Subsets II

shriyanshi24jindal

Mar 30th, 2026

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Java 6.03 KB | None | 0 0

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class Solution {
public List<List<Integer>> subsetsWithDup(int[] nums) {
// sorting to skip over duplicate elements during recursion
Arrays.sort(nums);
List<List<Integer>> result = new ArrayList<>();
List<Integer> temp = new ArrayList<>();
findSubsetsWithDuplicates(nums, 0, result, temp);
return result;
}
public void findSubsetsWithDuplicates(int[] nums, int index, List<List<Integer>> result, List<Integer> temp)
{
// base condition
if(index == nums.length)
{
result.add(new ArrayList<>(temp));
return;
}
// take the current element into the subset
temp.add(nums[index]);
findSubsetsWithDuplicates(nums, index+1, result, temp);
// do not take the current element into the subset
temp.remove(temp.size()-1);
// Skip all duplicates of the current number
while(index+1 < nums.length && nums[index] == nums[index+1])
index++;
findSubsetsWithDuplicates(nums, index+1, result, temp);
}
}

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class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        // sorting to skip over duplicate elements during recursion
        Arrays.sort(nums);
        List<List<Integer>> result = new ArrayList<>();
        List<Integer> temp = new ArrayList<>();

findSubsetsWithDuplicates(nums, 0, result, temp);
        return result;
    }

public void findSubsetsWithDuplicates(int[] nums, int index, List<List<Integer>> result, List<Integer> temp)
    {
        // base condition
        if(index == nums.length)
        {
            result.add(new ArrayList<>(temp));
            return;
        }

// take the current element into the subset
        temp.add(nums[index]);
        findSubsetsWithDuplicates(nums, index+1, result, temp);

// do not take the current element into the subset
        temp.remove(temp.size()-1);

// Skip all duplicates of the current number
        while(index+1 < nums.length && nums[index] == nums[index+1])
            index++;
        
        findSubsetsWithDuplicates(nums, index+1, result, temp);
    }
}