diff --git a/Problem1.java b/Problem1.java
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+++ b/Problem1.java
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+// 74. Search in a 2D Matrix
+// Time Complexity : O(log(m * n))
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes (Could not try since LC premium Question but ran on multiple example values)
+// Any problem you faced while coding this : No
+
+
+// Your code here along with comments explaining your approach in three sentences only
+// 1. Row 1 is sorted and the next row begans after first ends and is sorted again
+// 2. Row 1,2,3 can be flattened into one big single row of length, m*n
+// 3. Index of row can be calculated by i/n and columns as i%n where n would be the column. 
+class Solution {
+    public boolean searchMatrix(int[][] matrix, int target) {
+        int m = matrix.length;
+        int n = matrix[0].length;
+
+        int low = 0, high = m * n - 1;
+
+        while (low <= high) {
+            int mid = low + (high - low) / 2;
+            int midVal = matrix[mid / n][mid % n];
+
+            if (midVal == target) {
+                return true;
+            } else if (midVal < target) {
+                low = mid + 1;
+            } else {
+                high = mid - 1;
+            }
+        }
+
+        return false;
+    }
+}
diff --git a/Problem2.java b/Problem2.java
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+++ b/Problem2.java
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+// 33. Search in a Rotated Array
+
+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;
+
+            // Step 1: Identify which half is sorted
+            if (nums[low] <= nums[mid]) {  
+                // Step 2: Check if target lies in left half
+                if (target >= nums[low] && target < nums[mid]) {
+                    high = mid - 1;
+                } else {
+                    low = mid + 1;
+                }
+            } else {  // Right half is sorted
+                // Step 3: Check if target lies in right half
+                if (target > nums[mid] && target <= nums[high]) {
+                    low = mid + 1;
+                } else {
+                    high = mid - 1;
+                }
+            }
+        }
+        return -1;
+    }
+}
+
+
+// Time Complexity : O(Log n)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : I missed checking the value is pressent on the left side of mid and again checkin fthe righ side. 
+
+
+// Your code here along with comments explaining your approach in three sentences only
+// Set Low and High to start and ending of the array. Figure the Mid
+// Check which side is sorted from both, Low is less than mid then left side is shorter
+// After Getting Shorter side choose if the target value lies on which side and do binary search on same. 
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diff --git a/Problem3.java b/Problem3.java
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+++ b/Problem3.java
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+// 702. Search in a Sorted Array of Unknown Size
+import java.util.*;
+
+class Problem3{
+    public int search(ArrayReader reader, int target){
+        int low=0;
+        int high=1;// To be able to multiply twice everytime
+        while(reader.get(high)<target){
+            low=high;
+            high=high*2;
+        }
+        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;// If Value is not Present
+    }
+}
+
+// Time Complexity : O(N)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes (Could not try since LC premium Question but ran on multiple example values)
+// Any problem you faced while coding this : No
+
+
+// Your code here along with comments explaining your approach in three sentences only
+// 1. First I set the Low and HIgh pointer to 0 and 1 respectively and kep increasing the higher twice until its bigger than Mid.
+// 2. After getting a range I could Apply the normal binary search 
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