# Median of Two Sorted Arrays LeetCode Solution Best & Easiest

In this post, we will solve the Median of Two Sorted Arrays LeetCode Solution.

Problem Statement:

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`

## Median of Two Sorted Arrays LeetCode Solution

### Problem solution in JavaScript programming:

```function findMedianSortedArrays(nums1, nums2) {
const merged = nums1.concat(nums2).sort((a, b) => a - b);
const length = merged.length;
if (length % 2 === 0) {
return (merged[length / 2 - 1] + merged[length / 2]) / 2;
} else {
return merged[Math.floor(length / 2)];
}
}
```

### Problem solution in Python programming:

```class Solution:
def findMedianSortedArrays(self, nums1, nums2):
merged = sorted(nums1 + nums2)
length = len(merged)
if length % 2 == 0:
return (merged[length // 2 - 1] + merged[length // 2]) / 2
else:
return merged[length // 2]
```

### Problem solution in C++ programming:

```class Solution {
public:
double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
vector<int> merged;
int i = 0, j = 0;
while (i < nums1.size() && j < nums2.size()) {
if (nums1[i] <= nums2[j]) {
merged.push_back(nums1[i]);
i++;
} else {
merged.push_back(nums2[j]);
j++;
}
}
while (i < nums1.size()) {
merged.push_back(nums1[i]);
i++;
}
while (j < nums2.size()) {
merged.push_back(nums2[j]);
j++;
}
int length = merged.size();
if (length % 2 == 0) {
return (merged[length / 2 - 1] + merged[length / 2]) / 2.0;
} else {
return merged[length / 2];
}

}
};
```