binary Search 1 done

InfiniteSortedArray.java

@@ -0,0 +1,29 @@
+// Time Complexity : O(log n)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : 
+// Any problem you faced while coding this : No
+
+
+// Your code here along with comments explaining your approach
+class solution{
+	public int search(ArrayReader reader, int target){
+		int low =0,high=0;
+		if(reader.get(0)==Integer.MAX_VALUE) return -1;
+		while(reader.get(high)<target){
+			low=high+1;
+			high=low*2;
+		}
+
+		while(low<=high){
+			int mid = low +(high-low)/2;
+			if(reader(mid)==target) return mid;
+			if(reader(mid)>target){
+				high = mid-1;
+			}
+			else{
+				low = mid+1;
+			}
+		}
+		return -1;
+	}
+}

Search2DMatrix.java

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+// Time Complexity : O(log(m+n))
+// Space Complexity : O(1)  no extra space used
+// 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
+class Solution {
+    public boolean searchMatrix(int[][] matrix, int target) {
+        int m=matrix.length;
+        int n= matrix[0].length;
+        int i=0, j=m*n-1;
+        while(i<=j){
+            int mid = i+(j-i)/2;        
+            if(matrix[mid/n][mid%n]==target){
+                return true;
+            }
+            else if(matrix[mid/n][mid%n]>target){
+                j=mid-1;
+            }
+            else i=mid+1;
+        }
+        return false;
+
+    }
+}

SearchRotatedSortedArray.java

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+// Time Complexity : O(log n)
+// Space Complexity : O(1) // since we are not using additional we created apart from the original nums array given to us
+// 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
+class Solution {
+    public int search(int[] nums, int target) {
+        int low=0,high=nums.length-1;
+
+        while(low<=high){
+            int mid = low + (high - low)/2;
+            if(nums[mid]==target) return mid;
+            else{
+                if(nums[mid]<=nums[high]){ //if right part is sorted or not
+                    if(target>nums[mid] && target<=nums[high]){ // checking if the target is there in the sorted portion of the array or not
+                        low=mid+1;
+                    }
+                    else{
+                        high=mid-1;
+                    }
+                }
+                else{
+                    if(target>=nums[low] && target<nums[mid]){
+                        high=mid-1;
+                    }
+                    else{
+                        low=mid+1;
+                    }
+                }
+            }
+
+        }
+        return -1;
+    }
+}