Subsets II
Problem Statement :
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
Solution :
Solution in C :
void porcess(int** ans, int* idx, int* data, int pos, int** nums, int numsSize, int* colnums){
if(numsSize == 0){
colnums[*idx] = pos;
if(pos == 0){
*idx = *idx + 1;
return ;
}
ans[*idx] = malloc(pos * sizeof(int));
memcpy(ans[*idx], data, pos * sizeof(int));
*idx = *idx + 1;
return ;
}
for(int i = 0; i <= nums[0][1]; i++){
if(i == 0){
porcess(ans, idx, data, pos, &nums[1], numsSize-1, colnums);
}
else{
data[pos] = nums[0][0];
pos++;
porcess(ans, idx, data, pos, &nums[1], numsSize-1, colnums);
}
}
}
int** subsetsWithDup(int* nums, int numsSize, int* returnSize, int** returnColumnSizes){
int** ans = malloc(1024 * sizeof(int*));
returnColumnSizes[0] = malloc(1024 * sizeof(int));
int* data = malloc(10 * sizeof(int));
int* idx = calloc(1 , sizeof(int));
//>>>> create n_nums[val, count]
int* hash = calloc(21 , sizeof(int));
for(int i = 0; i < numsSize ; i++){
hash[ nums[i] + 10]++;
}
int** n_nums = malloc(numsSize * sizeof(int*));
int n_numsSize = 0;
for(int i = 0; i < 21; i++){
if(hash[i] != 0){
n_nums[n_numsSize] = malloc(2 * sizeof(int));
n_nums[n_numsSize][0] = i - 10;
n_nums[n_numsSize][1] = hash[i];
n_numsSize++;
}
}
free(hash);
//<<<<
porcess(ans, idx, data, 0, n_nums, n_numsSize, * returnColumnSizes);
ans = realloc(ans, (*idx)* sizeof(int*));
returnColumnSizes[0] = realloc(returnColumnSizes[0], (*idx) * sizeof(int));
* returnSize = *idx;
return ans;
}
Solution in C++ :
class Solution {
public:
void fun(vector<int>& nums, int index, vector<int>ds, set<vector<int>>& res) {
if(index == nums.size()) {
res.insert(ds);
return;
}
ds.push_back(nums[index]);
fun(nums, index+1, ds, res);
ds.pop_back();
fun(nums, index+1, ds, res);
}
vector<vector<int>> subsetsWithDup(vector<int>& nums) {
sort(nums.begin(), nums.end());
vector<vector<int>> ans;
set<vector<int>> res;
vector<int> ds;
fun(nums, 0, ds, res);
for(auto it = res.begin(); it!= res.end(); it++) {
ans.push_back(*it);
}
return ans;
}
};
Solution in Java :
class Solution {
public List<List<Integer>> subsetsWithDup(int[] nums) {
List<List<Integer>> ans = new ArrayList<>();
Set<List<Integer>> res = new HashSet<>();
List<Integer> ds = new ArrayList<>();
Arrays.sort(nums); // Sort the input array
fun(nums, 0, ds, res);
for (List<Integer> subset : res) {
ans.add(subset);
}
return ans;
}
private void fun(int[] nums, int index, List<Integer> ds, Set<List<Integer>> res) {
if (index == nums.length) {
res.add(new ArrayList<>(ds));
return;
}
ds.add(nums[index]);
fun(nums, index + 1, ds, res);
ds.remove(ds.size() - 1);
fun(nums, index + 1, ds, res);
}
}
Solution in Python :
class Solution:
def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
ans = []
res = set()
ds = []
nums.sort()
self.fun(nums, 0, ds, res)
for subset in res:
ans.append(list(subset))
return ans
def fun(self, nums, index, ds, res):
if index == len(nums):
sorted_ds = tuple(sorted(ds))
res.add(sorted_ds)
return
ds.append(nums[index])
self.fun(nums, index + 1, ds, res)
ds.pop()
self.fun(nums, index + 1, ds, res)
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