added mock interview problem

problem1.py

@@ -0,0 +1,66 @@
+"""
+1. Since we don't know which half of the mountain the element is present, we use two binary searches to find if element is present on left or right.
+2. First we find the peak to identify where to split the binary search. Then ascent is until peak, descent is from peak to end of mountain
+3. If we did find the element on the ascent, then we return, else we go search on the descent.
+
+TC: O(log(N))
+SC: O(1)
+"""
+
+# """
+# This is MountainArray's API interface.
+# You should not implement it, or speculate about its implementation
+# """
+#class MountainArray:
+#    def get(self, index: int) -> int:
+#    def length(self) -> int:
+
+class Solution:
+    def findInMountainArray(self, target: int, mountain_arr: 'MountainArray') -> int:
+            length = mountain_arr.length()
+
+            l = 0
+            r = length - 1
+
+            while l<r:
+                mid = l+(r-l)//2
+                mid_element = mountain_arr.get(mid)
+                mid_next_ele = mountain_arr.get(mid+1)
+                if mid_next_ele > mid_element:
+                    l = mid + 1
+                elif mid_element > mid_next_ele :
+                    r = mid
+            peak = l 
+
+            ans = self.binarySearch1(mountain_arr,0,peak,target)
+            if ans== -1:
+                ans = self.binarySearch2(mountain_arr,peak+1,length-1,target)
+            return ans
+
+
+    def binarySearch1(self,arr,l,r,target) -> int:
+        while l<=r:
+            mid = l + (r-l)//2
+            mid_ele = arr.get(mid)
+            if mid_ele == target:
+                return mid
+            elif mid_ele < target:
+                l = mid + 1
+            else:
+                r = mid - 1
+        return -1
+
+    def binarySearch2(self,arr,l,r,target) -> int:
+        while l<=r:
+            mid = l + (r-l)//2
+            mid_ele = arr.get(mid)
+            if mid_ele == target:
+                return mid
+            elif mid_ele > target:
+                l = mid + 1
+            else:
+                r = mid - 1
+        return -1
+
+
+