Solved Problem 1 - Paint House and Problem 2 - Coin Change 2

[anirudhv98]

Completed DP - 2, please review

[super30admin]

Paint House (Problem1.cs)

Strengths:

• Excellent use of dynamic programming to achieve O(n) time complexity vs the reference's O(2^n)
• Space-efficient with O(1) auxiliary space by modifying the input array in-place
• Clean, readable code with appropriate comments
• Correctly handles the constraint that adjacent houses cannot have the same color
• Well-structured with clear variable names

Areas for Improvement:

• The comment "where m is the total length of costs array" is slightly confusing; it should say "n is the number of houses" for consistency with the problem statement
• Consider edge case handling for empty input (though constraints guarantee at least 1 house)
• Could add a check for null or empty costs array for robustness

Overall: This is a solid, optimal solution that significantly outperforms the reference solution in both time and space complexity. The implementation is clean and follows best practices.

VERDICT: PASS

---

Coin Change II (Problem2.cs)

Strengths:

1. Excellent use of dynamic programming to solve the problem efficiently - much better than the exponential recursive solution
2. The 1D DP approach is space-optimized (O(n) vs potentially O(m*n) for 2D)
3. Code is clean, readable, and well-commented with complexity analysis
4. Correctly handles edge cases including amount=0
5. The inner loop correctly uses j >= coins[i] to avoid negative index access

Areas for Improvement:

1. The outer loop could start from 0 and include j=0 in the inner loop, which would make the logic more uniform. Currently, dp[0] is handled separately before the loops.
2. Consider adding input validation for edge cases like empty coins array or negative amounts (though constraints guarantee valid inputs).

Optimization Note:
The current solution is already optimal for this problem. The inner loop starting from 1 instead of 0 is a minor optimization that avoids unnecessary checks for j=0 (which would always be 0 since j >= coins[i] would be false for all coins[i] >= 1).

VERDICT: PASS