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# Median of Two Sorted Arrays

There are two sorted arrays nums1 and nums2 of size m and n respectively.

Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

You may assume nums1 and nums2 cannot be both empty.

Example 1:

```nums1 = [1, 3]
nums2 = [2]```

The median is 2.0

Example 2:

```nums1 = [1, 2]
nums2 = [3, 4]```

The median is (2 + 3)/2 = 2.5

Solution 1 in C++

```#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

class Solution {
public:
double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
nums1.insert( nums1.end(), nums2.begin(), nums2.end());
sort( nums1.begin(), nums1.end());
int num_size = nums1.size();
if (num_size)
{
int iMedian=(int)(nums1.size()/2);

cout << "count of numbers:" << num_size << endl;
for( int i=0; i<num_size; i++)
cout << nums1[i] << " ";
cout << endl;
cout << "Median index:" << iMedian << endl;

if (num_size%2==1) return (double)nums1[iMedian];
else
{
return (double)(nums1[iMedian]+nums1[iMedian-1])/2;
}
}
else return 0;
}
};

int main(void)
{
/*
vector<int> nums1{ 1, 3};
vector<int> nums2{ 2};
*/

vector<int> nums1{ 1, 2};
vector<int> nums2{ 3, 4};

Solution s;
cout << s.findMedianSortedArrays( nums1, nums2) << endl;
}```

Solution 2 in Java

```class Solution {
public double findMedianSortedArrays(int[] A, int[] B) {
int m = A.length;
int n = B.length;
if (m > n) { // to ensure m<=n
int[] temp = A; A = B; B = temp;
int tmp = m; m = n; n = tmp;
}
int iMin = 0, iMax = m, halfLen = (m + n + 1) / 2;
while (iMin <= iMax) {
int i = (iMin + iMax) / 2;
int j = halfLen - i;
if (i < iMax && B[j-1] > A[i]){
iMin = i + 1; // i is too small
}
else if (i > iMin && A[i-1] > B[j]) {
iMax = i - 1; // i is too big
}
else { // i is perfect
int maxLeft = 0;
if (i == 0) { maxLeft = B[j-1]; }
else if (j == 0) { maxLeft = A[i-1]; }
else { maxLeft = Math.max(A[i-1], B[j-1]); }
if ( (m + n) % 2 == 1 ) { return maxLeft; }

int minRight = 0;
if (i == m) { minRight = B[j]; }
else if (j == n) { minRight = A[i]; }
else { minRight = Math.min(B[j], A[i]); }

return (maxLeft + minRight) / 2.0;
}
}
return 0.0;
}
}```

Solution 3 in Python

```def median(A, B):
m, n = len(A), len(B)
if m > n:
A, B, m, n = B, A, n, m
if n == 0:
raise ValueError

imin, imax, half_len = 0, m, (m + n + 1) / 2
while imin <= imax:
i = (imin + imax) / 2
j = half_len - i
if i < m and B[j-1] > A[i]:
# i is too small, must increase it
imin = i + 1
elif i > 0 and A[i-1] > B[j]:
# i is too big, must decrease it
imax = i - 1
else:
# i is perfect

if i == 0: max_of_left = B[j-1]
elif j == 0: max_of_left = A[i-1]
else: max_of_left = max(A[i-1], B[j-1])

if (m + n) % 2 == 1:
return max_of_left

if i == m: min_of_right = B[j]
elif j == n: min_of_right = A[i]
else: min_of_right = min(A[i], B[j])

return (max_of_left + min_of_right) / 2.0```