### LeetCode Challenge #4. Median of Two Sorted Arrays

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`
##### Video Solution
###### Java Solution
```				```
class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
int[] ans = merge(nums1 , nums2);

if(ans.length %2 ==0 ){

double ans2 = (double)(ans[ans.length/2] + ans[ans.length/2 -1 ])/2;
return ans2 ;
}else{
double ans2 = (double) ( ans[ans.length/2]);

return ans2 ;
}
}

public int[] merge(int[] arr1 , int[] arr2 ){

int[] ans = new int[arr1.length + arr2.length];

int p1 = 0 ;
int p2 = 0 ;
int p3 = 0 ;

while(p1<arr1.length || p2<arr2.length){

int val1 = p1<arr1.length ? arr1[p1] : Integer.MAX_VALUE;
int val2 = p2<arr2.length ? arr2[p2] : Integer.MAX_VALUE ;

if(val1<val2){
ans[p3] = val1 ;
p1++;
}else{
ans[p3] = val2 ;
p2++;
}
p3++;
}
return ans ;
}
}
```
```