Add implementation for square root using Binary Search (#2509)

Co-authored-by: sahil.samantaray <sahil.samantaray@nymble.in>

Searches/SquareRootBinarySearch.java

@@ -0,0 +1,59 @@
+package Searches;
+
+import java.util.Scanner;
+
+/**
+ * Given an integer x, find the square root of x. If x is not a perfect square, then return floor(√x).
+ * <p>
+ * For example,
+ * if x = 5, The answer should be 2 which is the floor value of √5.
+ * <p>
+ * The approach that will be used for solving the above problem is not going to be a straight forward Math.sqrt().
+ * Instead we will be using Binary Search to find the square root of a number in the most optimised way.
+ *
+ * @author sahil
+ */
+public class SquareRootBinarySearch {
+
+    /**
+     * This is the driver method.
+     *
+     * @param args Command line arguments
+     */
+    public static void main(String args[]) {
+        Scanner sc = new Scanner(System.in);
+        System.out.print("Enter a number you want to calculate square root of : ");
+        int num = sc.nextInt();
+        long ans = squareRoot(num);
+        System.out.println("The square root is : " + ans);
+    }
+
+    /**
+     * This function calculates the floor of square root of a number.
+     * We use Binary Search algorithm to calculate the square root
+     * in a more optimised way.
+     *
+     * @param num Number
+     * @return answer
+     */
+    private static long squareRoot(long num) {
+        if (num == 0 || num == 1) {
+            return num;
+        }
+        long l = 1;
+        long r = num;
+        long ans = 0;
+        while (l <= r) {
+            long mid = l + (r - l) / 2;
+            if (mid == num / mid)
+                return mid;
+            else if (mid < num / mid) {
+                ans = mid;
+                l = mid + 1;
+            } else {
+                r = mid - 1;
+            }
+        }
+        return ans;
+    }
+}