Rotational sort and unknown range

problem2.py

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+# Time Complexity : O(log n)
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode : yes 
+# Any problem you faced while coding this : No
+
+class Solution(object):
+    def search(self, nums, target):
+        low = 0
+        high = len(nums) - 1
+
+        while low <= high:
+            mid = low + (high - low) // 2
+            #left sorted
+            if nums[mid] == target:
+                return mid
+            #left sorted
+            if nums[low] <= nums[mid]:
+               if nums[low] <= target < nums[mid]:
+                  high = mid - 1
+               else:
+                  low = mid + 1
+            else: #right sorted
+                if nums[mid] < target <= nums[high]:
+                    low = mid + 1
+                else:
+                    high = mid - 1
+        return -1
+
+

problem3.py

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+# Time Complexity : O(log m + log n)
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode : yes 
+# Any problem you faced while coding this : No, it was a bit tricky to understand the problem statement and how to approach it, but once I got the idea, it was straightforward.
+
+
+
+def search(self, reader: 'ArrayReader', target: int) -> int:
+    #this is to understand the range of the array, we will keep doubling the high until we find a value that is greater than the target
+    low = 0
+    high = 1
+    while reader.get(high) < target:
+        low = high
+        high = high * 2
+        #after this we will implement binary search on the range we found
+    while low <= high:
+         mid = low + (high - low) // 2
+    if reader.get(mid) == target:
+        return mid
+    if reader.get(low) <= reader.get(mid):
+        if reader.get(low) <= target < reader.get(mid):
+            high = mid - 1
+        else:
+            low = mid + 1
+    else:
+        if reader.get(mid) < target <= reader.get(high):
+            low = mid + 1
+        else:
+            high = mid - 1
+    return -1