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 = [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 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[0],b[0]);

        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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