Completed Binary Search 1 course

Search.java

@@ -0,0 +1,34 @@
+class Search {
+
+    /*
+    // Time Complexity : O(2log n) = O(log n)
+    // Space Complexity : O(1)
+        To serach in an array of unknown size, we implement binary search but not starting with first and last
+        index of the array as it is unknown. We start with the first two indices of the array and check if the value at high index
+        is less than the target, then we increment the serch size of array by high * 2. We do this as long as we get a high index which is 
+        less than the target value. Once we get the high index range, we do a regular binary search on the array to find index of the target element.
+    */
+
+    public int search(ArrayReader reader, int target) {
+        int low = 0;
+        int high = 1;
+
+        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;
+    }
+}

SearchMatrix.java

@@ -0,0 +1,30 @@
+public class SearchMatrix {
+
+    /*
+        // Time Complexity : O(log n + log m)
+        // Space Complexity : O(1)
+        To find a target element in a 2-D matrix which has elements sorted in ascending order, we are treating this 2D matrix as 
+        an array which will make it possible to find element using binary search. To find the mid element in the matrix between 
+        low and high indexes, we get the row index by dividing mid by column legth and column index by modulo operation with column length.
+        Then we compare if the value we are looking for is found return true. If its greater than mid element, we move the low pointer to mid + 1 index 
+        else we move the high pointer to mid - 1. If we do not find the element, return false.
+    */
+
+
+   public boolean searchMatrix(int[][] matrix, int target) {
+        int row = matrix.length;;
+        int col = matrix[0].length;
+        int low = 0;
+        int high = row * col - 1;
+        while (low <= high) {
+            int mid = low + (high - low)/2;
+            int r = mid/col;
+            int c = mid % col;
+            if (matrix[r][c] == target) return true;
+            else if (matrix[r][c] < target) low = mid + 1;
+            else high = mid - 1;
+        }
+
+        return false;
+    } 
+}

SearchRotatedSortedArray.java

@@ -0,0 +1,38 @@
+class SearchRotatedSortedArray {
+    /*
+        Time complexity - O(log n)
+        Space complexity = O(1)
+        to search in a rotated sorted array using binary search, we need to determine in which 
+        part of the array the target lies in. If mid element >= low element, it means the array 
+        is in sorted order. Check if mid element > target and low element <= target, if so we 
+        search in the left side lese we search in the right side of the array. When searching in 
+        right side if mid element < target and high ele is >= target then search in the right side of mid
+        else search on the left side. If mid element equals target we return index. Else we return -1.
+    */
+    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;
+            }
+            if (nums[mid] >= nums[low]) {
+                if (nums[mid] > target && nums[low] <= target) {
+                    high = mid - 1;
+                } else {
+                    low = mid + 1;
+                }
+            } else {
+                if (nums[mid] < target && nums[high] >= target) {
+                    low = mid + 1;
+                } else {
+                    high = mid - 1;
+                }
+            }
+        }
+
+        return -1;
+    }
+}