backtracking-1 done

[hvrgupta]

No description provided.

[super30admin]

Your solution for CombinationSum is well-written and correct. You have used a backtracking approach with a for loop and a pivot index to avoid duplicates. The code is clean and efficient.

Strengths:

• You avoid deep copies by using a single path list and backtracking by removing the last element.
• You skip candidates that would exceed the target, which prunes the recursion tree.

Areas for improvement:

1. Time Complexity Analysis: Your stated time complexity of O(2^N) is not accurate. The actual time complexity is exponential in the target value. A better way to describe it would be O(n^(target/min_candidate)) which is exponential. However, since the target is bounded by 40, it is acceptable.
2. Space Complexity: You stated O(n) which only accounts for the recursion depth. However, the space required to store the results is significant. You should mention that the space complexity is O(n * h) where h is the number of combinations.
3. Optimization: Consider sorting the candidates. Then, in the for loop, you can break (instead of continue) when the current candidate is greater than the remaining target because all subsequent candidates will be larger. This will prune the recursion further.

Example modification:

Arrays.sort(candidates);
// ... inside helper ...
for (int i = pivot; i < candidates.length; i++) {
    if (candidates[i] > target) break;  // break since sorted
    path.add(candidates[i]);
    helper(candidates, target - candidates[i], i, path);
    path.remove(path.size() - 1);
}

4. Redundant Check: You have an inner if condition if (target - candidates[i] >= 0) after the continue condition. This is redundant because the continue condition already ensures that we only proceed when target - candidates[i] >= 0. You can remove that inner if.

Overall, your solution is efficient and correct. With minor improvements, it can be even better.