Question

78. Subsets

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Description

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 = [0]
Output: [[],[0]]

 

Constraints:

• 1 <= nums.length <= 10
• -10 <= nums[i] <= 10
• All the numbers of nums are unique.

Solutions

Solution 1: DFS (Backtracking)

We design a function $dfs(i)$, which represents starting the search from the $i$th element of the array for all subsets. The execution logic of the function $dfs(i)$ is as follows:

• If $i = n$, it means the current search has ended. Add the current subset $t$ to the answer array $ans$, and then return.
• Otherwise, we can choose not to select the current element and directly execute $dfs(i + 1)$; or we can choose the current element, i.e., add the current element $nums[i]$ to the subset $t$, and then execute $dfs(i + 1)$. Note that we need to remove $nums[i]$ from the subset $t$ after executing $dfs(i + 1)$ (backtracking).

In the main function, we call $dfs(0)$, i.e., start searching all subsets from the first element of the array. Finally, return the answer array $ans$.

The time complexity is $O(n \times 2^n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array. There are a total of $2^n$ subsets, and each subset takes $O(n)$ time to construct.

class Solution:
  def subsets(self, nums: List[int]) -> List[List[int]]:
    def dfs(i: int):
      if i == len(nums):
        ans.append(t[:])
        return
      dfs(i + 1)
      t.append(nums[i])
      dfs(i + 1)
      t.pop()

    ans = []
    t = []
    dfs(0)
    return ans

class Solution {
  private List<List<Integer>> ans = new ArrayList<>();
  private List<Integer> t = new ArrayList<>();
  private int[] nums;

  public List<List<Integer>> subsets(int[] nums) {
    this.nums = nums;
    dfs(0);
    return ans;
  }

  private void dfs(int i) {
    if (i == nums.length) {
      ans.add(new ArrayList<>(t));
      return;
    }
    dfs(i + 1);
    t.add(nums[i]);
    dfs(i + 1);
    t.remove(t.size() - 1);
  }
}

class Solution {
public:
  vector<vector<int>> subsets(vector<int>& nums) {
    vector<vector<int>> ans;
    vector<int> t;
    function<void(int)> dfs = [&](int i) -> void {
      if (i == nums.size()) {
        ans.push_back(t);
        return;
      }
      dfs(i + 1);
      t.push_back(nums[i]);
      dfs(i + 1);
      t.pop_back();
    };
    dfs(0);
    return ans;
  }
};

func subsets(nums []int) (ans [][]int) {
  t := []int{}
  var dfs func(int)
  dfs = func(i int) {
    if i == len(nums) {
      ans = append(ans, append([]int(nil), t...))
      return
    }
    dfs(i + 1)
    t = append(t, nums[i])
    dfs(i + 1)
    t = t[:len(t)-1]
  }
  dfs(0)
  return
}

function subsets(nums: number[]): number[][] {
  const ans: number[][] = [];
  const t: number[] = [];
  const dfs = (i: number) => {
    if (i === nums.length) {
      ans.push(t.slice());
      return;
    }
    dfs(i + 1);
    t.push(nums[i]);
    dfs(i + 1);
    t.pop();
  };
  dfs(0);
  return ans;
}

impl Solution {
  fn dfs(i: usize, t: &mut Vec<i32>, res: &mut Vec<Vec<i32>>, nums: &Vec<i32>) {
    if i == nums.len() {
      res.push(t.clone());
      return;
    }
    Self::dfs(i + 1, t, res, nums);
    t.push(nums[i]);
    Self::dfs(i + 1, t, res, nums);
    t.pop();
  }

  pub fn subsets(nums: Vec<i32>) -> Vec<Vec<i32>> {
    let mut res = Vec::new();
    Self::dfs(0, &mut Vec::new(), &mut res, &nums);
    res
  }
}

Solution 2: Binary Enumeration

We can also use the method of binary enumeration to get all subsets.

We can use $2^n$ binary numbers to represent all subsets of $n$ elements. For the current binary number $mask$, if the $i$th bit is $1$, it means that the $i$th element is selected, otherwise it means that the $i$th element is not selected.

The time complexity is $O(n \times 2^n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array. There are a total of $2^n$ subsets, and each subset takes $O(n)$ time to construct.

class Solution:
  def subsets(self, nums: List[int]) -> List[List[int]]:
    ans = []
    for mask in range(1 << len(nums)):
      t = [x for i, x in enumerate(nums) if mask >> i & 1]
      ans.append(t)
    return ans

class Solution {
  public List<List<Integer>> subsets(int[] nums) {
    int n = nums.length;
    List<List<Integer>> ans = new ArrayList<>();
    for (int mask = 0; mask < 1 << n; ++mask) {
      List<Integer> t = new ArrayList<>();
      for (int i = 0; i < n; ++i) {
        if (((mask >> i) & 1) == 1) {
          t.add(nums[i]);
        }
      }
      ans.add(t);
    }
    return ans;
  }
}

class Solution {
public:
  vector<vector<int>> subsets(vector<int>& nums) {
    int n = nums.size();
    vector<vector<int>> ans;
    for (int mask = 0; mask < 1 << n; ++mask) {
      vector<int> t;
      for (int i = 0; i < n; ++i) {
        if (mask >> i & 1) {
          t.emplace_back(nums[i]);
        }
      }
      ans.emplace_back(t);
    }
    return ans;
  }
};

func subsets(nums []int) (ans [][]int) {
  n := len(nums)
  for mask := 0; mask < 1<<n; mask++ {
    t := []int{}
    for i, x := range nums {
      if mask>>i&1 == 1 {
        t = append(t, x)
      }
    }
    ans = append(ans, t)
  }
  return
}

function subsets(nums: number[]): number[][] {
  const n = nums.length;
  const ans: number[][] = [];
  for (let mask = 0; mask < 1 << n; ++mask) {
    const t: number[] = [];
    for (let i = 0; i < n; ++i) {
      if (((mask >> i) & 1) === 1) {
        t.push(nums[i]);
      }
    }
    ans.push(t);
  }
  return ans;
}