v1

[spencerkrebs]

No description provided.

[super30admin]

Interview Problem : Pascal's Triangle (pascals-triangle.py)

Your solution is correct for the given examples and has the right time and space complexity. However, there are a few areas where you can improve:

1. Variable Naming: Use more descriptive variable names. For instance, instead of l, r, and p, consider names like left_index, right_index, and current_index. This makes the code easier to understand.

2. Loop Structure: The while loop can be replaced with a for loop that iterates over the inner indices. This would make the code more straightforward. For example:

   for j in range(1, i-1):
        row[j] = res[-1][j-1] + res[-1][j]
   

   This avoids the need for manual index management and is more idiomatic.

3. Condition Check: The condition if i > 2 is acceptable, but it's better to check based on the row length. Since the row length is i, you can check if i > 2 or simply let the for loop run only when i-1 > 1 (i.e., i>2). Alternatively, you can remove the condition entirely and use a for loop that runs from 1 to i-2, which naturally does nothing when i<=2.

4. Efficiency: Your current approach is efficient, but the code can be simplified. You don't need to create variables l, r, and p and manage them in a while loop. A for loop is simpler.

Here's a revised version of your code that addresses these points:

class Solution:
    def generate(self, numRows: int) -> List[List[int]]:
        res = []
        for i in range(1, numRows+1):
            row = [1] * i
            if i > 2:
                prevRow = res[-1]
                for j in range(1, i-1):
                    row[j] = prevRow[j-1] + prevRow[j]
            res.append(row)
        return res

This version is cleaner and easier to read.

VERDICT: PASS

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Interview Problem: Pairs with K difference (k-diff-pairs.py)

Your solution is efficient and correct. You have used a Counter to count frequencies and then iterated over the unique numbers. For k != 0, you check if x + k exists in the counter, which ensures each pair is counted exactly once. For k=0, you check if the count is at least 2. This is a clean and efficient approach.

Strengths:

• Time and space complexity are optimal (O(n)).
• Code is concise and readable.
• Uses built-in Counter which is appropriate.

Areas for improvement:

• Although your solution is correct, it is worth noting why checking only x+k is sufficient. This might not be immediately obvious to someone reading the code. You could add a comment explaining that for k != 0, we only need to check x+k to avoid duplicate pairs and that this covers all cases because if (a, b) is a pair with a - b = k, then when we process b we will find a = b + k. But since the problem requires unique pairs and we are iterating over distinct numbers, each pair is counted exactly once.
• Alternatively, you could also check for x-k, but then you would have to avoid duplicates. Your current method is better.

Overall, this is a very good solution.

VERDICT: PASS