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LeetCode 90: Subsets II Solution

Master LeetCode problem 90 (Subsets II), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

Problem Explanation

Explanation

To solve this problem, we can use backtracking to generate all possible subsets while avoiding duplicates. By sorting the input array, we can ensure that duplicates are grouped together, making it easier to skip them during the backtracking process.

  1. Sort the input array to group duplicates together.
  2. Initialize an empty list to store the subsets.
  3. Use a recursive backtracking function to generate subsets:
    • Start with an empty subset.
    • For each element in the array, add it to the current subset and recursively generate subsets with and without the current element.
    • Skip duplicates to avoid duplicate subsets.
  4. Return the list of subsets.

The time complexity of this solution is O(2^n) as there are 2^n subsets. The space complexity is O(n) for the recursion stack.

Solution Code

class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        Arrays.sort(nums);
        List<List<Integer>> subsets = new ArrayList<>();
        generateSubsets(nums, 0, new ArrayList<>(), subsets);
        return subsets;
    }

    private void generateSubsets(int[] nums, int start, List<Integer> subset, List<List<Integer>> subsets) {
        subsets.add(new ArrayList<>(subset));
        for (int i = start; i < nums.length; i++) {
            if (i > start && nums[i] == nums[i - 1]) {
                continue; // skip duplicates
            }
            subset.add(nums[i]);
            generateSubsets(nums, i + 1, subset, subsets);
            subset.remove(subset.size() - 1);
        }
    }
}

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