DP-2 Solutions

[dbkuppagiri]

No description provided.

[super30admin]

Strengths:

• You have correctly identified the recursive structure of the problem: for each house and color, the minimum cost depends on the minimum cost of the next houses with different colors.
• You implemented memoization to avoid recomputation, which is essential for efficiency.
• Your code is mostly readable with comments explaining the intuition.

Areas for improvement:

1. Critical Error: In the initial calls to helper, you are passing three arguments: helper(0,0,0), but your helper function is defined to take only two parameters: idx and color. The third argument (0) is extra and will be ignored. This is a mistake that will cause your function to not work correctly. You should call helper(0,0), helper(0,1), and helper(0,2) without the third argument.
2. Memoization Initialization: Initializing the memo array with zeros is problematic because zero can be a valid value (for the base case, which is not stored in memo, but also for subproblems? Actually, in this problem, the cost is always at least 1, so the minimum cost for a subproblem will never be zero. However, it is a best practice to initialize memo with a value that indicates "not computed", such as null or undefined or -1. Using zero might work here by coincidence, but it is not robust. For example, if the problem allowed zero costs, your solution would fail. So you should initialize memo with -1 or another sentinel value.
3. Redundant Code: The variable result is declared and then assigned in the if-else blocks. You can directly return the value after storing in memo. Also, you have an unnecessary return 0 at the end of the helper function (which is never reached because all colors are covered). You can remove that.
4. Efficiency: Your solution is efficient with O(n) time and space, which is optimal for this problem. However, you can also solve it with iterative dynamic programming without recursion, which might be more efficient in terms of constant factors and avoid recursion stack overhead.

Corrected code snippet for memo initialization and helper calls:

var minCost = function (costs) {
    let n = costs.length;
    let memo = Array.from({ length: n }, () => Array(3).fill(-1)); // initialize with -1

    const helper = (idx, color) => {
        if (idx === n) return 0;
        if (memo[idx][color] !== -1) return memo[idx][color];
        let result;
        if (color === 0) {
            result = Math.min(helper(idx+1,1), helper(idx+1,2)) + costs[idx][0];
        } else if (color === 1) {
            result = Math.min(helper(idx+1,0), helper(idx+1,2)) + costs[idx][1];
        } else {
            result = Math.min(helper(idx+1,0), helper(idx+1,1)) + costs[idx][2];
        }
        memo[idx][color] = result;
        return result;
    }

    return Math.min(helper(0,0), helper(0,1), helper(0,2));
};

5. Note: The problem constraints (n up to 100) are manageable with recursion and memoization, but for larger n, iterative DP would be better. However, your solution is acceptable.