DEV Community: Nw3965

Corrected Code: List Typo

Nw3965 — Tue, 09 Jan 2024 12:59:51 +0000

1st try.

I remembered that I have to use From~ table name.

SELECT id
From Views WHERE Column Name 4 AND 7 AS id

2nd try with AI.

SELECT DISTINCT V.viewer_id AS id
FROM Views V
JOIN Views A ON V.viewer_id = A.author_id AND V.article_id = A.article_id
ORDER BY id;

Must study "JOIN ~ ON" and "ORDER BY".

Python Slice with For

Nw3965 — Tue, 09 Jan 2024 12:59:33 +0000

TASK 1 - To check if given word is like a Mirror

# e.g. #  [noon],[level],[deed],[civic] [dad] are mirror word, it's the same when read from right to the left.

1st tried by myself.

def check(str):
    # Get input words using "for" method/sentence one by one.
    x = []
    for str in range(x):
        #Check words the 1st character is the same as 4th character.
        if x[:0] == x[:3]:
            return True
        else:
            False

2nd.
Chat GPT

def check_mirror(words):
    # [noon], [level], [deed], [civic], [dad] are mirror words; they are the same when read from right to left.

    # Initialize a list to store the results for each word
    results = []

    # Loop through each word in the list
    for word in words:
        # Check if the word is a mirror word
        if word == word[::-1]:  # Use slicing to reverse the word
            results.append(True)
        else:
            results.append(False)

    return results

# Example usage:
word_list = ["noon", "level", "deed", "civic", "dad", "python"]
result_list = check_mirror(word_list)
print(result_list)

    #  while:
    #  [hello], [good], [bye] are not mirror.
    #   Now, try to check if the word is mirror image.

3rd.
Replit.com AI

def check(word):
    for x in word:
        # Check the word if the 1st character is the same as the 4th character.
        return x[0] == x[3]

#Test case.
word = ["noon", "mom", "dad", "racecar"]

4th

def check(words):
  # Initialize a list to store the results for each word
  x = []

  for word in words:
      # Check if the word is a mirror word (1st character is the same as the 4th character)
      x.append(word[0] == word[-1])

  return x

# Example usage:
word_list = ["noon", "mom", "dad", "free"]
result_list = check(word_list)
print(result_list)

Subsequence Check: Python Code

Nw3965 — Thu, 14 Dec 2023 18:09:41 +0000

1st.

class Solution:
    def isSubsequence(self, s: str, t: str) -> bool:
        for S in range(s):
            len(S)
            for T in range(t):
                len(T) 
                if S == T:
                    return True
                else:
                    return False

2nd with chatGPT

def isSubsequence(s: str, t: str) -> bool:
    i, j = 0, 0

    while i < len(s) and j < len(t):
        if s[i] == t[j]:
            i += 1
        j += 1

    return i == len(s)

# Example usage:
s = "abc"
t = "ahbgdc"
result = isSubsequence(s, t)
print(result)  # Output: True

SQL #1

Nw3965 — Thu, 14 Dec 2023 18:01:40 +0000

SQL
First

SELECT Tweets FROM
 content > 15;

Second with AI.

SELECT tweet_id, content
FROM Tweets
WHERE LENGTH(content) > 15;

Show only tweet_id

SELECT tweet_id
FROM Tweets
WHERE LENGTH(content) > 15;

Big Countries

Nw3965 — Mon, 04 Dec 2023 19:20:01 +0000

LeetCode

1st try.
SELECT area >= 3000000 and population >= 25000000 from World table

2nd try with AI.

SELECT*
FROM World
WHERE area >= 3000000 and population >= 25000000;

It shows unnecessarily column so I had to edit query again.

Message From ChatGPT as follows.

To find the name, population, and area of countries that meet the specified criteria (area at least 3 million km² or population at least 25 million), you can use the following SQL query:

SELECT name, population, area
FROM World
WHERE area >= 3000000 OR population >= 25000000;

This query selects the "name," "population," and "area" columns from the "World" table where the country has an area of at least 3 million km² or a population of at least 25 million. The result will include countries that meet either of these conditions.
Make sure to run this query in your SQL database to get the desired result.

I realized that it can be reduced by notating the columns to be displayed as follows
SELECT name, population, area

Count negatives in grid

Nw3965 — Mon, 04 Dec 2023 18:41:53 +0000

LeetCode

1st try.

class Solution:
    def countNegatives(self, grid: List[List[int]]) -> int:
        #Count negetive numbers.
        i = int<=-1
        negetive = count.List(i)
        return negetive

2nd try with AI.

from typing import List

class Solution:
    def countNegatives(self, grid: List[List[int]]) -> int:
        # Count negative numbers.
        negative_count = 0

        for row in grid:
            for num in row:
                if num < 0:
                    negative_count += 1

        return negative_count

What does this mean ?
for row in grid:
for num in row:

ChatGPT
for row in grid:
for num in row:
is a nested loop structure that iterates over each element in a 2D grid.

Here's a breakdown:

for row in grid:: This loop iterates over each row in the 2D grid. In each iteration, row represents a list of numbers (a row in the grid).

for num in row:: This inner loop iterates over each element (num) in the current row (row). In each iteration of the inner loop, num represents a specific number in the current row.

So, when you combine these loops, the entire structure iterates over each element in the 2D grid. This is commonly used to perform operations on each individual element or to extract information from the entire grid. In the context of your original code, it is used to count the number of negative numbers in the grid by checking each element (num) in each row (row).

Find Customer Referee

Nw3965 — Thu, 30 Nov 2023 20:32:44 +0000

Find Customer Referee

I wanted to write SQL statements to get the information except "referee_id == 2" .

SELECT   referee_id  FROM Customer
EXCEPT  referee_id == 2

I should select "name" to get the correct answer.

SELECT name FROM Customer
WHERE referee_id IS NULL OR referee_id <> 2;

I learned <> is for not equal.

Search Insert Binary Python

Nw3965 — Thu, 30 Nov 2023 20:01:29 +0000

Search Insert Position

I wanted to count numbers inside of "target" and I could not write what I wanted.

class Solution:
    def searchInsert(self, nums: List[int], target: int) -> int:
        i = count.target.index
        return i

AI checked this code and answered like this.

class Solution:
    def searchInsert(self, nums: List[int], target: int) -> int:
        try:
            i = nums.index(target)
        except ValueError:
            nums.append(target)
            nums.sort()
            i = nums.index(target)
        return i

I have to practice try except.

GCD of Strings Fix

Nw3965 — Thu, 30 Nov 2023 19:30:07 +0000

GDC, Greatest Common Divisor.

1st.

class Solution:
    def gcdOfStrings(self, str1: str, str2: str) -> str:
        #get the same string on str1 and str2 and print
        result =""
        i,j = 0,0
        while i < len(str1) and j < len(str2):
            result = str1[i]
            result = str2[j]
            i += 1
            j += 1
        result += str1[i:]
        result += str2[j:]
        return result

I thought I could get through this coding test by using the same code we used last time. I was not sure how to write the prefix part, so I asked AI, but I could not get the correct answer.

I tried editing the code several times but I could not pass this test case.

Input
str1 =
"ABABAB"
str2 =
"ABAB"
Output
"AB"
Expected
"AB"

This comment helped me.
https://leetcode.com/problems/greatest-common-divisor-of-strings/description/comments/1782570

I learned that it is better to divide a given string evenly.

class Solution:
    def gcdOfStrings(self, str1: str, str2: str) -> str:
        if str1 == str2:
            return str1

        # Find the length of the shorter string
        n = min(len(str1), len(str2))

        # Iterate over candidate string lengths
        for i in range(n, 0, -1):
            # Check if the candidate string is a divisor of both strings
            if len(str1) % i == 0 and len(str2) % i == 0:
                candidate = str1[:i]
                if candidate * (len(str1) // i) == str1 and candidate * (len(str2) // i) == str2:
                    return candidate

        return ""

SELECT DISTINCT

Nw3965 — Wed, 29 Nov 2023 22:31:10 +0000

From Leetcode Recyclable and Low Fat Products

I used SQL statement with C# and CSOM library but I have not used to SQL coding test so now I am practicing SQL coding test.

yesterday I learned SELECT DISTINCT statement. This statement can be used to pickup designated word like this.

SELECT DISTINCT product_id
From Products
  WHERE
  low_fats = 'Y' AND
  recyclable = 'Y';

It's very useful.

SQL SELECT DISTINCT

Binary Search

Nw3965 — Wed, 29 Nov 2023 22:25:32 +0000

Binary Search

The code I wrote.

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        result = count(nums)
        return  result

Checked by AI.


class Solution:
    def search(self, nums: List[int], target: int) -> int:
        result = nums.count(target)
        return result

I realized that I should write like this.
result = nums.count(target)

I could not pass the coding test so I asked AI again to check.

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        try:
            result = nums.index(target)
            return result
        except ValueError:
            return -1

I needed exception.

except ValueError: return -1

Question;User
what is the difference between nums.index(target) and nums.count(target)

Answer from ChatGPT

Let's use the list [-1, 0, 3, 5, 9, 12] and demonstrate the use of both nums.index(target) and nums.count(target) with this specific list:

nums = [-1, 0, 3, 5, 9, 12]
target = 9

# Using nums.index(target)
try:
    index = nums.index(target)
    print(f"The target {target} is at index {index}.")
except ValueError:
    print(f"The target {target} is not in the list.")

# Using nums.count(target)
count = nums.count(target)
print(f"The target {target} appears {count} times in the list.")

Output:

perl

The target 9 is at index 4.
The target 9 appears 1 times in the list.

In this example, nums.index(target) returns the index of the first occurrence of 9, which is 4. nums.count(target) returns the number of occurrences of 9, which is 1. Both methods are used to gather information about the presence and location of the target value in the list.

So I should use "index" to count index number for this test.

nums.count(target) returns the number of occurrences of 9,

I could not write this "Try Except" so I need to practice this also.
Python Try Except

Merge Words Algorithm

Nw3965 — Wed, 29 Nov 2023 19:17:43 +0000

Leetcode
Link

I need to input the word2 into the some of word1.I need to insert one word from word1 and the next should be word2 and word1 ...

I want to write picking up one word from word1 and insert between word2.

class Solution:
    def mergeAlternately(self, word1: str, word2: str) -> str:
        for i in range(word1):

I could not continue to write code from here.

class Solution:
    def mergeAlternately(self, word1: str, word2: str) -> str:
#Picking up one word from word2
        for i in range(word2): 
#Insertt word2 into word1.

Got this from AI.

class Solution:
    def mergeAlternately(self, word1: str, word2: str) -> str:
        result = ""
        i, j = 0, 0
        while i < len(word1) and j < len(word2):
            result += word1[i]
            result += word2[j]
            i += 1
            j += 1

        result += word1[i:]
        result += word2[j:]
        return result

Question; Can you use for loop to write that?

Answer from AI as follows.

class Solution:
    def mergeAlternately(self, word1: str, word2: str) -> str:
        result = ""

        min_length = min(len(word1), len(word2))
        for i in range(min_length):
            result += word1[i]
            result += word2[i]

        for i in range(min_length, len(word1)):
            result += word1[i]

        for i in range(min_length, len(word2)):  
            result += word2[i]

        return result

Using for loop makes longer.

Question; Could you explain the details of this code? How can I write this for programing beginner.
Answer from AI as follows.

The key steps are:
Initialize a result string to store the final merged string:
Copy code
result = ""
Initialize indexes i and j to keep track of current position in word1 and word2. Set both to 0 to start from beginning:
Copy code
i, j = 0, 0
Use a while loop to iterate through both words simultaneously. We want to merge characters until either word1 or word2 has reached its end:
Copy code
while i < len(word1) and j < len(word2):
In each iteration, append the next character from word1 to result:
Copy code
result += word1[i]
Then append the next character from word2:
Copy code
result += word2[j]
Increment both indexes to progress through the strings:
Copy code
i += 1
j += 1
After the while loop, one of the words may still have characters left. Append remainder of word1 using slicing:
Copy code
result += word1[i:]
Append remainder of word2:
Copy code
result += word2[j:]
Return the merged result string.
The key ideas are using indexes to keep track of current positions, mergingCharacter by character in a loop, and handling the remaining parts of longer string separately.
Let me know if any part is unclear! I'm happy to explain further for a beginner audience.

1 Declare the word "result" and have it recognized.
2 Add Initial number as zero.
3 Use while loop for this time to get words.
4 Add words one by one.