Solved DP-1

[pratul2789]

No description provided.

[super30admin]

Your DP solution for the coin change problem is correct and efficient. Well done! However, there are a few areas for improvement:

• Avoid using magic numbers like 999999. Instead, you can use float('inf') or initialize with amount+1 (since the maximum number of coins needed is at most amount if using coins of 1). For example, dp = [amount+1] * (amount+1) and then check if dp[-1] == amount+1 to return -1. This is more robust and readable.
• The comments in the code are a bit confusing. You have comments that refer to a 2D DP approach (e.g., "same column, one row above"), but you are using a 1D DP. It's better to remove outdated comments to avoid confusion.
• The inner loop in the DP solution can be optimized by starting from coins[i] to avoid the condition check inside. Instead of iterating from 1 to amount and then checking if j >= coins[i], you can iterate from coins[i] to amount. This might slightly improve performance.

Example improvement:

dp = [amount+1] * (amount+1)
dp[0] = 0
for coin in coins:
    for j in range(coin, amount+1):
        dp[j] = min(dp[j], 1 + dp[j-coin])

This is cleaner and avoids the inner condition.

Also, note that the problem states that if the amount is 0, return 0. Your solution handles that correctly because dp[0]=0.

Overall, your solution is good, but with minor improvements it can be more professional and readable.