Accepted

Author: gouthampradhan

problems/src/MedianOfTwoSortedArrays.java

@@ -0,0 +1,53 @@
+import com.sun.tools.internal.ws.wsdl.document.soap.SOAPUse;
+
+/**
+ * Created by gouthamvidyapradhan on 09/04/2017.
+ */
+public class MedianOfTwoSortedArrays
+{
+    /**
+     * Main method
+     * @param args
+     * @throws Exception
+     */
+    public static void main(String[] args) throws Exception
+    {
+        int[] A = {10};
+        int[] B = {12};
+        System.out.println(new MedianOfTwoSortedArrays().findMedianSortedArrays(A, B));
+    }
+
+    public double findMedianSortedArrays(int A[], int B[]) {
+        int n = A.length;
+        int m = B.length;
+        // the following call is to make sure len(A) <= len(B).
+        // yes, it calls itself, but at most once, shouldn't be
+        // consider a recursive solution
+        if (n > m)
+            return findMedianSortedArrays(B, A);
+
+        // now, do binary search
+        int k = (n + m - 1) / 2;
+        int l = 0, r = Math.min(k, n); // r is n, NOT n-1, this is important!!
+        while (l < r) {
+            int midA = (l + r) / 2;
+            int midB = k - midA;
+            if (A[midA] < B[midB])
+                l = midA + 1;
+            else
+                r = midA;
+        }
+
+        // if (n+m) is odd, the median is the larger one between A[l-1] and B[k-l].
+        // and there are some corner cases we need to take care of.
+        int a = Math.max(l > 0 ? A[l - 1] : Integer.MIN_VALUE, k - l >= 0 ? B[k - l] : Integer.MIN_VALUE);
+        if (((n + m) & 1) == 1)
+            return (double) a;
+
+        // if (n+m) is even, the median can be calculated by
+        //      median = (max(A[l-1], B[k-l]) + min(A[l], B[k-l+1]) / 2.0
+        // also, there are some corner cases to take care of.
+        int b = Math.min(l < n ? A[l] : Integer.MAX_VALUE, k - l + 1 < m ? B[k - l + 1] : Integer.MAX_VALUE);
+        return (a + b) / 2.0;
+    }
+}