Complete22_23

[dhruvil15]

No description provided.

[super30admin]

Paint House (Problem22.java)

Your solution is well-implemented and efficient. You correctly identified the dynamic programming approach, which is optimal for this problem. The code is straightforward and easy to understand. However, note that the space complexity can be further optimized to O(1) by using only three variables to store the previous row's costs instead of maintaining a full DP table. This would reduce the space usage without affecting the time complexity. For example:

class Solution {
    public int minCost(int[][] costs) {
        if (costs == null || costs.length == 0) return 0;
        int prevR = costs[0][0], prevB = costs[0][1], prevG = costs[0][2];
        for (int i = 1; i < costs.length; i++) {
            int currR = costs[i][0] + Math.min(prevB, prevG);
            int currB = costs[i][1] + Math.min(prevR, prevG);
            int currG = costs[i][2] + Math.min(prevR, prevB);
            prevR = currR;
            prevB = currB;
            prevG = currG;
        }
        return Math.min(prevR, Math.min(prevB, prevG));
    }
}

This optimization uses constant space while maintaining the same time complexity. Overall, your solution is correct and efficient, but consider this space optimization for even better performance.

VERDICT: PASS

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Coin Change II (Problem23.java)

Your solution shows a good understanding of dynamic programming and the intent to optimize space by using a 1D array. However, there are a few critical issues:

1. Incorrect DP Update Logic: Your inner loop updates dp[j] only when j >= coins[i-1]. This is correct for the unbounded knapsack problem, but you need to ensure that you are not counting duplicate combinations. The standard solution for this problem uses a 2D DP array or a 1D array updated in a specific order. For a 1D array, you should iterate the inner loop from the coin value to the amount to avoid using the same coin multiple times in the same combination. However, your inner loop starts from j=0, which is inefficient and unnecessary. Instead, you should start from j = coins[i-1] to n to avoid the condition check and update directly.

2. Missing Initialization for Coins: Your outer loop runs from i=1 to i<=m, which is correct for iterating over each coin. But you are not using a separate array to store the previous state. In the 1D DP approach, you should iterate over coins and for each coin, update the DP array from the coin's value up to the amount. This way, you are effectively counting combinations without duplicates.

3. Code Structure: Your code is concise and follows a standard DP approach. However, it lacks comments which would make it more readable. Consider adding comments to explain the DP state and the recurrence.

4. Correctness: The current code does not produce the correct output for all test cases. For example, with amount=5 and coins=[1,2,5], it should return 4, but your code might return an incorrect value due to the way the DP array is updated.

Here is a corrected version of your code using a 1D DP array:

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 version initializes dp[0] = 1 (one way to make amount 0). Then for each coin, it updates the DP array from coin to amount, adding the number of ways to form j - coin to dp[j]. This ensures that each coin is considered in the combinations without overcounting.

VERDICT: NEEDS_IMPROVEMENT