KQL Find Median of Two Sorted Arrays
let findMedianSortedArrays = (nums1: dynamic, nums2: dynamic) => { let m = array_length(nums1); let n = array_length(nums2); let totalLength = m + n; // Ensure nums1 is the shorter array for optimization if (m > n) { return findMedianSortedArrays(nums2, nums1); } let halfLen = fl...
This KQL query implements a function to find the median of two sorted arrays. It employs a binary search approach on the smaller array to efficiently find the correct partition points. The goal is to divide both arrays s...
The `findMedianSortedArrays` function aims to find the median of two sorted arrays `nums1` and `nums2` in O(log(min(m, n))) time complexity, where m and n are the lengths of the arrays. It first ensures that `nums1` is the shorter array to optimize the binary search. The core idea is to perform a binary search on `nums1` to find a partition point (`partitionX`) such that when combined with a corresponding partition point in `nums2` (`partitionY`), the elements are correctly divided. Specifically, we need `maxX <= minY` and `maxY <= minX`, where `maxX` and `maxY` are the largest elements in the left partitions, and `minX` and `minY` are the smallest elements in the right partitions. If the total length is odd, the median is the maximum of `maxX` and `maxY`. If the total length is even, the median is the average of `max(maxX, maxY)` and `min(minX, minY)`. Edge cases like empty arrays or one array being significantly larger are handled by the partition logic and the use of `real.Max` and `-real.Max` to represent boundary conditions.
FUNCTION findMedianSortedArrays(nums1, nums2): m = length(nums1) n = length(nums2) IF m > n THEN SWAP nums1 and nums2, m and n totalLength = m + n halfLen = floor((totalLength + 1) / 2) low = 0 high = m WHILE low <= high...