Backtracking-1 Done

[Sreeja-99]

No description provided.

[super30admin]

Strengths:

• You have provided two correct solutions for the Combination Sum problem, demonstrating understanding of different approaches.
• The code is clean, well-commented, and uses backtracking efficiently to save space.
• You have correctly avoided duplicates by using a pivot index in the for-loop approach.

Areas for Improvement:

• In Way1, the base case checks for i == candidates.length, which is correct. However, in Way2, you don't have an explicit check for the index beyond the array length, but it is not needed because the for loop condition i < candidates.length handles it. This is fine.
• In Way2, you might want to consider adding an early termination condition: if target < 0, we return. But note that in the for loop, we are adding a candidate and then recursing. However, if the candidates are positive (as per constraints), we can also break out of the loop if the current candidate makes the target negative? Actually, since the array is sorted? But the problem does not require sorted candidates. However, if we sort the array, we can break early when the current candidate is greater than the remaining target. But the problem does not require sorting. This is an optional optimization.
• The problem states that the candidates are distinct, but your solutions do not require sorting. However, if you sort the array, you can break early in the for loop when candidates[i] > target (since all subsequent candidates will be larger if sorted). This can improve efficiency. For example, in Way2, after path.add(candidates[i]), if candidates[i] > target, then we skip the recursive call? Actually, we are subtracting candidates[i] from target, so if candidates[i] > target, then target - candidates[i] becomes negative. So we can break out of the loop if candidates[i] > target only if the array is sorted. Otherwise, we cannot. So if you sort the array first, you can break early. But the problem does not require the output in any particular order.

Suggestion: You can sort the array at the beginning to enable early termination. For example:

Arrays.sort(candidates);

Then in the for loop of Way2, you can add:

if (candidates[i] > target) break;

This will prune the recursion tree.

But note: since the problem allows repeated elements, even if the current candidate is greater than the target, we break, but for the same candidate, we are adding only once? Actually, in the for loop, we are iterating from pivot to end. So if the array is sorted, and at a point candidates[i] > target, then all subsequent candidates are also greater, so we break the loop.

This optimization is optional but recommended.

Overall, your solutions are correct and efficient.