Complete DP-3 Assignment

[tejbharath]

No description provided.

[super30admin]

Minimum Falling Path Sum (MinimumFallingPathSum.java)

Strengths:

1. Excellent choice of dynamic programming approach, which is more efficient than the recursive approach in the reference solution
2. Clear and well-commented code that explains the logic
3. Proper handling of boundary conditions
4. Correct initialization and final answer extraction

Areas for Improvement:

1. Space Optimization: The dp matrix could be optimized to use only O(n) space by maintaining two rows (previous and current) instead of the full matrix. This would reduce memory usage while maintaining the same time complexity.
2. Variable Naming: Consider using more descriptive variable names like rows and cols instead of m and n for better readability.
3. Edge Case Handling: The current solution handles the case when matrix is null or empty, but it could also handle the case when matrix has only one row (though the current code would still work correctly).

Suggested Optimization:

// Use only two rows to save space
int[] prev = new int[n];
int[] curr = new int[n];

// Initialize first row
for (int j = 0; j < n; j++) {
    prev[j] = matrix[0][j];
}

// Process remaining rows
for (int i = 1; i < m; i++) {
    for (int j = 0; j < n; j++) {
        int middle = prev[j];
        int left = (j - 1 >= 0) ? prev[j - 1] : Integer.MAX_VALUE;
        int right = (j + 1 < n) ? prev[j + 1] : Integer.MAX_VALUE;
        curr[j] = matrix[i][j] + Math.min(left, Math.min(middle, right));
    }
    // Swap rows
    int[] temp = prev;
    prev = curr;
    curr = temp;
}

// Find minimum in the last processed row
int minSum = Integer.MAX_VALUE;
for (int j = 0; j < n; j++) {
    minSum = Math.min(minSum, prev[j]);
}

VERDICT: PASS

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Delete and Earn (DeleteAndEarn.java)

Excellent work! Your solution is essentially identical to the reference solution and correctly solves the problem. Here are some observations:

Strengths:

• Correctly identifies this as a House Robber variant and applies the appropriate DP approach
• Includes input validation for null/empty arrays (which the reference solution doesn't have)
• Clear comments explaining the approach
• Well-structured and readable code

Areas for Improvement:

1. Time Complexity Comment: You wrote O(n) but the actual complexity is O(n + max(nums)) where max(nums) is the maximum value in the array. The DP loop runs from 2 to max, not n.

2. Space Complexity Comment: Similarly, you wrote O(n) but the actual space complexity is O(max(nums)) because the arr and dp arrays are sized based on the maximum value, not the length of the input array.

3. Unused Variable: The variable int n = nums.length; is declared but never used. You can remove it.

4. Minor Optimization: You could reduce space to O(max(nums)) by using two variables instead of the full dp array, since you only need dp[i-1] and dp[i-2] at each step. However, this is optional and the current solution is already efficient.

VERDICT: PASS