ADD Median-of-Two-Sorted-Arrays problem #53

Author: khaled-alselwady

Hard/median-of-two-sorted-arrays.cs

@@ -0,0 +1,118 @@
+using System;
+using System.Collections.Generic;
+using System.Linq;
+
+public class Solution
+{
+    public static void Main(string[] args)
+    {
+        int[] nums1 = { };
+        int[] nums2 = { 1 };
+
+        Console.WriteLine(FindMedianSortedArrays(nums1, nums2));
+        Console.WriteLine(FindMedianSortedArraysUsingBinarySearch(nums1, nums2));
+
+        Console.ReadKey();
+    }
+
+    private static double _GetMedian(double min, double max)
+        => min + ((max - min) / 2);
+
+    private static bool _IsEven(int num)
+        => (num % 2 == 0);
+
+    #region Approach 1
+    public static double FindMedianSortedArrays(int[] nums1, int[] nums2)
+    {
+        if (nums1 == null || nums2 == null)
+        {
+            return double.NaN;
+        }
+
+        List<int> list1 = new List<int>(nums1);
+        List<int> list2 = new List<int>(nums2);
+
+        var list3 = list1.Concat(list2).OrderBy(x => x).ToList();
+
+        int listCount = list3.Count;
+
+        if (_IsEven(listCount))
+        {
+            int lastIndexInMedian = listCount / 2;
+            int firstIndexMedian = lastIndexInMedian - 1;
+
+            return _GetMedian(list3[firstIndexMedian], list3[lastIndexInMedian]);
+        }
+        else
+        {
+            return list3[listCount / 2];
+        }
+    }
+    #endregion
+
+    #region Approach 2
+    private static double _PerformBinarySearchToGetResult(int[] nums1, int[] nums2)
+    {
+        int m = nums1.Length;
+        int n = nums2.Length;
+        int total = m + n;
+        int half = (total + 1) / 2;
+        int left = 0;
+        int right = m;
+        double result = 0.0;
+        while (left <= right)
+        {
+            int middle = left + (right - left) / 2;
+            int j = half - middle;
+            int left1 = (middle > 0) ? nums1[middle - 1] : int.MinValue;
+            int right1 = (middle < m) ? nums1[middle] : int.MaxValue;
+            int left2 = (j > 0) ? nums2[j - 1] : int.MinValue;
+            int right2 = (j < n) ? nums2[j] : int.MaxValue;
+
+            if (left1 <= right2 && left2 <= right1)
+            {
+                if (_IsEven(total))
+                {
+                    result = (Math.Max(left1, left2) + Math.Min(right1, right2)) / 2.0;
+                }
+                else
+                {
+                    result = Math.Max(left1, left2);
+                }
+                break;
+            }
+            else if (left1 > right2)
+            {
+                right = middle - 1;
+            }
+            else
+            {
+                left = middle + 1;
+            }
+        }
+        return result;
+    }
+
+    public static double FindMedianSortedArraysUsingBinarySearch(int[] nums1, int[] nums2)
+    {
+        if (nums1.Length <= 0 && nums2.Length == 1)
+        {
+            return nums2[0];
+        }
+        if (nums2.Length <= 0 && nums1.Length == 1)
+        {
+            return nums1[0];
+        }
+
+        if (nums1.Length > nums2.Length)
+        {
+            // swapping
+            (nums1, nums2) = (nums2, nums1);
+            //return FindMedianSortedArrays(nums2, nums1);
+        }
+
+        return _PerformBinarySearchToGetResult(nums1, nums2);
+    }
+    #endregion
+}
+