grid47 Exploring patterns and algorithms
Oct 29, 2024
5 min read
Solution to LeetCode 90: Subsets II Problem
Given an integer array nums that may contain duplicates, return all possible subsets (the power set) of the array. The solution set should not contain duplicate subsets. The subsets should be returned in any order.
Problem
Approach
Steps
Complexity
Input: The input is an integer array nums, which may contain duplicate values.
Example: Input: nums = [3, 3, 1]
Constraints:
• 1 <= nums.length <= 10
• -10 <= nums[i] <= 10
Output: The output should be a list of lists, where each inner list represents a unique subset of the input array.
Goal: The goal is to generate all possible subsets of the input array while avoiding duplicates. This can be done by iterating through the array, considering each element either as part of a subset or not, and ensuring that subsets with duplicate elements are not included.
Steps:
• Sort the input array to ensure duplicates are adjacent.
• Use a backtracking approach to generate subsets, ensuring that if an element is included, the next occurrence of the same element is only included if it was included previously in the current subset.
Goal: The input array length is between 1 and 10, and elements are in the range from -10 to 10.
Steps:
• The input array may contain duplicate values.
• The result should not include duplicate subsets.
Assumptions:
• The input array is not empty.
• The input array length is at most 10, so it is feasible to generate all subsets.
• Input: Input: nums = [1, 2, 2]
• Explanation: For input [1, 2, 2], the possible unique subsets are: [], [1], [1, 2], [1, 2, 2], [2], and [2, 2]. The subset [2] appears twice, so only one of them is included in the result.
• Input: Input: nums = [0]
• Explanation: For input [0], the only subsets are the empty set [] and the set containing the element itself [0].
Approach: We can solve this problem using backtracking to generate all subsets. Sorting the array first helps ensure that duplicate subsets are avoided by making sure the same elements are adjacent.
Observations:
• The backtracking approach can help generate subsets by either including or excluding an element.
• The sorted array ensures that duplicate elements are adjacent and can be handled easily during the backtracking process.
• Handling duplicates requires careful management during the backtracking to avoid generating the same subset multiple times.
Steps:
• Sort the input array to bring duplicates together.
• Use a recursive backtracking function that explores all subsets by including or excluding elements.
• Track the subsets in a set or map to automatically discard duplicates.
Empty Inputs:
• An empty input array will result in a single subset: the empty set []
Large Inputs:
• For large inputs (n = 10), the solution must handle 2^n subsets efficiently.
Special Values:
• If the array contains duplicate values, the solution must ensure only unique subsets are returned.
Constraints:
• The constraints (1 <= nums.length <= 10) allow for generating all subsets directly, as 2^10 subsets is a manageable number.
vector<vector<int>> subsetsWithDup(vector<int>& nums) {
sort(nums.begin(), nums.end());
vector<vector<int>> res = {{}};
int startIndex =0, endIndex =0;
for (int i =0; i < nums.size(); i++) {
int n = res.size();
for (int j =0; j < n; j++) {
if (i ==0|| nums[i] != nums[i -1] || j >= startIndex) {
vector<int> temp = res[j];
temp.push_back(nums[i]);
res.push_back(temp);
}
}
startIndex = i +1;
}
return res;
}
Declares a function `subsetsWithDup` that takes a vector of integers `nums` as input and returns a vector of vectors representing all unique subsets.
2 : Sorting
sort(nums.begin(), nums.end());
Sorts the input array `nums` in ascending order to handle duplicates efficiently.
3 : Array Initialization
vector<vector<int>> res = {{}};
Initializes a vector `res` to store the subsets. The initial empty subset `{}` is added.
4 : Variable Initialization
int startIndex =0, endIndex =0;
Initializes `startIndex` and `endIndex` to keep track of the range of indices to consider for adding elements to subsets.
5 : Loop Iteration
for (int i =0; i < nums.size(); i++) {
Iterates through each element `nums[i]` in the sorted array.
6 : Size Calculation
int n = res.size();
Stores the current size of the `res` vector.
7 : Nested Loop Iteration
for (int j =0; j < n; j++) {
Iterates through the existing subsets in `res`.
8 : Conditional, Array Manipulation
if (i ==0|| nums[i] != nums[i -1] || j >= startIndex) {
Checks if it's the first iteration, or if the current element is different from the previous one, or if we've reached the `startIndex` for the current element. This ensures we avoid duplicates.
9 : Array Manipulation
vector<int> temp = res[j];
Creates a copy `temp` of the current subset `res[j]`.
10 : Array Manipulation
temp.push_back(nums[i]);
Adds the current element `nums[i]` to the `temp` subset.
11 : Array Manipulation
res.push_back(temp);
Adds the new subset `temp` to the `res` vector.
12 : Index Update
startIndex = i +1;
Updates `startIndex` to the next index for the next iteration, skipping duplicate elements.
13 : Return
return res;
Returns the final `res` vector containing all unique subsets.
Best Case: O(2^n), where n is the number of elements in the input array, since the solution involves generating all possible subsets.
Average Case: O(2^n)
Worst Case: O(2^n), as we must generate and check each subset once.
Description: The time complexity is exponential, as we are generating 2^n subsets.
Best Case: O(1), if no subsets are generated.
Worst Case: O(2^n), since we store all subsets.
Description: The space complexity depends on the number of subsets stored.
