add interview-3 solutions

[nikhylw]

Solution 1 (k-diff pairs) - W3_interview_k-diff-pairs.py
Solution 2 (pascals) - W3_interview_pascals.py

[super30admin]

Interview Problem : Pascal's Triangle (W3_interview_pascals.py)

Strengths:

• Clear, methodical approach with three distinct phases
• Good use of descriptive comments
• Clean variable naming
• Correctly handles edge cases (numRows=1 works correctly)

Areas for Improvement:

• The solution could be slightly more efficient by directly building rows instead of initializing with zeros and then overwriting. For example, directly pushing {1} for first row, then building subsequent rows by adding elements.
• The comment "Insert 0s for all the rows in increasing order" is slightly misleading - you're not inserting zeros, you're creating rows of zeros.
• Could add type hints for better code documentation (e.g., def generate(self, numRows: int) -> List[List[int]]:)

Minor Optimization Opportunity:
The reference solution is more space-efficient during construction as it builds rows incrementally. Your approach pre-allocates all rows with zeros, which is fine but slightly less elegant. However, both approaches have the same final space complexity.

VERDICT: PASS

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Interview Problem: Pairs with K difference (W3_interview_k-diff-pairs.py)

Strengths:

• Excellent single-pass solution that is more efficient than the reference
• Clean, readable code with good variable naming
• Correctly handles all edge cases including k=0
• Good use of sets for tracking seen numbers and unique pairs
• Well-commented with time/space complexity

Areas for Improvement:

• The solution is already well-optimized. One minor consideration: for very large arrays with small k values, the approach is already near-optimal.
• Could consider adding type hints for better code documentation (e.g., nums: List[int], k: int already has them, which is good)

The solution demonstrates a solid understanding of the problem and implements an efficient algorithm. The single-pass approach is superior to the reference solution's two-pass method.

VERDICT: PASS