Array - 1 Solutions

[AnupKulkarniVK]

No description provided.

[super30admin]

Product Except Self (Problem_1_Product_Except_Self.java)

Note: The verdict should be based on the overall quality and correctness of the solution.

Let's begin.

VERDICT: NEEDS_IMPROVEMENT

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Diagonal Traverse (Problem_2_Diagonal_traverse.java)

Your solution is on the right track and follows a similar approach to the reference solution. However, there are issues with the boundary conditions that need to be addressed.

Strengths:

• You have correctly identified the need to use a flag to toggle between upward and downward movements.
• The code is structured simply and is easy to follow.

Areas for improvement:

• The main issue is in the boundary conditions. For example, when moving upward (flag=true) and you are at the top-right corner (r=0, c=n-1), you should move down to the next row (r++). However, your code first checks if(r == 0 && c != n-1), which is false because c==n-1, so it moves to the next condition else if(c == n-1), which is true, and then sets r++ and flag=false. This is correct for this corner. But consider when you are at the top row but not the last column: your code sets c++ and flag=false. This is also correct. However, the problem arises when moving downward (flag=false). For instance, when you are at the bottom-left corner (r=m-1, c=0), you should move right (c++). Your code checks if(c == 0 && r != m-1), which is false because r==m-1, so it moves to else if(r == m-1), which is true, and then sets c++ and flag=true. This is correct. But if you are at the first column but not the last row, you set r++ and flag=true, which is correct.

Wait, after analyzing more carefully, it seems your conditions might actually work for the corners. However, there is a potential issue when the matrix has only one row or one column. For example, if the matrix has one row (m=1), then when flag=true, you are always in the top row. The condition if(r==0 && c != n-1) would be true until the last element. For the last element, you would hit c == n-1, so you do r++ which would go out of bounds since r=0 and m=1. Similarly, for a single column (n=1), when flag=false, you would have c==0 always, and the condition if(c==0 && r != m-1) would be true until the last row. Then when r=m-1, you would hit the else if(r==m-1) block and do c++, which would go out of bounds.

Therefore, your solution does not handle matrices with one row or one column correctly. You should test your code with these edge cases.

The reference solution handles these cases by checking boundaries in a different order. For upward movement (dir=true), it first checks if it is at the rightmost column (c==n-1), because if so, it must move down regardless of the row. Then it checks if it is at the top row (r==0) to move right. Similarly, for downward movement, it first checks if it is at the bottom row (r==m-1) to move right, then checks if it is at the leftmost column (c==0) to move down.

I recommend adjusting your conditions to match the reference solution's order: check the column boundary first when going up, and the row boundary first when going down. This will handle the edge cases correctly.

Also, consider adding comments to explain the direction changes for better readability.

VERDICT: NEEDS_IMPROVEMENT

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Spiral Matrix (Problem_3_Matrix_elements_spiral_order.java)

Your solution is correct and efficient, and you have successfully implemented the spiral order traversal. Well done! Here are a few points for improvement:

1. Redundant Condition Checks: You have added if(left <= right && top <= bottom) before every for-loop. This is not necessary for the first two directions (right and down) because the while loop condition already ensures that these boundaries are valid when the loop starts. Only the last two directions (left and up) require these checks because after updating top and right, the boundaries might have changed. You can remove the conditionals for the first two for-loops to make the code cleaner.

2. Code Readability: Consider using more descriptive variable names. For example, l could be renamed to result or spiralList to make the code more readable.

3. Consistency: The reference solution uses a similar structure but with fewer conditionals. You can simplify your code by following the same pattern:

   • First, traverse from left to right (without an extra if check).
   • Then, traverse from top to bottom (without an extra if check).
   • Then, for the leftward traversal, check if top <= bottom (because we updated top and might have exhausted the rows).
   • Finally, for the upward traversal, check if left <= right (because we updated right and might have exhausted the columns).

Here is a slightly optimized version of your code for clarity:

class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        int m = matrix.length;
        int n = matrix[0].length;
        int top = 0, bottom = m-1, left = 0, right = n-1;
        List<Integer> result = new ArrayList<>();
        
        while (left <= right && top <= bottom) {
            // Traverse from left to right
            for (int j = left; j <= right; j++) {
                result.add(matrix[top][j]);
            }
            top++;
            
            // Traverse from top to bottom
            for (int i = top; i <= bottom; i++) {
                result.add(matrix[i][right]);
            }
            right--;
            
            if (top <= bottom) {
                // Traverse from right to left
                for (int j = right; j >= left; j--) {
                    result.add(matrix[bottom][j]);
                }
                bottom--;
            }
            
            if (left <= right) {
                // Traverse from bottom to top
                for (int i = bottom; i >= top; i--) {
                    result.add(matrix[i][left]);
                }
                left++;
            }
        }
        return result;
    }
}

Overall, your solution is correct and efficient. With minor adjustments for readability and removing redundant checks, it can be even better.

VERDICT: PASS