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.
<ul> <li>For example, the median of <code>[2,3,1,4]</code> is <code>2</code>, and the median of <code>[8,4,3,5,1]</code> is <code>4</code>.</li> </ul> </li> <li>A subarray is a contiguous part of an array.</li>
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.length1 <= n <= 1051 <= nums[i], k <= n- The integers in
numsare 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 $[0, 2n]$.
Python3
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
Java
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;
}
}
C++
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;
}
};
Go
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
}
TypeScript
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;
}