# Median of Two Sorted Arrays Leetcode Solution

#### ByBrokenprogrammers

Nov 30, 2022

In this post, you will know how to solve the Median of Two Sorted Arrays Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Leetcode by yourself. If you find any difficulty after trying several times, then you can look for solutions.

## 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`

### Median of Two Sorted Arrays Leetcode Solutions in Python

```class Solution:
def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float:
n1 = len(nums1)
n2 = len(nums2)
if n1 > n2:
return self.findMedianSortedArrays(nums2, nums1)
l = 0
r = n1
while l <= r:
partition1 = (l + r) // 2
partition2 = (n1 + n2 + 1) // 2 - partition1
maxLeft1 = -2**31 if partition1 == 0 else nums1[partition1 - 1]
maxLeft2 = -2**31 if partition2 == 0 else nums2[partition2 - 1]
minRight1 = 2**31 - 1 if partition1 == n1 else nums1[partition1]
minRight2 = 2**31 - 1 if partition2 == n2 else nums2[partition2]
if maxLeft1 <= minRight2 and maxLeft2 <= minRight1:
return (max(maxLeft1, maxLeft2) + min(minRight1, minRight2)) * 0.5 if (n1 + n2) % 2 == 0 else max(maxLeft1, maxLeft2)
elif maxLeft1 > minRight2:
r = partition1 - 1
else:
l = partition1 + 1
```

### Median of Two Sorted Arrays Leetcode Solutions in CPP

```class Solution {
public:
double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
const int n1 = nums1.size();
const int n2 = nums2.size();
if (n1 > n2)
return findMedianSortedArrays(nums2, nums1);
int l = 0;
int r = n1;
while (l <= r) {
const int partition1 = (l + r) / 2;
const int partition2 = (n1 + n2 + 1) / 2 - partition1;
const int maxLeft1 = partition1 == 0 ? INT_MIN : nums1[partition1 - 1];
const int maxLeft2 = partition2 == 0 ? INT_MIN : nums2[partition2 - 1];
const int minRight1 = partition1 == n1 ? INT_MAX : nums1[partition1];
const int minRight2 = partition2 == n2 ? INT_MAX : nums2[partition2];
if (maxLeft1 <= minRight2 && maxLeft2 <= minRight1)
return (n1 + n2) % 2 == 0
? (max(maxLeft1, maxLeft2) + min(minRight1, minRight2)) * 0.5
: max(maxLeft1, maxLeft2);
else if (maxLeft1 > minRight2)
r = partition1 - 1;
else
l = partition1 + 1;
}
throw;
}
};
```

### Median of Two Sorted Arrays Leetcode Solutions in Java

```class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
final int n1 = nums1.length;
final int n2 = nums2.length;
if (n1 > n2)
return findMedianSortedArrays(nums2, nums1);
int l = 0;
int r = n1;
while (l <= r) {
final int partition1 = (l + r) / 2;
final int partition2 = (n1 + n2 + 1) / 2 - partition1;
final int maxLeft1 = partition1 == 0 ? Integer.MIN_VALUE : nums1[partition1 - 1];
final int maxLeft2 = partition2 == 0 ? Integer.MIN_VALUE : nums2[partition2 - 1];
final int minRight1 = partition1 == n1 ? Integer.MAX_VALUE : nums1[partition1];
final int minRight2 = partition2 == n2 ? Integer.MAX_VALUE : nums2[partition2];
if (maxLeft1 <= minRight2 && maxLeft2 <= minRight1)
return (n1 + n2) % 2 == 0
? (Math.max(maxLeft1, maxLeft2) + Math.min(minRight1, minRight2)) * 0.5
: Math.max(maxLeft1, maxLeft2);
else if (maxLeft1 > minRight2)
r = partition1 - 1;
else
l = partition1 + 1;
}
throw new IllegalArgumentException();
}
}
```

Note: This problem Median of Two Sorted Arrays is generated by Leetcode but the solution is provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.

NEXT: Pascal’s Triangle Leetcode Solution