Done dp-2

[sainathek1999]

No description provided.

[super30admin]

Paint House (problem1.java)

Your solution is excellent! You correctly identified the overlapping subproblems and applied dynamic programming to reduce the time complexity from exponential (in the recursive brute-force) to linear. Your code is clear and well-commented, which makes it easy to understand.

Strengths:

• Correct implementation of DP with bottom-up approach.
• Proper handling of the constraint that adjacent houses must have different colors.
• Good comments explaining the time and space complexity.

Areas for Improvement:

• While your space complexity is O(n), you could optimize it to O(1) by using only three variables to store the costs for the next house instead of a full DP table. This would be more memory efficient without sacrificing readability. However, given the constraints (n <= 100), your current solution is perfectly acceptable.
• Consider adding more comments to explain the DP recurrence relation for future reference.

Example of space-optimized solution (for your reference):

public int minCost(int[][] costs) {
    int n = costs.length;
    int cost0 = costs[n-1][0], cost1 = costs[n-1][1], cost2 = costs[n-1][2];
    for (int i = n-2; i >=0; i--) {
        int new0 = costs[i][0] + Math.min(cost1, cost2);
        int new1 = costs[i][1] + Math.min(cost0, cost2);
        int new2 = costs[i][2] + Math.min(cost0, cost1);
        cost0 = new0;
        cost1 = new1;
        cost2 = new2;
    }
    return Math.min(cost0, Math.min(cost1, cost2));
}

Overall, great job! Your solution is efficient and correct.

VERDICT: PASS

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Coin Change II (problem2.java)

Your solution has a good intention to use dynamic programming with a 1D array to optimize space. However, there are a few critical issues:

1. Indexing Error: Your inner loop runs from j=0 to j<n (which is amount+1). When j < coins[i-1], the condition j >= coins[i-1] fails, so you skip the update. This is correct, but when j is less than the coin value, you should not update. However, the problem is that you are not initializing the DP array for each coin? Actually, the standard solution for this problem uses:

   dp[0] = 1;
   for (int coin : coins) {
        for (int j = coin; j <= amount; j++) {
            dp[j] += dp[j - coin];
        }
   }
   

   Notice that the inner loop starts from coin to amount to avoid negative indices and to correctly accumulate combinations.

2. Incorrect Loop Structure: Your outer loop runs from i=1 to i < m (which is coins.length+1). This is acceptable because you are iterating over each coin. However, you should use for (int coin : coins) for clarity. Also, the inner loop should start from coin to amount to avoid checking the condition and to prevent negative indices.

3. Base Case: You correctly set dp[0]=1, which is good.

4. Code Clarity: Use more descriptive variable names. For example, instead of n and m, use dpSize and numCoins. Also, the outer loop variable i is not very clear. Consider iterating directly over the coins.

5. Correctness: Your current code would throw an ArrayIndexOutOfBoundsException because when j < coins[i-1], you try to access dp[j - coins[i-1]] which is negative. You need to change the inner loop to start from coins[i-1] to n-1 (which is amount).

To fix your code:

• Change the inner loop to start from j = coins[i-1] to j < n (i.e., from the current coin value to the amount).
• Alternatively, you can keep the inner loop from 0 to amount but add a condition to avoid negative indices. However, starting from the coin value is more efficient.

Here is a corrected version of your code:

class Solution {
    public int change(int amount, int[] coins) {
        int[] dp = new int[amount + 1];
        dp[0] = 1;
        for (int coin : coins) {
            for (int j = coin; j <= amount; j++) {
                dp[j] += dp[j - coin];
            }
        }
        return dp[amount];
    }
}

This solution is efficient and correct.

VERDICT: NEEDS_IMPROVEMENT