4. Median of Two Sorted Arrays Leetcode Solution in JavaScript (JS)

Problem:

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.

Constraints:

• nums1.length == m
• nums2.length == n
• 0 <= m <= 1000
• 0 <= n <= 1000
• 1 <= m + n <= 2000
• -106 <= nums1[i], nums2[i] <= 106

Solution:

Solution with separate function to merge two arrays in O(n) complexity.

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number}
 */
var findMedianSortedArrays = function(nums1, nums2) {
   
    let newarray = new Array();
    let lengthh = (nums1.length > nums2.length)?nums1.length:nums2.length;
    for(let i = 0; i<lengthh; i++){
        if(nums1[i] || nums2[i]){
            if(nums1[i] > nums2[i]){
                if(newarray[newarray.length - 1] > nums2[i]){
                    let buff = newarray[newarray.length - 1];
                    newarray.pop();
                    newarray.push(nums2[i]);
                    newarray.push(buff);
                }else{
                    newarray.push(nums2[i]);
                }
                if(newarray[newarray.length - 1] > nums1[i]){
                    let buff = newarray[newarray.length - 1];
                    newarray.pop();
                    newarray.push(nums1[i]);
                    newarray.push(buff);
                }else{
                    newarray.push(nums1[i]);
                }
                
            }
            else{
                if(newarray[newarray.length - 1] > nums1[i]){
                    let buff = newarray[newarray.length - 1];
                    newarray.pop();
                    newarray.push(nums1[i]);
                    newarray.push(buff);
                }else{
                    newarray.push(nums1[i]);
                }
                if(newarray[newarray.length - 1] > nums2[i]){
                    let buff = newarray[newarray.length - 1];
                    newarray.pop();
                    newarray.push(nums2[i]);
                    newarray.push(buff);
                }else{
                    newarray.push(nums2[i]);
                }
            }
            
        }
    }
    newarray = mergeTwo(nums1, nums2)
    newarray = newarray.filter(function( element ) {
   return element !== undefined;
});
    
    if(newarray.length % 2){
        if(newarray.length){
            return newarray[Math.ceil(newarray.length/2 -1)];
        }
        else{
            newarray[0];
        }
        
    }
    else{
        if(newarray[newarray.length/2 - 1] || newarray[newarray.length/2]){
            return (newarray[newarray.length/2 - 1] + newarray[newarray.length/2])/2;
        }
        else {
            return 0;
        }
    } 
};
// O(n) time & O(n) space
function mergeTwo(arr1, arr2) {
  let merged = [];
  let index1 = 0;
  let index2 = 0;
  let current = 0;

  while (current < (arr1.length + arr2.length)) {

    let isArr1Depleted = index1 >= arr1.length;
    let isArr2Depleted = index2 >= arr2.length;

    if (!isArr1Depleted && (isArr2Depleted || (arr1[index1] < arr2[index2]))) {
      merged[current] = arr1[index1];
      index1++;
    } else {
      merged[current] = arr2[index2];
      index2++;
    }

    current++;
  }

  return merged;
}