Submit Binay Search -1

.project

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+<?xml version="1.0" encoding="UTF-8"?>
+<projectDescription>
+	<name>Binary-Search-1</name>
+	<comment></comment>
+	<projects>
+	</projects>
+	<buildSpec>
+	</buildSpec>
+	<natures>
+	</natures>
+</projectDescription>

BinarySearch1.java

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+// Search a 2D Matrix
+// Time Complexity : O(log (r*c))
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this :
+// 
+
+// Your code here along with comments explaining your approach in three sentences only
+
+//Approach 
+// 1) Get rows from size of matrix and columns from each row of matrix
+// 2) Binary search on indexes
+// 3) convert into row and column
+
+public class BinarySearch1 {
+    public boolean searchMatrix(int[][] matrix, int target) {
+        int rows = matrix.length;
+        int cols = matrix[0].length;
+
+        int left = 0;
+        int right = rows * cols - 1;
+
+        while (left <= right) {
+            int mid = left + (right - left) / 2;
+
+            int row = mid / cols;
+            int col = mid % cols;
+
+            int value = matrix[row][col];
+
+            if (value == target) {
+                return true;
+            } 
+            else if (value < target) {
+                left = mid + 1;
+            } 
+            else {
+                right = mid - 1;
+            }
+        }
+
+        return false;
+    }
+}

BinarySearch2.java

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+// 33. Search in Rotated Sorted Array
+// Time Complexity : O(log n)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this :
+// 
+
+// Your code here along with comments explaining your approach in three sentences only
+
+//Approach 
+// 1) find the mid element
+// 2) Sort the left oarray of the mid element
+// 3) Sort the right array of mid element
+// 4) return if found the target 
+
+public class BinarySearch2 {
+	public int search(int[] nums, int target) {
+        int left = 0, right = nums.length - 1;
+
+    while (left <= right) {
+        int mid = left + (right - left) / 2;
+
+        if (nums[mid] == target) return mid;
+
+        // Left half sorted
+        if (nums[left] <= nums[mid]) {
+            if (target >= nums[left] && target < nums[mid])
+                right = mid - 1;
+            else
+                left = mid + 1;
+        }
+        // Right half sorted
+        else {
+            if (target > nums[mid] && target <= nums[right])
+                left = mid + 1;
+            else
+                right = mid - 1;
+        }
+    }
+    return -1;
+    }
+}

BinarySearch3.java

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+// 702. Search in a Sorted Array of Unknown Size
+// Time Complexity : O(log n)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this :
+// 
+
+// Your code here along with comments explaining your approach in three sentences only
+
+//Approach 
+// 1) no fixed length given in this
+// 2) find the range of search 
+// 3) Search using binary searchx
+// 4) return if found the target 
+
+public class BinarySearch3 {
+	ppublic int search(ArrayReader reader, int target) {
+        int left = 0;
+        int right = 1;
+
+        while (reader.get(right) < target) {
+            left = right;
+            right = right * 2;
+        }
+
+        while (left <= right) {
+            int mid = left + (right - left) / 2;
+            int val = reader.get(mid);
+
+            if (val == target)
+                return mid;
+            else if (val < target)
+                left = mid + 1;
+            else
+                right = mid - 1;
+        }
+
+        return -1;
+    }
+}