Missing Number in a Sorted Array

MissingNumbers.py

@@ -0,0 +1,30 @@
+# Time Complexity --> O(logN)
+# Space Complexity --> O(1)
+# Explanation --> Using a binary search, check if the difference between the middle number and any of its adjacent numbers is greater than 1 and if so then return the output accordingly. 
+# If the missing number not adjacent to middle index, go towards the half where the difference between high index and middle index is lower than the difference bewteen corresponding values.
+
+class Solution:
+    def missingNumber(self, arr):
+        # code here
+        n = len(arr)+1 
+        low = 0
+        high = len(arr)-1
+
+        if arr[0]!=1:
+            return 1
+        if arr[-1]!=n:
+            return n
+
+        while low<=high:
+            mid = low + (high-low)//2
+            if arr[mid+1]-arr[mid]>1:
+                return arr[mid]+1
+            if arr[mid]-arr[mid-1]>1:
+                return arr[mid]-1
+            if arr[high]-arr[mid] > high-mid:
+                low = mid
+            else:
+                high = mid
+
+
+