Given an integer array nums of unique elements, return all possible subsets (the power set).
The solution set must not contain duplicate subsets. Return the solution in any order.
Example 1
Input: nums = [1,2,3]
Output: [[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]
Example 2
Input: nums = [1]
Output: [[],[1]]
Constraints
Solution
You might need to find the relationship between the result of the array nums with the result of itself without the last element.
Example 3
Input: nums = [1,2]
Output: [[],[1],[2],[1,2]]
You can see the powerset of Example 3 was obtained from the one in Example 2 with additional subsets [2], [1,2]. These new subsets were constructed from subsets [], [1] of Example 2 appended with the new element 2.
Similarly, the powerset of Example 1 was obtained from the one in Example 3 with the additional subsets [3], [1,3], [2,3], [1,2,3]. These new subsets were constructed from the ones of Example 3 appended with the new element 3.
Code
#include <vector>
#include <iostream>
using namespace std;
vector<vector<int>> subsets(vector<int>& nums) {
vector<vector<int>> powerset = {{}};
int i = 0;
while (i < nums.size()) {
vector<vector<int>> newSubsets;
for (auto subset : powerset) {
subset.push_back(nums[i]);
newSubsets.push_back(subset);
}
powerset.insert(powerset.end(), newSubsets.begin(), newSubsets.end());
i++;
}
return powerset;
}
void print(vector<vector<int>>& powerset) {
for (auto& set : powerset ) {
cout << "[";
for (auto& element : set) {
cout << element << ",";
}
cout << "]";
}
cout << endl;
}
int main() {
vector<int> nums{1,2,3};
auto powerset = subsets(nums);
print(powerset);
nums = {1};
powerset = subsets(nums);
print(powerset);
}
Output:
[][1,][2,][1,2,][3,][1,3,][2,3,][1,2,3,]
[][1,]
Code explanation
The code initializes an empty vector of vectors called powerset to store the subsets. It starts with an initial subset containing an empty vector, representing the empty set: {{}}.
Within the loop, it creates a new vector of vectors called newSubsets to store subsets that include the i-th element of nums.
It iterates through the existing subsets in powerset. For each subset, it creates a copy of it (represented by the subset variable), adds the i-th element from nums to the copied subset, and pushes the modified subset into the newSubsets vector.
After processing all existing subsets, the code adds the contents of newSubsets to the powerset vector. This effectively combines the current subsets with subsets that include the i-th element of nums.
It repeats steps 2-4 until all elements in nums have been considered.
Finally, the code returns the powerset, which now contains all possible subsets of nums.
Complexity
This code essentially generates subsets by iteratively adding each element of nums to the existing subsets and accumulating the results. The time complexity of this code is O(2^N), where N is the number of elements in nums, as it generates all possible subsets. The space complexity is also O(2^N) due to the space required to store the subsets.
Runtime: O(2^N).
Extra space: O(2^N).
Thanks for reading. Feel free to share your thought about my content and check out my FREE book 10 Classic Coding Challenges.
