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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;
}

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