2 Problems completed

[BharathVuppala96]

No description provided.

[super30admin]

The student has implemented a dynamic programming solution that is generally correct (except for the initial check). The time complexity is optimal for a DP solution. However, there are a few issues:

• The initial check should be: if amount==0, return 0. Then if coins is None or empty, and amount>0, return -1.
• The condition inside the outer loop to check for -1 is placed incorrectly. It should be after the entire DP table is filled. Actually, the student checks at the end of each row, but only the last row matters. So it would be better to move the check outside the loops.
• The code can be optimized in space by using a 1D array.

Also, the student submitted another solution for "House Robber", which is not relevant to this problem. We should only evaluate the "Coin Change" solution.

Given that the core DP

[super30admin]

Strengths:

• The student has chosen a dynamic programming approach which is suitable for this problem.
• The initialization of the DP table is mostly correct (using amount+1 as a substitute for infinity).

Areas for Improvement:

1. Handling edge cases:

   • The initial check for empty coins should return 0 only if amount is 0; otherwise, it should return -1. For example, if coins is empty and amount is 5, it should return -1, but the code returns 0.
   • Also, handle the case when amount is 0: the code should return 0. Currently, the code would work for amount=0 because the double loops are skipped (since they start from 1) and then it returns dp[n][0] which is 0. But the initial check might interfere: if coins is empty and amount=0, the initial check returns 0 which is correct. However, if coins is not empty and amount=0, the code returns 0 correctly.

2. Syntax error: There is an extra closing bracket in the min function. Remove it.

3. Logic error: The condition to return -1 is placed inside the outer loop. It should be after both loops. Also, note that the condition should be: if dp[n][amount] > amount then return -1, else return dp[n][amount].

4. Code readability: Use more descriptive variable names. For example, n can be num_coins, and dp is acceptable but well-known.

5. Space optimization: Consider using a 1D DP array to reduce space complexity to O(amount).

Corrected code should look like:

class Solution:
    def coinChange(self, coins: List[int], amount: int) -> int:
        if amount == 0:
            return 0
        if not coins:
            return -1
        n = len(coins)
        dp = [[0] * (amount+1) for _ in range(n+1)]
        # Initialize first row: with no coins, we need infinity for any amount >0
        for j in range(1, amount+1):
            dp[0][j] = amount + 1
            
        for i in range(1, n+1):
            for j in range(1, amount+1):
                if j < coins[i-1]:
                    dp[i][j] = dp[i-1][j]
                else:
                    dp[i][j] = min(dp[i-1][j], 1 + dp[i][j - coins[i-1]])
        # Check after filling the entire table
        if dp[n][amount] > amount:
            return -1
        return dp[n][amount]

But note: the above corrected code still uses 2D DP. Alternatively, we can use 1D DP:

def coinChange(self, coins: List[int], amount: int) -> int:
    dp = [amount + 1] * (amount + 1)
    dp[0] = 0
    for coin in