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# Median of Two Sorted Arrays LeetCode Solution

## Problem – Median of Two Sorted Arrays LeetCode Solution

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 = 
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 in Java

``````public double findMedianSortedArrays(int[] A, int[] B) {
int m = A.length, n = B.length;
int l = (m + n + 1) / 2;
int r = (m + n + 2) / 2;
return (getkth(A, 0, B, 0, l) + getkth(A, 0, B, 0, r)) / 2.0;
}

public double getkth(int[] A, int aStart, int[] B, int bStart, int k) {
if (aStart > A.length - 1) return B[bStart + k - 1];
if (bStart > B.length - 1) return A[aStart + k - 1];
if (k == 1) return Math.min(A[aStart], B[bStart]);

int aMid = Integer.MAX_VALUE, bMid = Integer.MAX_VALUE;
if (aStart + k/2 - 1 < A.length) aMid = A[aStart + k/2 - 1];
if (bStart + k/2 - 1 < B.length) bMid = B[bStart + k/2 - 1];

if (aMid < bMid)
return getkth(A, aStart + k/2, B, bStart,       k - k/2);// Check: aRight + bLeft
else
return getkth(A, aStart,       B, bStart + k/2, k - k/2);// Check: bRight + aLeft
}
``````

## Median of Two Sorted Arrays LeetCode Solution in Python

``````def findMedianSortedArrays(self, A, B):
l = len(A) + len(B)
if l % 2 == 1:
return self.kth(A, B, l // 2)
else:
return (self.kth(A, B, l // 2) + self.kth(A, B, l // 2 - 1)) / 2.

def kth(self, a, b, k):
if not a:
return b[k]
if not b:
return a[k]
ia, ib = len(a) // 2 , len(b) // 2
ma, mb = a[ia], b[ib]

# when k is bigger than the sum of a and b's median indices
if ia + ib < k:
# if a's median is bigger than b's, b's first half doesn't include k
if ma > mb:
return self.kth(a, b[ib + 1:], k - ib - 1)
else:
return self.kth(a[ia + 1:], b, k - ia - 1)
# when k is smaller than the sum of a and b's indices
else:
# if a's median is bigger than b's, a's second half doesn't include k
if ma > mb:
return self.kth(a[:ia], b, k)
else:
return self.kth(a, b[:ib], k)
``````

## Median of Two Sorted Arrays LeetCode Solution in C++

``````class Solution {
public:
int kth(int a[], int m, int b[], int n, int k) {
if (m < n) return kth(b,n,a,m,k);
if (n==0) return a[k-1];
if (k==1) return min(a,b);

int j = min(n,k/2);
int i = k-j;
if (a[i-1] > b[j-1]) return kth(a,i,b+j,n-j,k-j);
return kth(a+i,m-i,b,j,k-i);
}

double findMedianSortedArrays(int a[], int m, int b[], int n) {
int k = (m+n)/2;
int m1 = kth(a,m,b,n,k+1);
if ((m+n)%2==0) {
int m2 = kth(a,m,b,n,k);
return ((double)m1+m2)/2.0;
}
return m1;
}
};
``````
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