Question

2488. Count Subarrays With Median K

中文文档

Description

You are given an array nums of size n consisting of distinct integers from 1 to n and a positive integer k.

Return _the number of non-empty subarrays in _nums_ that have a median equal to _k.

Note:

• The median of an array is the middle element after sorting the array in ascending order. If the array is of even length, the median is the left middle element.
  • For example, the median of [2,3,1,4] is 2, and the median of [8,4,3,5,1] is 4.
• A subarray is a contiguous part of an array.

 

Example 1:

Input: nums = [3,2,1,4,5], k = 4
Output: 3
Explanation: The subarrays that have a median equal to 4 are: [4], [4,5] and [1,4,5].

Example 2:

Input: nums = [2,3,1], k = 3
Output: 1
Explanation: [3] is the only subarray that has a median equal to 3.

 

Constraints:

• n == nums.length
• 1 <= n <= 105
• 1 <= nums[i], k <= n
• The integers in nums are distinct.

Solutions

Solution 1: Traversal + Counting

First, we find the position $i$ of the median $k$ in the array, and then start traversing from $i$ to both sides, counting the number of subarrays with a median of $k$.

Define an answer variable $ans$, which represents the number of subarrays with a median of $k$. Initially, $ans = 1$, which means that there is currently a subarray of length $1$ with a median of $k$. In addition, define a counter $cnt$, used to count the number of differences between the "number of elements larger than $k$" and the "number of elements smaller than $k$" in the currently traversed array.

Next, start traversing to the right from $i + 1$. We maintain a variable $x$, which represents the difference between the "number of elements larger than $k$" and the "number of elements smaller than $k$" in the current right subarray. If $x \in [0, 1]$, then the median of the current right subarray is $k$, and the answer variable $ans$ is incremented by $1$. Then, we add the value of $x$ to the counter $cnt$.

Similarly, start traversing to the left from $i - 1$, also maintaining a variable $x$, which represents the difference between the "number of elements larger than $k$" and the "number of elements smaller than $k$" in the current left subarray. If $x \in [0, 1]$, then the median of the current left subarray is $k$, and the answer variable $ans$ is incremented by $1$. If $-x$ or $-x + 1$ is also in the counter, it means that there is currently a subarray that spans both sides of $i$, with a median of $k$, and the answer variable $ans$ increases the corresponding value in the counter, that is, $ans += cnt[-x] + cnt[-x + 1]$.

Finally, return the answer variable $ans$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array.

In coding, we can directly open an array of length $2 \times n + 1$, used to count the difference between the "number of elements larger than $k$" and the "number of elements smaller than $k$" in the current array. Each time we add the difference by $n$, we can convert the range of the difference from $[-n, n]$ to $[0, 2n]$.

class Solution:
  def countSubarrays(self, nums: List[int], k: int) -> int:
    i = nums.index(k)
    cnt = Counter()
    ans = 1
    x = 0
    for v in nums[i + 1 :]:
      x += 1 if v > k else -1
      ans += 0 <= x <= 1
      cnt[x] += 1
    x = 0
    for j in range(i - 1, -1, -1):
      x += 1 if nums[j] > k else -1
      ans += 0 <= x <= 1
      ans += cnt[-x] + cnt[-x + 1]
    return ans

class Solution {
  public int countSubarrays(int[] nums, int k) {
    int n = nums.length;
    int i = 0;
    for (; nums[i] != k; ++i) {
    }
    int[] cnt = new int[n << 1 | 1];
    int ans = 1;
    int x = 0;
    for (int j = i + 1; j < n; ++j) {
      x += nums[j] > k ? 1 : -1;
      if (x >= 0 && x <= 1) {
        ++ans;
      }
      ++cnt[x + n];
    }
    x = 0;
    for (int j = i - 1; j >= 0; --j) {
      x += nums[j] > k ? 1 : -1;
      if (x >= 0 && x <= 1) {
        ++ans;
      }
      ans += cnt[-x + n] + cnt[-x + 1 + n];
    }
    return ans;
  }
}

class Solution {
public:
  int countSubarrays(vector<int>& nums, int k) {
    int n = nums.size();
    int i = find(nums.begin(), nums.end(), k) - nums.begin();
    int cnt[n << 1 | 1];
    memset(cnt, 0, sizeof(cnt));
    int ans = 1;
    int x = 0;
    for (int j = i + 1; j < n; ++j) {
      x += nums[j] > k ? 1 : -1;
      if (x >= 0 && x <= 1) {
        ++ans;
      }
      ++cnt[x + n];
    }
    x = 0;
    for (int j = i - 1; ~j; --j) {
      x += nums[j] > k ? 1 : -1;
      if (x >= 0 && x <= 1) {
        ++ans;
      }
      ans += cnt[-x + n] + cnt[-x + 1 + n];
    }
    return ans;
  }
};

func countSubarrays(nums []int, k int) int {
  i, n := 0, len(nums)
  for nums[i] != k {
    i++
  }
  ans := 1
  cnt := make([]int, n<<1|1)
  x := 0
  for j := i + 1; j < n; j++ {
    if nums[j] > k {
      x++
    } else {
      x--
    }
    if x >= 0 && x <= 1 {
      ans++
    }
    cnt[x+n]++
  }
  x = 0
  for j := i - 1; j >= 0; j-- {
    if nums[j] > k {
      x++
    } else {
      x--
    }
    if x >= 0 && x <= 1 {
      ans++
    }
    ans += cnt[-x+n] + cnt[-x+1+n]
  }
  return ans
}

function countSubarrays(nums: number[], k: number): number {
  const i = nums.indexOf(k);
  const n = nums.length;
  const cnt = new Array((n << 1) | 1).fill(0);
  let ans = 1;
  let x = 0;
  for (let j = i + 1; j < n; ++j) {
    x += nums[j] > k ? 1 : -1;
    ans += x >= 0 && x <= 1 ? 1 : 0;
    ++cnt[x + n];
  }
  x = 0;
  for (let j = i - 1; ~j; --j) {
    x += nums[j] > k ? 1 : -1;
    ans += x >= 0 && x <= 1 ? 1 : 0;
    ans += cnt[-x + n] + cnt[-x + 1 + n];
  }
  return ans;
}