Bot: Prettified Java code!

Author: ARajan1084

src/com/codefortomorrow/advanced/chapter13/examples/BinarySearch.java

@@ -1,46 +1,42 @@
 package com.codefortomorrow.advanced.chapter13.examples;
 
 public class BinarySearch {
-        // Java implementation of recursive Binary Search
-        // Returns index of x if it is present in arr[l..
-        // r], else return -1
-        public static int binarySearch(int arr[], int l, int r, int x)
-        {
-            if (r >= l) {
-                int mid = l + (r - l) / 2;
-
-                // If the element is present at the
-                // middle itself
-                if (arr[mid] == x)
-                    return mid;
-
-                // If element is smaller than mid, then
-                // it can only be present in left subarray
-                if (arr[mid] > x)
-                    return binarySearch(arr, l, mid - 1, x);
-
-                // Else the element can only be present
-                // in right subarray
-                return binarySearch(arr, mid + 1, r, x);
-            }
-
-            // We reach here when element is not present
-            // in array
-            return -1;
-        }
 
-        // Driver method to test above
-        public static void main(String args[])
-        {
-            int arr[] = { 2, 3, 4, 10, 40 };
-            int n = arr.length;
-            int x = 10;
-            int result = binarySearch(arr, 0, n - 1, x);
-            if (result == -1)
-                System.out.println("Element not present");
-            else
-                System.out.println("Element found at index " + result);
+    // Java implementation of recursive Binary Search
+    // Returns index of x if it is present in arr[l..
+    // r], else return -1
+    public static int binarySearch(int arr[], int l, int r, int x) {
+        if (r >= l) {
+            int mid = l + (r - l) / 2;
+
+            // If the element is present at the
+            // middle itself
+            if (arr[mid] == x) return mid;
+
+            // If element is smaller than mid, then
+            // it can only be present in left subarray
+            if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
+
+            // Else the element can only be present
+            // in right subarray
+            return binarySearch(arr, mid + 1, r, x);
         }
+
+        // We reach here when element is not present
+        // in array
+        return -1;
+    }
+
+    // Driver method to test above
+    public static void main(String args[]) {
+        int arr[] = { 2, 3, 4, 10, 40 };
+        int n = arr.length;
+        int x = 10;
+        int result = binarySearch(arr, 0, n - 1, x);
+        if (result == -1) System.out.println(
+            "Element not present"
+        ); else System.out.println("Element found at index " + result);
+    }
     /* This code is contributed by Rajat Mishra
     https://www.geeksforgeeks.org/binary-search/ */