LeetCode-in-Java

2488. Count Subarrays With Median K

Hard

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.

Solution

import java.util.HashMap;
import java.util.Map;

public class Solution {
    public int countSubarrays(int[] nums, int k) {
        int idx;
        int n = nums.length;
        int ans = 0;
        for (idx = 0; idx < n; idx++) {
            if (nums[idx] == k) {
                break;
            }
        }
        int[][] arr = new int[n - idx][2];
        int j = 1;
        for (int i = idx + 1; i < n; i++) {
            if (nums[i] < k) {
                arr[j][0] = arr[j - 1][0] + 1;
                arr[j][1] = arr[j - 1][1];
            } else {
                arr[j][1] = arr[j - 1][1] + 1;
                arr[j][0] = arr[j - 1][0];
            }
            j++;
        }
        Map<Integer, Integer> map = new HashMap<>();
        for (int[] ints : arr) {
            int d2 = ints[1] - ints[0];
            map.put(d2, map.getOrDefault(d2, 0) + 1);
        }
        int s1 = 0;
        int g1 = 0;
        for (int i = idx; i >= 0; i--) {
            if (nums[i] < k) {
                s1++;
            } else if (nums[i] > k) {
                g1++;
            }
            int d1 = g1 - s1;
            int cur = map.getOrDefault(-d1, 0) + map.getOrDefault(1 - d1, 0);
            ans += cur;
        }
        return ans;
    }
}

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