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@Rutvik17
Last active August 9, 2021 02:08
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Median Of Two Sorted Arrays LeetCode
/*
Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays.
The overall run time complexity should be O(log (m+n)).
Example 1:
Input: nums1 = [1,3], nums2 = [2]
Output: 2.00000
Explanation: merged array = [1,2,3] and median is 2.
Example 2:
Input: nums1 = [1,2], nums2 = [3,4]
Output: 2.50000
Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5.
Example 3:
Input: nums1 = [0,0], nums2 = [0,0]
Output: 0.00000
Example 4:
Input: nums1 = [], nums2 = [1]
Output: 1.00000
Example 5:
Input: nums1 = [2], nums2 = []
Output: 2.00000
Constraints:
nums1.length == m
nums2.length == n
0 <= m <= 1000
0 <= n <= 1000
1 <= m + n <= 2000
-106 <= nums1[i], nums2[i] <= 106
*/
/**
* @param {number[]} nums1
* @param {number[]} nums2
* @return {number}
*/
var findMedianSortedArrays = function(nums1, nums2) {
if(nums1.length > nums2.length) {
return findMedianSortedArrays(nums2, nums1)
};
let x = nums1.length;
let y = nums2.length;
let low = 0, high = x;
while(low <= high) {
const partitionX = (high + low) >> 1;
const partitionY = ((x + y + 1) >> 1) - partitionX;
console.log((7 + 0) >> 1)
const maxX = partitionX == 0 ? Number.NEGATIVE_INFINITY : nums1[partitionX - 1];
const maxY = partitionY == 0 ? Number.NEGATIVE_INFINITY : nums2[partitionY - 1];
const minX = partitionX == nums1.length ? Number.POSITIVE_INFINITY : nums1[partitionX];
const minY = partitionY == nums2.length ? Number.POSITIVE_INFINITY : nums2[partitionY ];
if(maxX <= minY && maxY <= minX) {
const lowMax = Math.max(maxX, maxY);
if( (x + y) % 2 == 1) {
return lowMax;
}
return (lowMax + Math.min(minX, minY)) / 2
} else if(maxX < minY) {
low = partitionX + 1;
} else {
high = partitionX - 1;
}
}
};
const nums1 = [1,2,3,4,5,6,7];
const nums2 = [4,5,6,7,8,9];
console.log(findMedianSortedArrays(nums1, nums2));
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