diff --git a/SearchInRotatedArray.java b/SearchInRotatedArray.java
new file mode 100644
index 00000000..0f83c7eb
--- /dev/null
+++ b/SearchInRotatedArray.java
@@ -0,0 +1,41 @@
+// 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
+
+
+// Your code here along with comments explaining your approach in three sentences only
+// 1: For a rotated sorted array, at least either left or right half of the array will be sorted
+// 2: We use this property to determine which pointers to move to find the target
+// 3: For a left sorted array, we move the high pointer closer, for a right sorted array, we move the low pointer closer
+class Solution {
+    public int search(int[] nums, int target) {
+        int low = 0;
+        int high = nums.length - 1;
+
+        while(low <= high){
+            int mid = low + ((high-low)/2);
+            if(nums[mid] == target){
+                return mid;
+            }
+            if(nums[low] <= nums[mid]){
+                // left sorted
+                if(nums[low] <= target && nums[mid] > target){
+                    high = mid - 1;
+                }
+                else{
+                    low = mid + 1;
+                }
+            }
+            else{
+                // right sorted
+                if(nums[high] >= target && nums[mid] < target){
+                    low = mid + 1;
+                }
+                else
+                    high = mid - 1;
+            }
+        };
+        return -1;
+    }
+}
\ No newline at end of file
diff --git a/SearchInUnknownArray.java b/SearchInUnknownArray.java
new file mode 100644
index 00000000..1aa10e79
--- /dev/null
+++ b/SearchInUnknownArray.java
@@ -0,0 +1,35 @@
+// Time Complexity : O(log n)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : No - premium question
+// Any problem you faced while coding this : No
+
+
+// Your code here along with comments explaining your approach in three sentences only
+// 1: We set a fixed range to search within and then use ArrayReader to double the range in which we're searching
+// 2: We use the ArrayReader to get the mid element
+// 3: And use Binary Search as normal to find the target element
+class Solution{
+    public int search(ArrayReader reader, int target){
+        int low = 0;
+        int high = 1;
+        while(reader.get(high) < target){
+            low = high;
+            high = 2 * high;
+        };
+
+        while(low <= high){
+            int mid = low + (high-low)/2;
+
+            if(reader.get(mid) == target){
+                return mid;
+            }
+            else if(reader.get(mid) > target){
+                high = mid - 1;
+            }
+            else {
+                low = mid + 1;
+            }
+        }
+        return -1;
+    }
+}
\ No newline at end of file