Question

2597. The Number of Beautiful Subsets

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Description

You are given an array nums of positive integers and a positive integer k.

A subset of nums is beautiful if it does not contain two integers with an absolute difference equal to k.

Return _the number of non-empty beautiful subsets of the array_ nums.

A subset of nums is an array that can be obtained by deleting some (possibly none) elements from nums. Two subsets are different if and only if the chosen indices to delete are different.

 

Example 1:

Input: nums = [2,4,6], k = 2
Output: 4
Explanation: The beautiful subsets of the array nums are: [2], [4], [6], [2, 6].
It can be proved that there are only 4 beautiful subsets in the array [2,4,6].

Example 2:

Input: nums = [1], k = 1
Output: 1
Explanation: The beautiful subset of the array nums is [1].
It can be proved that there is only 1 beautiful subset in the array [1].

 

Constraints:

• 1 <= nums.length <= 20
• 1 <= nums[i], k <= 1000

Solutions

Solution 1: Counting + Backtracking

We use a hash table or an array $cnt$ to record the currently selected numbers and their counts, and use $ans$ to record the number of beautiful subsets, initially $ans = -1$, indicating that the empty set is excluded.

For each number $x$ in the array $nums$, we have two choices:

• Do not choose $x$, and then directly recurse to the next number;
• Choose $x$, then we need to check whether $x + k$ and $x - k$ have appeared in $cnt$ before, if neither has appeared before, then we can choose $x$, at this time we add one to the number of $x$, and then recurse to the next number, and finally subtract one from the number of $x$.

Finally, we return $ans$.

Time complexity $O(2^n)$, space complexity $O(n)$, where $n$ is the length of the array $nums$.

class Solution:
  def beautifulSubsets(self, nums: List[int], k: int) -> int:
    def dfs(i: int) -> None:
      nonlocal ans
      if i >= len(nums):
        ans += 1
        return
      dfs(i + 1)
      if cnt[nums[i] + k] == 0 and cnt[nums[i] - k] == 0:
        cnt[nums[i]] += 1
        dfs(i + 1)
        cnt[nums[i]] -= 1

    ans = -1
    cnt = Counter()
    dfs(0)
    return ans

class Solution {
  private int[] nums;
  private int[] cnt = new int[1010];
  private int ans = -1;
  private int k;

  public int beautifulSubsets(int[] nums, int k) {
    this.k = k;
    this.nums = nums;
    dfs(0);
    return ans;
  }

  private void dfs(int i) {
    if (i >= nums.length) {
      ++ans;
      return;
    }
    dfs(i + 1);
    boolean ok1 = nums[i] + k >= cnt.length || cnt[nums[i] + k] == 0;
    boolean ok2 = nums[i] - k < 0 || cnt[nums[i] - k] == 0;
    if (ok1 && ok2) {
      ++cnt[nums[i]];
      dfs(i + 1);
      --cnt[nums[i]];
    }
  }
}

class Solution {
public:
  int beautifulSubsets(vector<int>& nums, int k) {
    int ans = -1;
    int cnt[1010]{};
    int n = nums.size();

    function<void(int)> dfs = [&](int i) {
      if (i >= n) {
        ++ans;
        return;
      }
      dfs(i + 1);
      bool ok1 = nums[i] + k >= 1010 || cnt[nums[i] + k] == 0;
      bool ok2 = nums[i] - k < 0 || cnt[nums[i] - k] == 0;
      if (ok1 && ok2) {
        ++cnt[nums[i]];
        dfs(i + 1);
        --cnt[nums[i]];
      }
    };
    dfs(0);
    return ans;
  }
};

func beautifulSubsets(nums []int, k int) int {
  ans := -1
  n := len(nums)
  cnt := [1010]int{}
  var dfs func(int)
  dfs = func(i int) {
    if i >= n {
      ans++
      return
    }
    dfs(i + 1)
    ok1 := nums[i]+k >= len(cnt) || cnt[nums[i]+k] == 0
    ok2 := nums[i]-k < 0 || cnt[nums[i]-k] == 0
    if ok1 && ok2 {
      cnt[nums[i]]++
      dfs(i + 1)
      cnt[nums[i]]--
    }
  }
  dfs(0)
  return ans
}

function beautifulSubsets(nums: number[], k: number): number {
  let ans: number = -1;
  const cnt: number[] = new Array(1010).fill(0);
  const n: number = nums.length;
  const dfs = (i: number) => {
    if (i >= n) {
      ++ans;
      return;
    }
    dfs(i + 1);
    const ok1: boolean = nums[i] + k >= 1010 || cnt[nums[i] + k] === 0;
    const ok2: boolean = nums[i] - k < 0 || cnt[nums[i] - k] === 0;
    if (ok1 && ok2) {
      ++cnt[nums[i]];
      dfs(i + 1);
      --cnt[nums[i]];
    }
  };
  dfs(0);
  return ans;
}