DP1

[YogeshPardeshi]

No description provided.

[super30admin]

The Coin Change (Problem1.java)

Strengths:

• Correctly implements the coin change DP solution
• Good use of bottom-up dynamic programming
• Clear code structure and readable logic
• Significantly better time complexity than the reference recursive solution

Areas for Improvement:

• Use Integer.MAX_VALUE instead of 99999 as the sentinel value for "impossible" states - this is more idiomatic and handles edge cases better
• Consider adding comments explaining the DP recurrence relation for better readability
• Could potentially optimize space from O(m×n) to O(n) using a 1D array, though current approach is correct

VERDICT: PASS

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House Robber (Problem2.java)

EVALUATION

1. Correctness

The student's solution correctly solves the House Robber problem. It uses dynamic programming to find the maximum amount of money that can be robbed without alerting the police. The approach is sound:

• Base cases are correctly handled (n=1 returns nums[0])
• The recurrence relation dp[i] = Math.max(dp[i-1], nums[i]+dp[i-2]) correctly captures the choice between robbing or skipping house i
• The final answer is correctly returned

2. Time Complexity

• Student's solution: O(n) - iterates through the array once
• Reference solution: O(2^n) - exponential due to recursive tree without memoization

The student's solution is significantly more efficient. The reference solution's time complexity is actually incorrect for the provided code - it should be O(2^n) without memoization, but with proper memoization it would be O(n).

3. Space Complexity

• Student's solution: O(n) - uses a DP array of size n
• Reference solution: O(n) - recursion stack depth

The student's solution uses the same space complexity as the reference. However, the space can be optimized to O(1) by using only two variables instead of the full array.

4. Code Quality

The code is well-structured and readable:

• Clear variable names (n, dp, i)
• Proper initialization of base cases
• Clean loop structure
• Good use of Math.max() for the decision logic

5. Efficiency

The solution is efficient but has room for optimization:

• Space can be reduced from O(n) to O(1) by using two variables instead of the full DP array
• The current approach is already optimal in terms of time complexity

FEEDBACK

Strengths:

• Correctly implements the dynamic programming approach
• Clean, readable code with good variable naming
• Handles edge cases properly (n=1)
• Time complexity is optimal (O(n))

Areas for Improvement:

• Space optimization: The DP array can be replaced with just two variables since we only need the previous two values at any point. This reduces space from O(n) to O(1).
• Edge case handling: Consider adding a check for n=0 (empty array), though the problem constraints guarantee n>=1.

Suggested optimization:

public int rob(int[] nums) {
    int n = nums.length;
    if (n == 1) return nums[0];
    
    int prev2 = nums[0];
    int prev1 = Math.max(nums[0], nums[1]);
    
    for (int i = 2; i < n; i++) {
        int current = Math.max(prev1, nums[i] + prev2);
        prev2 = prev1;
        prev1 = current;
    }
    
    return prev1;
}

The student's solution correctly solves the problem with good code quality and optimal time complexity. The space complexity can be further optimized, but the current solution is valid and demonstrates a solid understanding of dynamic programming.

VERDICT: PASS