78. Subsets

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;
}

doocs / leetcode

78. Subsets

Description

Solutions

Solution 1: DFS (Backtracking)

Solution 2: Binary Enumeration

简介

发行版

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78. Subsets

Description

Solutions

Solution 1: DFS (Backtracking)

Solution 2: Binary Enumeration

简介

发行版

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