class Solution:
    def lengthOfLastWord(self, s: str) -> int:
        word_list = s.split()
        last_word = word_list[-1]
        lw_count = len(last_word)
        return(lw_count)

Explanation:

• First we split the words by using split. This will remove all of the spaces if there are any in s
• We then grab the last word by indexing the last word word_list[-1]
• We then get the count of the last word by using len and return the count of it. This is just a simple solution to the problem. We can actually trim the answer if needed to the following:

  class Solution:
    def lengthOfLastWord(self, s: str) -> int:
        s = s.rstrip()  #or strip
        return len(s.split()[-1])
  

class Solution:
    def addBinary(self, a: str, b: str) -> str:

Explanation:

• We use int to convert the string to an integer and use base 2
• We use format to convert the integer back into binary without the "0b" Another way that we can solve this is by using bin Similarly we just remove the first 2 strings.

class Solution:
    def addBinary(self, a: str, b: str) -> str:
        return bin(int(a , 2) + int(b,2))[2:]

class Solution:
    def getRow(self, rowIndex: int) -> List[int]:
        result = []
        for i in range(0, rowIndex+1):
            result.append([])

            for x in range(0, i + 1):
                if x == 0:
                    result[i].append(1)
                elif i == x:
                    result[i].append(1)
                else:
                    result[i].append((result[i-1][x-1]) + (result[i-1][x]))
        return result[rowIndex]

Explanation: This problem asks you to return the rowIndex-th (0-based index) row of Pascal's Triangle as a list of integers. Pascal's Triangle is a triangular array of binomial coefficients, where each number is the sum of the two numbers directly above it. The first few rows of Pascal's Triangle look like this:

Row 0: [1]
Row 1: [1, 1]
Row 2: [1, 2, 1]
Row 3: [1, 3, 3, 1]

1. We create an empty list called result to store the rows of Pascal's Triangle. Create an empty list called result to store the rows of Pascal's Triangle.

2. We then iterate over the rows from 0 to rowIndex, inclusive, using the variable i to represent the current row index.

3. For each row, append an empty list to the result list. This empty list will be used to store the elements of the current row.

4. Inside the inner loop, iterate over the elements in the current row from 0 to i, using the variable x to represent the current column index.

5. For each element, check if it is the first element in the row (i.e., x == 0) or the last element in the row (i.e., i == x). If it is either the first or last element, append a 1 to the current row because the first and last elements of each row in Pascal's Triangle are always 1.

6. If the element is not the first or last element, calculate its value by adding the element from the previous row in the same column (result[i-1][x-1]) and the element from the previous row in the next column (result[i-1][x]). This follows the rule of Pascal's Triangle where each element is the sum of the two elements above it.

7. Append the calculated value to the current row.

8. Repeat steps 5-7 for all elements in the current row.

9. Once the inner loop finishes, the current row is complete, and you move on to the next row.

10. Finally, return the row at the rowIndex index from the result list.

        class Solution:
    def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
        while m > 0 and n > 0:
            if nums1[m-1] > nums2[n-1]:
                nums1[m+n-1] = nums1[m-1]
                m -= 1
            else:
                nums1[m+n-1] = nums2[n-1]
                n -= 1
        for i in range(n):
            nums1[i] = nums2[i]

Explanation: To go through this solution, we will go through line by line.

1. starting with the while loop. It will continue as long as both m and n are greater than 0. This will help merge the two arrays.

2. Inside the loop, we then compare the last element of nums1 at a index of m-1 with the last element of nums2 at index n-1. These are the largest elements of each array.

3. If the element in nums1 is greater, it means that this element should be placed at the end of the merged array, which is at index m+n-1 in nums1. So, it assigns the value of nums1[m-1] to nums1[m+n-1]. This merges the element from nums1 into the array.

4. After merging an element from nums1 it decreases m by 1 to move the m pointer to the previous element in nums1

5. If the element in nums2 is greater or equal to nums1 then that means that this element should be placed at the end of the merged array.

6. From the line aboves if statement , we then assign the value of nums2[n-1] to nums1[m+n-1] to merge the elements from nums2 to nums1

7. After merging an element from nums2 it decrements n by 1 to move the n pointer to the previous element in nums2 n -= 1

8. After the while loop exits, there might be remaining elements in nums2 that were not merged. This for loop iterates over the remaining elements of nums2 from index 0 to n-1 for i in range(n)

9. In the for loop, it copies the remaining elements from nums2 into nums1 by merging the remaining elements from nums2 to into nums1 nums1[i] = nums2[i]

The key idea here is to work from the end of the arrays towards the beginning, which avoids overwriting elements in nums1 before they are compared and merged. This approach ensures that the merged array is sorted without the need for extra space or creating a new array.

        class Solution:
    def convertToTitle(self, columnNumber: int) -> str:
        alphabet="ABCDEFGHIJKLMNOPQRSTUVWXYZ"
        result=""
        while columnNumber:
            columnNumber=columnNumber-1
            result=alphabet[columnNumber%26]+result
            columnNumber=columnNumber//26
        return result

Explanation: This problem asks you to convert a positive integer, columnNumber, into an Excel sheet column title. Excel column titles are represented using uppercase English letters, and they follow a pattern similar to base 26 numbering, where the digits are represented by the English alphabet (A=1, B=2, ..., Z=26), and when the column number exceeds 26, it starts using two-letter combinations (AA=27, AB=28, ..., ZZ=702, AAA=703, and so on).

1. We first define alphabet that contains all uppercase letters from A-Z. These strings will be used to map column numbers to titles.

2. We then initializes an empty string result to store the Excel column title

3. We start a while loop that continues as long as columnNumber is not zero. The loop will gradually convert the column number to column title.

4. Inside of the loop, it subtracts 1 from the columnNumber. This is done to handle the fact that the Excel column numbering starts from 1, but our algorithm will work with 0-based indexing.

5. result = alphabet[columnNumber % 26] + result

This calculates the remainder when columnNumber is divided by 26. The remainder corresponds to a letter in the alphabet. It then takes that letter at the position in the alphabet string and appends it to the beginning of the result string. This builds the Excel column title from right to left.

1. columnNumber = columnNumber // 26

It updates columnNumber by performing an integer division by 26 (columnNumber // 26). This reduces columnNumber to the next lower place value.

The loop will continue with the reduced columnNumber and the next letter is added to the result string.

This process continues until columnNumber becomes zero, at which point we have constructed the complete Excel column title in the result string.

1. We finally return the result string which contains the Excel column title corresponding to the input columnNumber.

This algorithm effectively converts a decimal number into a base 26 representation using the English alphabet letters and builds the Excel column title accordingly.

class Solution:
    def strStr(self, haystack: str, needle: str) -> int:
        while needle in haystack:
            res = haystack.index(needle)
            return res
        else:
            return -1

Explanation

        class Solution:
    def strStr(self, haystack: str, needle: str) -> int:
        while needle in haystack:
            res = haystack.index(needle)
            return res
        else:
            return -1

Explanation

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        if sorted(s) == sorted(t):
            return True
        else:
            return False

Explanation

class Solution:
    def isIsomorphic(self, s: str, t: str) -> bool:
        if len(s) != len(t):
            return False
        else:
            d1, d2 = {}, {}
            for i in range(len(s)):
                str1, str2 = s[i], t[i]
                if str1 not in d1:
                    d1[str1] = str2
                if str2 not in d2:
                    d2[str2] = str1
                if d1[str1] != str2 or d2[str2] != str1:
                    return False
        return True

Explanation

        class Solution:
    def generate(self, numRows: int) -> List[List[int]]:
        result = []
        for i in range(0, numRows):
            result.append([])

            for x in range(0, i + 1):
                if x == 0:
                    result[i].append(1)
                elif i == x:
                    result[i].append(1)
                else:
                    result[i].append((result[i-1][x-1]) + (result[i-1][x]))
        return result

Explanation

        class Solution:
    def moveZeroes(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        for i in nums:
            if i == 0:
                nums.remove(i)
                nums.append(0)

Explanation

        def reverseVowels(self, s: str) -> str:
    vowels = re.findall('[aeiouAEIOU]', s)
    return re.sub('[aeiouAEIOU]', lambda _ : vowels.pop(), s)

Explanation

        class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        if not prices:
            return 0

        maxProfit = 0
        minPurchase = prices[0]
        for i in range(1, len(prices)):        
            maxProfit = max(maxProfit, prices[i] - minPurchase)
            minPurchase = min(minPurchase, prices[i])
        return maxProfit

Explanation

        class Solution:
    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:

        merged = ListNode() # Create an empty ListNode to serve as the dummy node for the merged list.
        current = merged # Initialize a pointer 'current' to the dummy node.
        # The while loop iterates as long as there are elements in both list1 and list2. Inside the loop, we compare the values of the current nodes in both lists
        while list1 and list2: 
            if list1.val <= list2.val:  # Append the smaller value from list1.
                current.next = list1  # Move the pointer in list1 to the next node.
                list1 = list1.next # Move the pointer in list1 to the next node.
            else:
                current.next = list2 # # Append the smaller value from list2.
                list2 = list2.next # Move the pointer in list2 to the next node.
            current = current.next  # Move the 'current' pointer to the newly appended node.
        # Append any remaining elements from list1 or list2
        if list1:
            current.next = list1 
        elif list2:
            current.next = list2
        return merged.next  # Return the merged list starting from the first node.

Explanation merged = ListNode(): We create an empty ListNode named merged to serve as the dummy node for the merged list. This dummy node helps simplify the merging process by providing a starting point for the merged list.

current = merged: We initialize a pointer current to the dummy node merged. This pointer will be used to traverse and build the merged list.

The while loop iterates as long as there are elements in both list1 and list2. Inside the loop, we compare the values of the current nodes in both lists and append the node with the smaller value to the merged list.

current.next = list1 or current.next = list2: We update the next attribute of the current node to point to the smaller node between list1 and list2. This effectively appends the smaller node to the merged list.

list1 = list1.next or list2 = list2.next: We move the pointer in the corresponding input list (list1 or list2) to the next node, as we've already appended the current node to the merged list.

current = current.next: We move the current pointer to the newly appended node in the merged list so that the next iteration appends the next smallest node.

After the loop completes, we check if there are any remaining elements in list1 or list2 and append them to the merged list if necessary.

Finally, we return the merged list starting from the node after the dummy node (merged.next) to exclude the dummy node from the final result.

        class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        res=[]
        for i in range(len(nums)):
            for a in range(i+1,len(nums)):
                if (nums[i]+nums[a]==target):
                    res.append(i)
                    res.append(a)
                    break     
        return res

Explanation

        class Solution:
    def isPalindrome(self, x: int) -> bool:
        a = ''.join(reversed(str(x)))
        if a == str(x):
            return True
        else:
            return False

Explanation

        class Solution:
    def isPalindrome(self, x: int) -> bool:
        a = ''.join(reversed(str(x)))
        if a == str(x):
            return True
        else:
            return False

Explanation

        class Solution:
    def isValid(self, s: str) -> bool:
        while '()' in s or '[]'in s or '{}' in s:
            s = s.replace('()','').replace('[]','').replace('{}','')
        if len(s) !=0:
            return False
        else:
            return True

Explanation

class Solution:
    def floodFill(self, image: List[List[int]], sr: int, sc: int, color: int) -> List[List[int]]:
        if image[sr][sc] == color:
            return image
        def dfs(row, col, startColor):
            if row < 0 or row >= len(image) or col < 0 or col >= len(image[0]) or image[row][col] != startColor:
                return
            image[row][col] = color

            dfs(row + 1 , col, startColor)
            dfs(row - 1, col, startColor)
            dfs(row, col + 1, startColor)
            dfs(row, col - 1, startColor)
        startColor = image[sr][sc]
        dfs(sr, sc, startColor)
        return image

Explanation

1. Check if the newColor is the same as the color at the starting point (sr, sc). If they are the same, there's no need to perform the flood fill, so return the original image.

2. Define a depth-first search (DFS) function, dfs, which takes the current row and column and the starting color as parameters. This function recursively explores neighboring pixels and changes their color to the newColor.

3. Inside the dfs function:

   • Check if the current pixel is out of bounds or if it doesn't have the same color as the starting color. If any of these conditions are met, return, as there's no need to visit or change the color of this pixel.
   • Change the color of the current pixel to the newColor.
   • Recursively call dfs on the neighboring pixels in all four directions (up, down, left, right).

4. Set the startColor variable to the color of the starting pixel (sr, sc).

5. Call the dfs function to start the flood fill operation from the (sr, sc) point.

6. Return the modified image after the flood fill is complete.

In this problem, you're required to design a simple HashSet data structure, supporting basic operations like insertion, deletion, and checking for the presence of an element. While this problem might seem straightforward, it has some underlying connections to system design and architecture concepts:

1. Data Modeling and Storage: When designing a HashSet, you need to think about how to organize and store the data efficiently. This can relate to database schema design in a larger system, where you would consider how to store and access data optimally.

2. Data Access Optimization: In a larger system, quick access to data is crucial. When designing a HashSet, you're challenged to implement efficient lookup and manipulation operations, similar to how efficient data retrieval is a key consideration in designing systems.

3. Collision Handling: Hash collisions can occur when multiple elements map to the same hash value. This connects to concepts of distributed systems and hashing techniques, which are relevant when handling data across multiple servers.

4. Concurrency and Consistency: While the problem might not explicitly require it, in a distributed system, you would need to consider issues like concurrent access and maintaining data consistency. This mirrors challenges faced in distributed databases and systems.

5. Scalability: While the problem doesn't specifically touch on this, when designing a HashSet for a large-scale system, you'd need to think about how to make the data structure scalable, possibly by sharding or partitioning data across different nodes.

6. Data Integrity and Error Handling: Ensuring that your HashSet functions correctly and handles errors gracefully is akin to ensuring data integrity and handling exceptions in a real-world system.

While this problem isn't a full-blown system design challenge, it provides a microcosm of considerations that come into play when designing and implementing data structures in a larger system. It's about making design choices that optimize for performance, reliability, and scalability—core tenets of system design and architecture.

Design a HashSet without using any built-in hash table libraries.

Implement MyHashSet class:

• void add(key) Inserts the value key into the HashSet.
• bool contains(key) Returns whether the value key exists in the HashSet or not.
• void remove(key) Removes the value key in the HashSet. If key does not exist in the HashSet, do nothing

Example 1:

Input

["MyHashSet", "add", "add", "contains", "contains", "add", "contains", "remove", "contains"]
[[], [1], [2], [1], [3], [2], [2], [2], [2]]

Output

[null, null, null, true, false, null, true, null, false]

Explanation

MyHashSet myHashSet = new MyHashSet();
myHashSet.add(1);      // set = [1]
myHashSet.add(2);      // set = [1, 2]
myHashSet.contains(1); // return True
myHashSet.contains(3); // return False, (not found)
myHashSet.add(2);      // set = [1, 2]
myHashSet.contains(2); // return True
myHashSet.remove(2);   // set = [1]
myHashSet.contains(2); // return False, (already removed)

Constraints:

• 0 <= key <= 106
• At most 10⁴ calls will be made to add, remove, and contains.