# 2488. Count Subarrays With Median K

## 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

• 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;
}
};

• 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


• 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;
}