Median of two sorted array of different length

Author: Tushar Roy

src/com/interview/binarysearch/MedianOfTwoSortedArrayOfDifferentLength.java

@@ -0,0 +1,62 @@
+package com.interview.binarysearch;
+
+/**
+ * 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)).
+ *
+ * Solution
+ * Take minimum size of two array. Possible number of cuts are from 0 to m in m size array.
+ * Try every cut in binary search way. When you cut first array at i then you cut second array at (m + n + 1)/2 - i
+ * Now try to find the i where a[i-1] <= b[j] and b[j-1] <= a[i]. So this i is the answerr
+ *
+ * Time complexity is O(log(min(m,n))
+ *
+ * https://leetcode.com/problems/median-of-two-sorted-arrays/
+ * https://discuss.leetcode.com/topic/4996/share-my-o-log-min-m-n-solution-with-explanation/4
+ */
+public class MedianOfTwoSortedArrayOfDifferentLength {
+    public double findMedianSortedArrays(int input1[], int input2[]) {
+        if (input1.length > input2.length) {
+            return findMedianSortedArrays(input2, input1);
+        }
+        if (input1.length == 0) {
+            return getMedian(input2);
+        }
+        int m = input1.length;
+        int n = input2.length;
+
+        int low = 0;
+        int high = m;
+        while (low <= high) {
+            int cut1 = (low + high)/2;
+            int cut2 = (m + n + 1)/2 - cut1;
+
+            int a1 = (cut1 == 0) ? Integer.MIN_VALUE : input1[cut1 - 1];
+            int a2 = (cut1 == m) ? Integer.MAX_VALUE : input1[cut1];
+
+            int b1 = (cut2 == 0) ? Integer.MIN_VALUE : input2[cut2 - 1];
+            int b2 = (cut2 == n) ? Integer.MAX_VALUE : input2[cut2];
+
+            if (a1 <= b2 && b1 <= a2) {
+                if ((m + n) % 2 == 0) {
+                    return ((double)Math.max(a1, b1) + Math.min(a2, b2))/2;
+                } else {
+                    return (double)Math.max(a1, b1);
+                }
+            } else if (a1 > b2) {
+                high = cut1 - 1;
+            } else {
+                low = cut1 + 1;
+            }
+        }
+        throw new IllegalArgumentException();
+    }
+
+    private double getMedian(int arr[]){
+        if(arr.length % 2 == 0){
+            return ((double)arr[arr.length/2 - 1] + arr[arr.length/2])/2;
+        } else {
+            return arr[arr.length/2];
+        }
+    }
+}