5 -> one fifth”**
2/5 → “two fifths”
3/5 → “three fifths”
4/5 → “four fifths”

This style (ordinal denominator with “-th” or “-ths”) is standard for fractions where the denominator is 5 or higher (except for 2 and 4, which are commonly “half” and “quarter”).

标点符号

1. 标点符号英文表达

中文名称	英文名称	符号	英文示例及用法说明
冒号	Colon	:	用于引出解释、列表或引用。 Example: She said: “I love coding.”
感叹号	Exclamation `/ˌeks.kləˈmeɪ.ʃən/` mark	!	表示强烈情感或强调。 Example: What a beautiful day!
问号	Question mark	?	用于疑问句结尾。 Example: Where are you going?
分号	Semicolon	;	分隔并列分句或复杂列表项。 Example: I need to buy apples, bananas; and oranges.
逗号	Comma	,	分隔句子成分、列举项或引导从句。 Example: He is tall, smart, and kind.

2. 关键字的英文表达

关键字 的英文是 Keyword（单数）或 Keywords（复数）。
- 用法：在编程、搜索引擎优化（SEO）、写作中，指具有特定含义或功能的词。
- 示例：
  - In Python, “if”, “for”, and “while” are keywords.（在Python中，“if”、“for”和“while”是关键字。）
  - Use relevant keywords in your essay to improve searchability.（在文章中使用相关关键字以提高可搜索性。）

补充：常见符号的英文读法

@：At sign / At symbol（如邮箱地址中）
Example: user.name@domain.com → “user dot name at domain dot com”
#：Hash(In programming) / Number sign（美式英语）/ Pound sign（英式英语）
Example: #topic → “hash topic” (in Python/Java)
$：Dollar sign
*: asterisk
%：Percent sign
&：Ampersand(/ ˈæmpərsænd /)
/: slash
-: dash
_: underscore
~: tilde (/ˈtɪl.də/)
()：Parentheses /pəˈren.θə.sɪs/（括号，单数为 parenthesis）
- ( : left parenthesis
[]：Brackets / Square brackets
{}：Braces / Curly brackets / Braces
>=: “greater than or equal to”

英文加减乘除怎么说

英文中表示加减乘除的常见词汇和表达方式如下，以具体例子说明：

1. 加法（Addition）

词汇：add（动词）、plus（介词）、addition（名词）
例子：
- 8 + 3
  - 读作：Eight plus three 或 Eight add three
  - 结果表达：Eight plus three equals eleven（8 + 3 = 11）。
  - 结果表达：Eight plus three is equal to eleven（8 + 3 = 11）。

2. 减法（Subtraction）

词汇：subtract（动词）、minus（介词）、subtraction（名词）
例子：
- 8 - 3
  - 读作：Eight minus three 或 Three subtracted from eight
  - 结果表达：Eight minus three equals five（8 - 3 = 5）。

3. 乘法（Multiplication）

词汇：multiply（动词）、times（介词）、multiplication（名词）
例子：
- 8 × 3
  - 读作：Eight times three 或 Eight multiplied by three
  - 结果表达：Eight times three equals twenty-four（8 × 3 = 24）。

4. 除法（Division）

词汇：divide（动词）、divided by（介词短语）、division（名词）
例子：
- 8 ÷ 3
  - 读作：Eight divided by three
  - 结果表达：
    - 若为整数除法：Eight divided by three is two with a remainder of two（8 ÷ 3 = 2 余 2）。
    - 若为小数结果：Eight divided by three equals approximately two point six six six*（8 ÷ 3 ≈ 2.666）。
    - Eight divided by three is about 2.667.

其他常用表达

等于：equals / is equal to / is
- Eight plus three equals eleven（8 + 3 = 11）。
- Eight plus three is equal to eleven（8 + 3 = 11）。
- Eight plus three is eleven（8 + 3 = 11）。
总和：the sum of
- 例：The sum of eight and three is eleven（8 + 3 = 11）。
总和：the difference between
- 例：The difference between eight and three is five（8 - 3 = 5）。
乘积：the product of
- 例：The product of eight and three is twenty-four（8 × 3 = 24）。
商：the quotient(/ˈkwoʊʃ(ə)nt/) of
- 例：The quotient of eight and three is approximately two point six six six（8 ÷ 3 ≈ 2.666）。

真题-java代码表达式

有个java代码表达式是这样的: i *= i /= i , 英语应该怎么读?
- i times-equals i divided-by-equals i
x += y -= x *= y
- x plus equals y minus equals x times equals y.
a +-*/ b
- we should say: Here, the plus, minus, asterisk and slash operators are used.
a === b
- a triple equals b
a = b == c ? d : e
- a equals b double equals c question mark d colon e
i++
- i plus plus
'5' + '2' = '52'
- We concatenate(/kənˈkæt(ə)nˌeɪt/) strings ‘5’ and ‘2’ to get ‘52’.
- String ‘5’ is concatenated with string ‘2’ to get ‘52’
- String ‘5’ concatenated with string ‘2’ becomes ‘52’
- The string ‘5’ and ‘2’ are concatenated to get ‘52’.
- The values ‘5’ and ‘2’ are concatenated into ‘52’.
i++ + ++i
- i post-increment plus pre-increment i.
x = (y = (z = 5))
- x equals y equals z equals five.
a &= b |= c ^= d
- a bitwise and equals b bitwise or equals c bitwise xor equals d.
i = (j = (k = (l = 0)))
- i equals j equals k equals l equals zero.

ask interviewer some questions

What is your typical day like?
Can I ask what your typical day looks like on this team?
I wonder, like, what’s the daily routine work like? ‘Cause I remember it’s a hybrid, right? So I’m kind of interested in, like, the daily routine of the team.
- Like all the things you said, I’m pretty familiar with them. Like a daily stand-up — at Costco, we do it every day, so we can align our expectations among all the people, so that we can keep each other on track with our work. So everything you said is pretty familiar to me. And I think once I can get on board, I can start working really easily and quickly.

ending

Yeah I really appreciate your time today and uh yeah thank you for taking your time to interview with me.
it was really great to meet you.
great meeting you.
Have a good rest of your day.
(if it’s only a vendor interview)
- thank u for your time and what’s the next step for the interview?