Completed Backtracking-3

[sarvanibaru]

No description provided.

[super30admin]

N-Queens (N-Queens.java)

Strengths:

• The solution correctly implements the backtracking algorithm with optimizations using boolean arrays to track constraints, which improves efficiency by avoiding linear scans for validity checks.
• The code is well-structured with clear comments explaining the approach.
• The use of helper functions and meaningful variable names enhances readability.

Areas for Improvement:

• The arrays colSet, diagSet, and antiDiagSet are initialized with size 2*n, which is correct for diagonals and anti-diagonals (since row+col ranges from 0 to 2*(n-1) and row-col+n ranges from 0 to 2*(n-1)), but the variable names could be more descriptive. For example, colSet might be better named as cols or colUsed, and similarly for the diagonal arrays.
• The solution does not include a base case for n=0, but since n>=1 per constraints, this is acceptable.
• The code could benefit from a brief inline comment in the helper function to explain the backtracking step.

Overall, the solution is efficient and correct. The optimizations using boolean arrays are a good practice for reducing time complexity.

VERDICT: PASS

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Word Search (WordSearch.java)

Strengths:

• The solution uses DFS with backtracking correctly, marking visited cells and restoring them.
• The code is well-commented and explains the approach clearly.
• The frequency check is a good optimization idea to quickly prune impossible cases.

Areas for Improvement:

• The frequency check does not handle case sensitivity correctly. The problem states that the board and word consist of both lowercase and uppercase English letters. Therefore, the frequency check should be case-insensitive or should not be used at all if case sensitivity is required. Alternatively, you could convert all characters to the same case (e.g., lowercase) before counting, but note that the problem does not specify that 'A' and 'a' should be considered the same. Actually, the problem says "the same letter cell", which implies that case matters: 'A' and 'a' are different. Therefore, the frequency check must be case-sensitive. However, your current frequency check uses an array of size 128, which is sufficient for ASCII, but you are comparing characters exactly (including case). So the frequency check itself is case-sensitive, which is correct. But wait: in the frequency check, you are counting the characters in the board exactly as they are, and then subtracting the occurrences in the word. This should be correct for case-sensitive matching. However, there is a logical error: you are decrementing the count for each character in the word, but if the word has a character that is not in the board (or not enough), you return false. This is correct. But note: the board might have both cases, and the word might have mixed cases. So if the word has 'A' but the board has only 'a', then the count for 'A' in the board is 0, so when you subtract, it becomes negative -> return false. This is correct because the board does not have the required character. So the frequency check is actually correct for case sensitivity.

Wait, let me double-check: the problem says "consists of only lowercase and uppercase English letters", and the search is exact. So if the word has "A", the board must have "A" (not "a"). Therefore, the frequency check is correct. However, there is another issue: the frequency check counts all characters in the board, but when we do DFS, we cannot use the same cell more than once. The frequency check does not account for the fact that we might have duplicate characters but we are limited by distinct cells. For example, if the word has two 'A's, and the board has two 'A's, the frequency check passes. But if the board has two 'A's that are the same cell? No, each cell is distinct. Actually, the frequency check counts the number of occurrences in the board, which is exactly the number of distinct cells. So it is correct to use frequency check to ensure the board has at least the number of each character required by the word. This is a valid optimization.

But wait: what if the board has more than enough of a character? The frequency check only ensures that for each character in the word, the board has at least the same count. However, it is possible that the board has more, so we subtract the word's characters and check for negative. This is correct.

So the frequency check is correct. However, there is a subtle bug: the frequency array is of size 128, which covers ASCII characters. But the board and word only have letters, so it is safe. But note: the array is indexed by the char value. For example, 'A' is 65, 'a' is 97. So it will correctly distinguish cases.

Therefore, the frequency check is correctly implemented. But wait: in the code, the frequency array is named wordCount, but it is actually used to count the board first. Then for each character in the word, it decrements. If it becomes negative, it returns false. This is correct.

So why did I initially think there was an issue? Actually, I misread. The frequency check is correct.

But there is another issue: the frequency check might be too strict. Consider a word "aaa" and the board has 3 'a's. But what if the word requires 3 'a's but they are not connected? The frequency check will pass, but the DFS might fail. So the frequency check is a valid pre-check to quickly return false if the board doesn't have enough of a character. It is correct.

So the solution is correct.

However, one more thing: the student's code uses char[][] for board, but the problem says the grid is of characters. In Java, this is correct.

Also, the DFS function has the right base cases: index out of bounds, character mismatch, and reaching the end of the word.

Therefore, the solution is correct.

But note: the frequency check adds an extra O(m*n + L) time, which is acceptable since it can prune many cases quickly.

Code Quality: The code is clean. However, the direction array is initialized with four directions. The order of directions might affect performance in some cases, but it is acceptable.

One minor improvement: the frequency check could be done in a single loop. Currently, it first counts the board, then iterates the word. This is efficient.

VERDICT: PASS