working solution

[avcode3]

No description provided.

[super30admin]

Strengths:

• You have implemented an efficient dynamic programming solution with O(n) time and O(1) space.
• The code is concise and uses meaningful variable names.
• The bottom-up approach is well-executed.

Areas for Improvement:

• The solution in problem1.py is for a different problem and should not be included when submitting for the Paint House problem. Make sure to only submit the relevant code.
• Consider adding comments to explain the logic, especially for the dynamic programming transition. For example, you could comment that for each house, the cost for red is the current cost for red plus the minimum cost of the next house for blue and green.
• Although not necessary for this problem, if the number of colors were to change, the code would need to be modified. Using an array of size 3 instead of three separate variables could make the code more adaptable. For example, you could use an array dp of length 3 and update it in a loop.

Example of a more generalized approach (optional):

def minCost(self, costs):
    n = len(costs)
    dp = costs[n-1][:]  # copy the last row
    for i in range(n-2, -1, -1):
        temp = dp[:]  # create a copy of current dp
        dp[0] = costs[i][0] + min(temp[1], temp[2])
        dp[1] = costs[i][1] + min(temp[0], temp[2])
        dp[2] = costs[i][2] + min(temp[0], temp[1])
    return min(dp)

This approach uses a list for dp and avoids the need for temporary variables. However, your current solution is also correct and efficient.

Overall, your solution for the Paint House problem is excellent.