diff --git a/Searchin2DArray.java b/Searchin2DArray.java
new file mode 100644
index 00000000..7f8517ef
--- /dev/null
+++ b/Searchin2DArray.java
@@ -0,0 +1,41 @@
+/*There are multiple ways to solve this problem
+
+Approach 1: Check if the elements belongs to that row, if it does perform BS on that row
+Approach 2: Start from either top right or bottom left corner because from there we will have clear idea about where to go
+Approach 3: Imagine the array as a 1D array and perform BS on that
+// Time Complexity : O(logn)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No
+*/
+
+class Solution {
+    public boolean searchMatrix(int[][] matrix, int target) {
+
+        int m = matrix.length; //number of rows
+        int n = matrix[0].length; // number of columns
+
+        int start = 0;
+        int end = m * n - 1; // total number of elements minus one
+
+        while(start <= end){
+            int mid = start + (end - start) / 2;
+
+            //translating 1D array mid to a 2D array mid
+            // number of rows is dependent on number of column hence using n as the denominator 
+            // division will help with chunking the rows, modulo will help with finding the element
+            int r = mid / n;
+            int c = mid % n;
+
+            if(matrix[r][c] == target){
+                return true;
+            }
+            else if(matrix[r][c] > target){
+                end = mid - 1;
+            }
+            else
+                start = mid + 1;
+        }
+        return false;
+    }
+}
diff --git a/SearchinInfiniteArray.java b/SearchinInfiniteArray.java
new file mode 100644
index 00000000..602d7021
--- /dev/null
+++ b/SearchinInfiniteArray.java
@@ -0,0 +1,41 @@
+/**
+ * // This is ArrayReader's API interface.
+ * // You should not implement it, or speculate about its implementation
+ * interface ArrayReader {
+ *     public int get(int index) {}
+ * }
+ */
+
+ /*
+Approach: In a rotated sorted array one side of the mid will always be sorted! 
+// Time Complexity : O(logn)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No
+*/
+
+class Solution {
+    public int search(ArrayReader reader, int target) {
+        int low = 0;
+        int high = 1;
+
+        // loop through until you find and element greater than target
+        while(reader.get(high) < target){
+            low = high;
+            high = high * 2; // to have balanced operations
+        }
+
+        // binary search to find target
+        while(low <= high){
+            int mid = low + (high - low) / 2;
+            if(reader.get(mid) == target)
+                return mid;
+            else if(reader.get(mid) < target){
+                low = mid + 1;
+            }
+            else 
+                high = mid - 1;
+        }
+        return - 1;
+    }
+}
\ No newline at end of file
diff --git a/SearchinRotatedSortedArray.java b/SearchinRotatedSortedArray.java
new file mode 100644
index 00000000..f8de4db4
--- /dev/null
+++ b/SearchinRotatedSortedArray.java
@@ -0,0 +1,40 @@
+/*
+Approach: In a rotated sorted array one side of the mid will always be sorted! 
+// Time Complexity : O(logn)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No
+
+*/
+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;
+            //check which side of the array is sorted
+
+            if(nums[mid] == target)
+                return mid;
+
+            //check if left half is sorted
+            if(nums[low] <= nums[mid]){
+                // condition to validate where to go
+                if(nums[low] <= target && nums[mid] > target)
+                    high = mid - 1;
+                else
+                    low = mid + 1;
+            }
+
+            //check if right half is sorted
+            else{
+                if(nums[high] >= target && nums[mid] < target)
+                    low = mid + 1;
+                else
+                    high = mid - 1; 
+            }
+        }
+        return -1;
+    }
+}
\ No newline at end of file