I knew LC 39 well enough but the reversed traversal direction threw me off more than it should have.
Use a backtracking algorithm that iterates through the array from the last index to the first, allowing reuse of each element by staying at the same index after including it. At each step, subtract the chosen element from the remaining target and recurse; when the target reaches zero, record the combination. This naturally produces combinations in reverse order and ensures uniqueness by only considering elements from the current index onward.
Pro tip: Emphasize that iterating from the end doesn't change the algorithm's complexity but demonstrates adaptability; mention that sorting the array first can enable early pruning if elements are positive, and clarify that the output order is reversed but combinations are still unique.
Restate the problem to ensure understanding: find all unique combinations summing to target, with unlimited reuse, but iterate from the end. Ask about constraints like array size, element range, and whether negative numbers are allowed.
Explain that you'll use recursion with a start index, but loop from the last index down to the start index. At each recursive call, include the current element and recurse with the same index (for reuse) and reduced target.
Describe base cases: if target == 0, add the current combination to results; if target < 0 or index out of bounds, return. Ensure uniqueness by only considering elements from the current index onward, avoiding permutations.
Write pseudocode or code, clearly showing the reverse loop. Trace through a small example (e.g., [2,3,6,7], target=7) to demonstrate correctness and the order of combinations.
State time complexity O(N^(T/M)) where T is target and M is minimal element, space O(T/M) for recursion depth. Mention that sorting can help prune if elements are positive, but note that reverse iteration doesn't affect complexity.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This is the LC 40 variant and the key move is sorting first, then skipping over repeated values at the same recursion depth.
First, clarify that the core algorithm remains backtracking, but we need to avoid generating duplicate combinations/permutations by skipping duplicate elements at the same recursion level. Then, explain the two key modifications: sorting the input to bring duplicates together, and using a 'used' boolean array or a 'start index' to skip duplicates during iteration. Finally, emphasize that the skipping logic must only apply at the same depth, not across different depths, to allow legitimate duplicates in different positions.
Pro tip: Mention that sorting is O(n log n) and doesn't affect overall complexity, but it's crucial for efficient duplicate skipping. Also, note that the same technique applies to both combinations (e.g., combination sum II) and permutations (e.g., permutations II), but the skip condition differs slightly.
Sorting groups identical elements together, making it easy to detect and skip duplicates during recursion. This is a prerequisite for the duplicate-skipping logic.
For permutations, maintain a boolean 'used' array to track which elements are already in the current path. For combinations, use a start index to avoid reusing earlier elements.
During iteration, if the current element equals the previous element and the previous element is not used (or we are at the same level), skip the current element to avoid duplicate branches.
The condition must check that the duplicate is not part of the current path (e.g., for permutations: i > 0 && nums[i] == nums[i-1] && !used[i-1]). This ensures duplicates are allowed in different branches but not in the same level.
Sorting adds O(n log n) time, but the overall complexity remains O(n * 2^n) for combinations or O(n * n!) for permutations. Skipping duplicates reduces the number of recursive calls, improving practical performance.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.