DP-1

[nagasai67]

No description provided.

[super30admin]

The Coin Change (coinChange.py)

Your solution has the right idea of using dynamic programming with memoization to avoid recalculating states. However, there are several issues that need to be addressed:

• Base Cases: You need to handle negative amounts correctly. When amt < 0, you should return infinity (no solution). Also, when i == len(coins) and amt > 0, you should return infinity.

• Memoization: Your memo table should store results for all states (i, amt). Currently, you are initializing the memo table with -1, but you are storing floats (infinity). This might work, but it's better to initialize with None or a value that indicates not computed. Also, you are not storing the result for the state (i, amt) when amt < 0 or when i == len(coins). You should store all states to avoid recomputation.

• Recursive Logic: The logic for "pick" and "not pick" is almost correct, but you need to ensure that when you pick a coin, you stay at the same index (because coins can be reused), and when you don't pick, you move to the next index. However, your code for "not pick" is helper(i+1, amt, memo), which is correct. But for "pick", you call helper(i, amt - coins[i], memo), which is also correct. But you must check that amt - coins[i] >=0 before calling, otherwise return infinity.

• Implementation: Here is a corrected version of your recursive function:

def helper(i, amt, memo):
    if amt < 0:
        return float('inf')
    if i == len(coins):
        return float('inf') if amt > 0 else 0
    if memo[i][amt] != -1:
        return memo[i][amt]
    pick = float('inf')
    if amt >= coins[i]:
        pick = 1 + helper(i, amt - coins[i], memo)
    not_pick = helper(i+1, amt, memo)
    memo[i][amt] = min(pick, not_pick)
    return memo[i][amt]

But note: this recursive function may not be efficient because it does not prune unnecessary states. Alternatively, you can use an iterative DP approach:

dp = [float('inf')] * (amount+1)
dp[0] = 0
for coin in coins:
    for i in range(coin, amount+1):
        dp[i] = min(dp[i], dp[i-coin] + 1)
return dp[amount] if dp[amount] != float('inf') else -1

This iterative solution is simpler and more efficient.

• Code Quality: Use more descriptive variable names. For example, use index instead of i, remaining instead of amt. Also, add comments to explain each step.

VERDICT: NEEDS_IMPROVEMENT

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House Robber (houseRob.py)

Your solution is correct and efficient. You have successfully implemented a bottom-up dynamic programming approach that runs in linear time. The code is clean and well-commented, which is good practice.

One improvement you can make is to optimize the space usage. Currently, you are using an array of size n+2, but you only need the last two computed values at any step. You can replace the DP array with two variables to track the maximum amount starting from the next house (dp[i+1]) and the house after that (dp[i+2]). This will reduce the space complexity from O(n) to O(1). Here's how you can do it:

class Solution:
    def rob(self, nums: List[int]) -> int:
        # Initialize two variables to represent dp[i+1] and dp[i+2]
        next_house = 0
        next_next_house = 0
        for i in range(len(nums)-1, -1, -1):
            # The current house: take nums[i] + next_next_house, or skip and take next_house
            current = max(nums[i] + next_next_house, next_house)
            # Update for the next iteration: shift the values
            next_next_house = next_house
            next_house = current
        return next_house

This approach uses constant space and is just as efficient in time. It's a common optimization for DP problems where only the previous states are needed.

Overall, your solution is excellent, and with this small optimization, it can be even better. Keep up the good work!

VERDICT: PASS