Explanation betterment and some minor changes

Author: Aditya Mukherjee

Leetcode/Median_of_Two_Sorted_Arrays.cpp

@@ -6,13 +6,27 @@
     This will have a time Complexity of O(n1 + n2), Space Complexity of O(n1 + n2)
 
     Now, lets optimise it.
-    So, the sorted form of the given array is [1, 2, 3, 3, 4, 6, 7, 10, 12, 15]. To find the median of the array, we need to select the 2 mid elements and average it out. Now suppose, on partitioning the above merged-sorted array in the mid-point, we get 5 elements on left and 5 elements on the right.
-    Now, we can get 5 elements by selecting {4 from left, 1 from right}, {3 from left, 2 from right}, {2 from left, 3 from right} and {1 from left, 4 from right}.
-    Lets analyse case-wise:
-        case 1: 4 from left, 1 from right
-
+    So, the sorted form of the given array is arr = [1, 2, 3, 3, 4, 6, 7, 10, 12, 15]. To find the median of the array, we need to select the 2 mid elements and average it out.
+    If we observe the sorted array carefully, then we notice that the 2 middle elements are arr[4] = 4 and arr[5] = 6, => arr[4] <= arr[5].
+    
+    Thought process:
+    Now, since the arrays are sorted, so the binary searh may be able to solve the problem.
+    Observation 1: If we partition the sorted array arr into 2 halves, then for sure we know that there would be 5 elements on left half and 5 elements on right half.
+        Now, we can select 5 elements for right half from nums1 and nums2 combinedly and similarly the rest of the elements for the left half.
+        Example:
+            1. left => [1, 3, 4, 7, 2], right => [10, 12, 3, 6, 15]
+            2. left => [1, 3, 4, 2, 3], right => [7, 10, 12, 6, 15]
+            3. left => [1, 3, 2, 3, 6], right => [4, 7, 10, 12, 15]
+    
+    Observation 2: All the elements on left half is lesser than all the elements on the right half.
+        Now, according to the observation, I have to check that all the elements in the left <= all the elements in the right. This can be done by just comparing the maximum of left half <= minimum of right half.
+    
+    Hence, the problem boils down to a searching problem; but how to identify the binary search approach??
+        Suppose, we partition the element as example 1, then the max of left[] = 7 and min of right[] = 3, but according to the 2nd observation, it is not valid. So, in order to have a correct max in left, I need to reduce its value and consequestly, the min of right[] should be increased. That means, I have to move leftwards in nums1 to have a correct max value in left half.
 */
 
+// Leetcode link: https://leetcode.com/problems/median-of-two-sorted-arrays/
+
 #include<bits/stdc++.h>
 
 class Solution {
@@ -28,7 +42,6 @@ class Solution {
         while(lo <= hi){
             int mid1 = (lo+hi)/2; // mid of nums1
             int mid2 = (n1 + n2 + 1)/2 - mid1;
-            // why?? => suppose nums1[] is partitioned at index 3 => nums1[3] = 7.
 
             std::pair<int, int> maxleft, minright;
             maxleft.first = mid1 == 0 ? INT_MIN : nums1[mid1-1];