LeetCode 2488: Count Subarrays With Median K

Explanation

To solve this problem, we can iterate over all subarrays of the given array nums. For each subarray, we calculate the median and check if it equals the given integer k. To efficiently calculate the median, we sort the subarray and find the middle element(s) based on the length of the subarray. By keeping track of the count of subarrays with median equal to k, we can return the final result.

Algorithm:

1. Initialize a variable result to keep track of the count of subarrays with median equal to k.
2. Iterate over all subarrays of nums.
3. For each subarray, sort it and calculate the median.
4. If the median is equal to k, increment the result.
5. Return the final result.

Time Complexity: The time complexity of this approach is O(n^3) where n is the length of the input array nums.

Space Complexity: The space complexity is O(1) as we are not using any extra space that scales with the input size.

class Solution {
    public int countSubarraysWithMedianK(int[] nums, int k) {
        int result = 0;
        for (int i = 0; i < nums.length; i++) {
            for (int j = i; j < nums.length; j++) {
                int[] subarray = Arrays.copyOfRange(nums, i, j + 1);
                Arrays.sort(subarray);
                int median = (subarray.length % 2 == 0) ? subarray[subarray.length / 2 - 1] : subarray[subarray.length / 2];
                if (median == k) {
                    result++;
                }
            }
        }
        return result;
    }
}