Array 2 Completed

[arjunsahas]

No description provided.

[super30admin]

Disappeared Numbers (DisappearedNumbers1.java)

Your solution is efficient and correct. You have successfully implemented an O(n) time and O(1) space solution (excluding the output list). The approach of using the array itself to mark presence is clever and meets the problem constraints.

Strengths:

• Efficient time and space complexity.
• The code is concise and to the point.

Areas for improvement:

• Consider using more descriptive variable names. For example, mList could be renamed to result or missingNumbers to improve readability.
• Add comments to explain the algorithm briefly, especially the marking step, to make the code more understandable for others.
• Ensure that the solution handles all edge cases, such as an empty array or arrays with all elements present. Your solution should handle these, but it's good to test them.

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

VERDICT: PASS

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max and min (GameOfLife.java)

Dear student,

It seems you have submitted a solution for the "Game of Life" problem instead of the "max and min" problem. Please double-check the problem statement and ensure your code addresses the correct problem.

For the "max and min" problem, you need to find the minimum and maximum values in an array with fewer than 2*(n-2) comparisons. The reference solution uses an efficient method that processes elements in pairs to reduce the number of comparisons.

If you were actually working on the "Game of Life" problem, here are some pointers:

• You need to preserve the original state of the board when checking neighbors. One common approach is to use two passes: first to compute the next state without altering the original, and then update. Alternatively, you can use state encoding (like you did with 2 and 3) but you must ensure that when counting neighbors, you only consider the original live cells (state 1 or state 2? Actually, in your encoding, state 2 means "was alive but will die", so for neighbor count, you should consider both 1 and 2 as alive for the current state). However, your neighbor counting logic is incorrect because it counts cells that are dead in the next state as alive. You should count only cells that are currently alive (which are 1 or 2, since 2 is currently alive but will die). But note: when you update a cell to 2, it is still alive for the purpose of neighbor counts for other cells? Actually, yes, because the rules require all updates to be simultaneous. So your encoding is correct, but the implementation has flaws.
• The rules of the game are:
  1. Any live cell with fewer than two live neighbors dies (underpopulation).
  2. Any live cell with two or three live neighbors lives on.
  3. Any live cell with more than three live neighbors dies (overpopulation).
  4. Any dead cell with exactly three live neighbors becomes alive.
• You should first iterate over each cell and count its live neighbors (considering states 1 and 2 as alive). Then, based on the count and the current state, set the new state (using 2 for death and 3 for new life). After processing all cells, update the board to the next state (convert 2 to 0 and 3 to 1).

Your current code tries to do everything in one pass, but the neighbor checks are affected by the changes you make. For example, when you set a cell to 2 (dead in next state), it is still counted as alive for neighbors, which is correct for the current generation. However, your logic for applying the rules is incorrect: you are checking the neighbor count and then immediately updating the cell, which might be okay if you are using the encoded states correctly, but your conditions are not aligned with the rules.

I recommend rewriting the solution for "Game of Life" by first iterating to count live neighbors for each cell (without changing the board), then determining the next state for each cell and storing it in a separate array or using the encoding scheme. Then, in a second pass, update the board.

For the "max and min" problem, please refer to the reference solution provided.

VERDICT: NEEDS_IMPROVEMENT

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Life Game

Your solution shows a good attempt at solving the problem in-place with constant space. You correctly identified the need to mark cells that change state without immediately updating them, so that the original state can be used for other computations. However, there is a critical flaw: when you count the neighbors for a cell, you are checking if the neighbor is 1 or 2. But note that in your encoding, 2 means "currently alive but will die" (so it should be counted as alive for the current state), and 3 means "currently dead but will become alive" (so it should be counted as dead for the current state). Your code currently counts 2 as alive (which is correct) but does not count 3 as dead. However, the bigger issue is that you are updating the board as you traverse it. This means that when you process a cell, you change its value to 2 or 3, and then when you process its neighbors, you are using these updated values. This breaks the requirement of simultaneous update because the neighbors are being computed based on a mix of original and updated states.

To fix this, you should follow the reference solution's approach:

• First, traverse the board and for each cell, count the number of live neighbors (using the original state: only 1 and 2 are live? Actually, in the reference solution, 2 is used to mark a cell that is currently alive but will die, so when counting neighbors, it considers 1 and 2 as alive (because they were alive in the original state). Similarly, you should count only the cells that are originally alive (1) or are marked to die (2) as alive. The 3 (new born) should not be counted as alive in the current state because they are currently dead.
• But note: in your code, you are using board[i][j] == 1 || board[i][j] == 2 to count alive neighbors. This is correct for the current state because 2 represents a live cell that will die. However, you are also updating the current cell to 2 or 3 during the same pass. This means that when you later process a neighbor of this cell, you will see the updated value (2 or 3) and count it correctly? Actually, for a neighbor that is updated to 2 (which was alive), it will be counted as alive. But for a neighbor that is updated to 3 (which was dead), it will not be counted as alive because your condition only counts 1 and 2. So that part is correct.

However, the main issue is that you are updating the board in the same pass. This causes the state of the board to change while you are still processing other cells. For example, if you update a cell to 2 (meaning it will die) early in the traversal, then when you process a neighbor of that cell later, you will see 2 and count it as alive (which is correct for the current state). But what if you update a cell to 3 (new born) early? Then when you process a neighbor, you will see 3 and your condition does not count it (because 3 is not 1 or 2). This is correct because 3 represents a dead cell in the current state.

So actually, your counting condition is correct: you count only cells that are 1 or 2 (which are alive in the current state). And you do not count 3 (which is dead). Therefore, the neighbor counting might be correct. However, the problem is that you are applying the rules and updating the cell immediately. This means that when you check the rules for a cell, you are using the current state of its neighbors (which might have been updated to 2 or 3). But note: if a neighbor was updated to 2, it was originally alive and should be counted. If it was updated to 3, it was originally dead and should not be counted. So your counting condition is correct.

But wait: what if the neighbor hasn't been processed yet? Then it has value 0 or 1. So the counting condition is consistent.

However, there is another issue: your rules are applied incorrectly. Let me check:

For a live cell (1 or 3? Actually, at the start, the board has only 0 and 1. So when you first encounter a live cell (1), you check the rules. But note: you also have condition "if (board[i][j] == 1 || board[i][j] == 3)". Why are you including 3? Because you might have updated a dead cell to 3 earlier? But 3 represents a dead cell that will become alive. So it should not be treated as a live cell in the current state. In fact, when you start, there are no 3's. So this condition is redundant and incorrect.

Actually, your code has a logical error: you are checking for board[i][j] == 1 OR board[i][j] == 3 to determine if the cell is currently alive. But 3 is not alive! It is a dead cell that will become alive in the next state. So you should only consider board[i][j] == 1 (or 2) for currently alive. Similarly, for dead cells, you should consider board[i][j] == 0 (or 3) for currently dead.

This mistake leads to incorrect rule application. For example, if a cell has been marked as 3 (new born) in this pass, then when you process it later, you will treat it as a live cell (because of the condition board[i][j] == 1 || board[i][j] == 3) and apply the rules for live cells. But it is actually dead in the current state! So you should not do that.

The correct approach is to only use the original state when applying the rules. The reference solution does not update the board until after the entire first pass. In the first pass, it only reads the board and writes markers (2 and 3) without reading the markers for rule application. But in your code, you are reading the markers (2 and 3) when counting neighbors and when determining the current state of the cell.

To fix this, you should:

• In the first pass, only read the board and write markers. But do not use the markers to determine the current state of the cell or its neighbors. Instead, when counting neighbors, only consider values 1 and 0 (but wait, you need to know the original state). The reference solution uses: for a neighbor, if it is 1 or 2, it was alive (because 2 is a marker for "alive now but will die", so it was alive). Similarly, if it is 0 or 3, it was dead.

So your counting function is correct: you count board[r][c] == 1 or 2 as alive. But when determining the current state of the cell (i,j), you should not use the marker. You should use only the original value. However, you have already overwritten the original value with 2 or 3. So you need to interpret the markers correctly.

Actually, your code for counting neighbors is correct. But for the cell (i,j) itself, you are using board[i][j] == 1 || board[i][j] == 3 to mean it is alive. This is wrong because 3 means it is dead. You should have:
if the cell is currently alive (board[i][j] == 1 or board[i][j] == 2) then apply live rules.
if the cell is currently dead (board[i][j] == 0 or board[i][j] == 3) then apply dead rules.

So you should change your condition to:

VERDICT: NEEDS_IMPROVEMENT