Subsets II – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
Given an integer array nums that may contain duplicates, 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 = [1,2,2] Output: [[],[1],[1,2],[1,2,2],[2],[2,2]]
Example 2:
Input: nums = [0] Output: [[],[0]]
Constraints:
1 <= nums.length <= 10-10 <= nums[i] <= 10
C++ Subsets II LeetCode Solution
class Solution {
public:
vector<vector<int> > subsetsWithDup(vector<int> &S) {
vector<vector<int> > totalset = {{}};
sort(S.begin(),S.end());
for(int i=0; i<S.size();){
int count = 0; // num of elements are the same
while(count + i<S.size() && S[count+i]==S[i]) count++;
int previousN = totalset.size();
for(int k=0; k<previousN; k++){
vector<int> instance = totalset[k];
for(int j=0; j<count; j++){
instance.push_back(S[i]);
totalset.push_back(instance);
}
}
i += count;
}
return totalset;
}
};
Java Subsets II LeetCode Solution
class Solution {
public List<List<Integer>> subsetsWithDup(int[] nums) {
Arrays.sort(nums);
List<List<Integer>> res = new ArrayList<>();
helper(res,new ArrayList<>(),nums,0);
return res;
}
public void helper(List<List<Integer>> res, List<Integer> ls, int[] nums, int pos) {
res.add(new ArrayList<>(ls));
for(int i=pos;i<nums.length;i++) {
if(i>pos&&nums[i]==nums[i-1]) continue;
ls.add(nums[i]);
helper(res,ls,nums,i+1);
ls.remove(ls.size()-1);
}
}
}
Python 3 Subsets II LeetCode Solution
class Solution:
# @param num, a list of integer
# @return a list of lists of integer
def subsetsWithDup(self, S):
res = [[]]
S.sort()
for i in range(len(S)):
if i == 0 or S[i] != S[i - 1]:
l = len(res)
for j in range(len(res) - l, len(res)):
res.append(res[j] + [S[i]])
return res
Array-1180
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Bit Manipulation-140
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Prefix Sum-96
Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
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Game Theory-24
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Interactive-18
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Shell-4
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