Completed Binary-Search-1

search_2D_matrix.java

@@ -0,0 +1,24 @@
+public class search_2D_matrix {
+    public boolean searchMatrix(int[][] matrix, int target) {
+        int m = matrix.length;
+        int n = matrix[0].length;
+
+        int low = 0;
+        int high = m * n - 1;
+
+        while (low <= high) {
+            int mid = low + (high-low)/2;
+            int r = mid / n;
+            int c = mid % n;
+
+            if (matrix[r][c] == target) {
+                return true;
+            } else if (matrix[r][c] > target) {
+                high = mid - 1;
+            } else {
+                low = mid + 1;
+            }
+        }
+        return false;
+    }
+}

search_in_infinte_sorted_array.java

@@ -0,0 +1,37 @@
+// Time Complexity : O(log k), where k is the index of the target
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No
+
+ /**
+ * // This is ArrayReader's API interface.
+ * // You should not implement it, or speculate about its implementation
+ * interface ArrayReader {
+ *     public int get(int index) {}
+ * }
+ */
+
+class Solution {
+    public int search(ArrayReader reader, int target) {
+        int low = 0;
+        int high = 1;
+
+        // First expand the search boundary exponentially until high is large enough to include the target.
+        // Then perform binary search within that range to find the target efficiently.
+        // This works well for unknown-sized sorted arrays because it avoids needing the array length.
+        while (reader.get(high) < target) {
+            low = high;
+            high = 2 * high;
+        }
+
+        while (low <= high) {
+            int mid = low + (high - low) / 2;
+            int value = reader.get(mid);
+
+            if (value == target) return mid;
+            if (value > target) high = mid - 1;
+            else low = mid + 1;
+        }
+        return -1;
+    }
+}