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In this approach, the problem of subsets with duplicates is solved by first sorting the array. This assists in easily identifying and omitting duplicates. The backtracking method is utilized to construct all possible subsets, ensuring no duplicate subsets are generated by skipping duplicate numbers during recursive backtracking.
The time complexity of this solution is O(2^n * n) due to the number of subsets (2^n) and time taken to convert each subset to the result. The space complexity is O(n), where n is the length of the array, due to the space required for the subset building.
1function subsetsWithDup(nums) {
2 nums.sort((a, b) => a - b);
3 const result = [];
4
5 function backtrack(start, path) {
6 result.push([...path]);
7 for (let i = start; i < nums.length; i++) {
8 if (i > start && nums[i] === nums[i - 1]) continue;
9 path.push(nums[i]);
10 backtrack(i + 1, path);
11 path.pop();
12 }
13 }
14
15 backtrack(0, []);
16 return result;
17}
18
19// Example Usage
20console.log(subsetsWithDup([1, 2, 2]));JavaScript implements the same strategy of ensuring sorted order and applying backtracking to traverse subsets. The method efficiently avoids duplicate paths by ensuring that duplicate elements are skipped when necessary.
This approach iteratively constructs the power set by extending previously generated sets with each new number. During the process, duplicates are skipped by comparing the current number with the previous one. This ensures subsets generated in each iteration are unique.
The time complexity is O(2^n * n); space is O(2^n * n) as it iteratively builds the power set using subsets directly.
1
In this Java solution, subsets are iteratively expanded. The sorting makes managing the duplicates straightforward. Subset collections dynamically grow, maintaining control over redundancy by observing and limiting duplicate origins.