Binary Search - 1 Completed

search_in_2d_matrix.java

@@ -0,0 +1,37 @@
+// Time Complexity : log(mn)
+// Space Complexity : O(mn)
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : yes mid calcaution is a bit weird to wrap my head around as i ma seeing it for the first time
+
+/*
+ * we are treating 2D matrix as a 1d matrix as in the problem statment is it given that last number of row will be smaller then the
+ * first number of the next row
+ * 
+ * we are calculating mid as if it was a 1d array and after after words we are converting that mid to row and column index using row size
+ * divide by row size for row postion and modlus by row size to get column postion
+ * 
+ * we will use standard binary search 
+ */
+
+class Solution {
+    public boolean searchMatrix(int[][] matrix, int target) {
+        int left = 0;
+        int right = matrix[0].length * matrix.length - 1;
+        int n = matrix[0].length;
+        while (left <= right) {
+            int mid = left + (right - left) / 2;
+            int r = mid / n;
+            int c = mid % n;
+            if (matrix[r][c] == target) {
+                return true;
+            } else if (matrix[r][c] > target) {
+                right = mid - 1;
+            } else {
+                left = mid + 1;
+            }
+        }
+
+        return false;
+
+    }
+}

search_in_a_sorted_array_of_unknown_size.java

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+class Solution {
+    public int search(ArrayReader reader, int target) {
+        int low = 0;
+        int high = 1;
+        while (reader.get(high) <= target) {
+            if (reader.get(high) == target) {
+                return high;
+            } else {
+                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;
+    }
+}

search_in_rotated_sorted_array.java

@@ -0,0 +1,34 @@
+// Time Complexity : log n
+// Space Complexity : O(n)
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : keeping in mind that every time we will get one side as sorted and one side a not was deficult 
+
+/*
+ * Approach
+ * standard binary search 
+ * first we check check which side is sorted and then check if the target is in that side or not, if not check the other side
+ */
+
+class Solution {
+    public int search(int[] nums, int target) {
+        int left = 0;
+        int right = nums.length - 1;
+        while (left <= right) {
+            int mid = left + (right - left) / 2;
+            if (nums[mid] == target) {
+                return mid;
+            } else if (nums[left] <= nums[mid]) {
+                if (nums[left] <= target && nums[mid] > target) {
+                    right = mid - 1;
+                } else {
+                    left = mid + 1;
+                }
+            } else if (nums[mid] < target && nums[right] >= target) {
+                left = mid + 1;
+            } else {
+                right = mid - 1;
+            }
+        }
+        return -1;
+    }
+}