Complete backtracking-2

[paridhimalviya]

Subsets (https://leetcode.com/problems/subsets/)
Palindrome Partitioning(https://leetcode.com/problems/palindrome-partitioning/)

[super30admin]

Subsets (Subsets.swift)

Strengths:

• You have implemented three different approaches, showing a good understanding of the problem and multiple techniques.
• The code is clean and well-commented.
• You have used backtracking effectively in the recursive solutions to avoid unnecessary copies.

Areas for Improvement:

• In the iterative approach, the print statements should be removed as they are not needed in the final code and can affect performance for large inputs (though the constraints are small).
• For the 0-1 recursion, you might consider adding a comment to explain why you remove the last element after the recursive call (to backtrack).
• The time and space complexity analysis in the comments is partially correct. For the 0-1 recursion, the time complexity is O(n * 2^n) because each subset is built in O(n) time in the worst case. The space complexity for the recursion stack is O(n), but the output space is O(n * 2^n). You should note that.

Overall, your solutions are correct and efficient. Keep up the good work!

VERDICT: PASS

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Palindrome Partitioning (PalindromePartitions.swift)

Your solution is correct and follows the backtracking approach appropriately. The time and space complexity are optimal for this problem. Here are some suggestions for improvement:

1. Optimize Palindrome Check: Instead of creating a new array for each substring (like Array(sArr[start...i])), you can check the palindrome directly on the original sArr by passing the start and end indices to the isPalindrome function. This avoids unnecessary array copies and improves performance. For example, you can change the isPalindrome function to take the array and the start and end indices: isPalindrome(sArr: [Character], start: Int, end: Int) -> Bool. Then, inside the loop, you would call isPalindrome(sArr: sArr, start: start, end: i).

2. Variable Naming: The variable sArr is a bit unclear. Consider renaming it to characters or sArray for clarity.

3. Commenting: The comments in the code are helpful, but you could add more comments to explain the backtracking step (like the action, recurse, and backtrack steps) to make it even clearer.

4. Initialization: The init method in the class is empty and not necessary. You can remove it unless you plan to add initialization logic later.

5. Consistency: The class name PalindromePartition does not match the problem name exactly (it's "Palindrome Partitioning"), but this is minor. However, the problem expects a function partition, which you have implemented correctly.

Here is how you could refactor the palindrome check:

func helper(sArr: [Character], start: Int, path: inout [String], result: inout [[String]]) {
    if start == sArr.count {
        result.append(path)
        return
    }
    for i in start..<sArr.count {
        if isPalindrome(sArr: sArr, left: start, right: i) {
            let substring = String(sArr[start...i])
            path.append(substring)
            helper(sArr: sArr, start: i+1, path: &path, result: &result)
            path.removeLast()
        }
    }
}

private func isPalindrome(sArr: [Character], left: Int, right: Int) -> Bool {
    var l = left, r = right
    while l < r {
        if sArr[l] != sArr[r] {
            return false
        }
        l += 1
        r -= 1
    }
    return true
}

This change avoids creating intermediate arrays for palindrome checks and should improve performance.

Overall, your solution is solid and only requires minor optimizations for better efficiency.

VERDICT: PASS