diff --git a/Problem1.py b/Problem1.py
new file mode 100644
index 00000000..9cd876db
--- /dev/null
+++ b/Problem1.py
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+# Time Complexity : O(log m*n)
+# Space Complexity : O(1)
+#Did this code successfully run on Leetcode : Yes
+#Any problem you faced while coding this : No
+# Three line explanation of solution in plain english:
+#1. We use binary search to find the target in a 2D matrix for this we flatten the 2d matrix to an array.
+#2. We treat the 2D matrix as a 1D array by calculating the row and column indices from the mid index.
+#we use r and c by dividing mid by n (number of columns) and taking the modulus with n respectively.
+#3. If the target is found, we return True; otherwise, we adjust the search range based on the comparison with the mid element.
+
+class Solution(object):
+    def searchMatrix(self, matrix, target):
+        """
+        :type matrix: List[List[int]]
+        :type target: int
+        :rtype: bool
+        """
+        #get the number of rows and columns in the matrix
+        m=len(matrix)
+        n=len(matrix[0])
+        #low will be the starting index and high will be the last index of the flattened matrix
+        low=0
+        high=m*n-1
+        #binary search to find the target in the flattened matrix
+        while low<=high:
+            mid=low+(high-low)/2
+            #calculate the row and column indices from the mid index
+            #if you divide mid by number of columns n, you get the row index r
+            #and if you take mid modulo n, you get the column index c
+            r=mid/n
+            c=mid%n
+            #if the target is found at the mid position, return True
+            if matrix[r][c]==target:
+                return True
+            #if the target is less than the mid element, adjust the high pointer to mid-1 and search in the left half
+            # if the target is greater than the mid element, adjust the low pointer to mid+1 and search in the right half
+            if matrix[r][c]<target:
+                low=mid+1
+            else:
+                high=mid-1
+        #if the target is not found, return False
+        return False
\ No newline at end of file
diff --git a/Problem2.py b/Problem2.py
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+++ b/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
+# Three line explanation of solution in plain english:
+#1. We use binary search to find the target in a rotated sorted array.
+#2. We check if the left side of the array is sorted and if the target lies
+#3. If the right side is sorted, we check if the target lies in that range and adjust the pointers of high and low accordingly.
+
+
+
+
+class Solution(object):
+    def search(self, nums, target):
+        """
+        :type nums: List[int]
+        :type target: int
+        :rtype: int
+        """
+        low=0
+        high=len(nums)-1
+        while low<=high:
+            mid=low+(high-low)/2
+            #if the target is at mid, return mid position 
+            if target==nums[mid]:
+                return mid
+            #left side sorted
+            elif nums[low]<=nums[mid]:
+                #check if target lies in the left side sorted array
+                #if it does, reduce the high pointer to mid-1, else increase the low pointer to mid+1
+                if nums[low]<=target and nums[mid]>target:
+                    high=mid-1
+                else:
+                    low=mid+1
+            #right side sorted
+            else:
+                #if it does, increase the low pointer to mid+1, else reduce the high pointer to mid
+                if nums[mid]<target and nums[high]>=target:
+                    low=mid+1
+                else:
+                    high=mid-1
+        #if the target is not found, return -1
+        return -1
diff --git a/Problem3.py b/Problem3.py
new file mode 100644
index 00000000..39f5237f
--- /dev/null
+++ b/Problem3.py
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+# Time Complexity : O(log n)
+# Space Complexity : O(1)
+#Did this code successfully run on Leetcode : No
+#Any problem you faced while coding this : No
+# Three line explanation of solution in plain english:
+# 1. The solution uses an exponential search to find the range where the target might be located.
+# 2. It then applies binary search within that range to find the exact index of the target.
+# 3. The reader.get method is used to access elements, simulating a scenario where the array size is unknown.
+
+
+class Solution:
+    def search(self, reader, target):
+        low = 0
+        high = 1
+
+        # Expand the search range exponentially
+        # until the value at high index is greater than or equal to the target
+        while reader.get(high) < target:
+            low = high
+            high *= 2
+
+        return self.binarySearch(reader, target, low, high)
+    #
+    # Perform binary search within the determined range
+    def binarySearch(self, reader, target, low, high):
+        while low <= high:
+            mid = low + (high - low) // 2
+            # Use reader.get to access the value at mid
+            val = reader.get(mid)
+
+            # Check if the value at mid is equal to the target
+            if val == target:
+                return mid
+            elif val > target:
+                high = mid - 1
+            else:
+                low = mid + 1
+        # If the target is not found, return -1
+        return -1