Complete24_25

[dhruvil15]

No description provided.

[super30admin]

Least Falling Path Sum (Problem24.java)

Your solution is correct and efficient. You have successfully implemented a dynamic programming approach that runs in O(n^2) time, which is optimal for this problem. The code is clean and easy to understand.

One improvement you could make is to reduce the space complexity from O(n^2) to O(n). Instead of storing the entire DP table, you can use two arrays: one for the previous row and one for the current row. This is because when computing the values for row i, you only need the values from row i-1. Here's how you can do it:

public int minFallingPathSum(int[][] matrix) {
    if (matrix == null || matrix.length == 0) {
        return 0;
    }
    int n = matrix.length;
    int[] prev = new int[n];
    // Initialize the first row
    for (int j = 0; j < n; j++) {
        prev[j] = matrix[0][j];
    }
    for (int i = 1; i < n; i++) {
        int[] curr = new int[n];
        for (int j = 0; j < n; j++) {
            if (j == 0) {
                curr[j] = matrix[i][j] + Math.min(prev[j], prev[j+1]);
            } else if (j == n-1) {
                curr[j] = matrix[i][j] + Math.min(prev[j-1], prev[j]);
            } else {
                curr[j] = matrix[i][j] + Math.min(prev[j], Math.min(prev[j-1], prev[j+1]));
            }
        }
        prev = curr; // update prev to be the current row for the next iteration
    }
    int min = Integer.MAX_VALUE;
    for (int num : prev) {
        min = Math.min(min, num);
    }
    return min;
}

This change reduces the space complexity to O(n) while maintaining the same time complexity.

Another minor point: in your inner loop, when j is not at the boundary, you can write the expression as Math.min(Math.min(prev[j-1], prev[j]), prev[j+1]) for clarity. However, your current approach is also correct.

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

VERDICT: PASS

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Delete and Earn (Problem25.java)

Your solution is excellent. You have correctly implemented the dynamic programming approach with state compression, which reduces the space complexity from O(max(n)) to O(1) for the DP part. This is an improvement over the reference solution that uses an array for storing all DP states.

Strengths:

• You correctly identified the need to create an array that aggregates points for each number.
• You used an efficient iterative DP approach with only two variables to store the necessary states.
• The code is concise and easy to understand.

Areas for Improvement:

• While your solution is already optimal, one minor consideration is that if the maximum value in nums is very large but the number of distinct values is small, the arr array might be inefficient in terms of space. However, given the constraints (nums[i] <= 10^4), this is acceptable.
• You might want to add a comment explaining the logic briefly, especially the DP part, for better maintainability.

Overall, this is a very good solution.

VERDICT: PASS