ModularExp code added

Author: AayushBangroo

Number Theory/ModularExp.java

@@ -0,0 +1,60 @@
+/*
+Problem Statement: Given three numbers 'a', 'b' and 'm' calculate a^b under modulo m.
+Constraints: a<=10^9, b<=10^9, m<=10^9
+
+Idea: Any number can be represented in sum of powers of 2. The idea is to represent the power 'b' in sum of
+powers of 2.
+
+Example: 3^11 can be written as 3^(1011) where power is in binary.
+        = 3^(1*(2^3) + 0*(2^2) + 1*(2^1) + 1*(2^0))
+        = 3^(1*8 + 0*4 + 1*2 + 1*1)
+        = 3^8 * 3^0 * 3^2 * 3^1
+        = 6561 * 1 * 9 * 3
+        = 177417
+
+Why use modulo m?
+Answer: Since the multiplication operation in lines 48 and 51 can easily exceed 'm' modulo is taken.
+
+Complexity: O(Log(b))
+Reason: The while loop will run for number of bits in 'b' which is Log(b)
+
+Modulo properties used:
+(a * b * c *d) % m = ((( a * b ) % m) * c) % m) * d) % m
+*/
+
+import java.util.Scanner;
+
+public class ModularExp {
+    public static void main(String[] args) {
+        Scanner sc = new Scanner(System.in);
+
+        long a = sc.nextLong();
+        long b = sc.nextLong();
+        long m = sc.nextLong();
+
+        long result = modularExp(a, b, m);
+
+        System.out.println(result);
+    }
+
+    public static long modularExp(long a, long b, long m) {
+        long ans = 1;
+
+        //handle case where 'a' is large
+        a = a % m;
+
+        while (b > 0) {
+            //LSB is set
+            if ((b & 1) == 1) {
+                //multiply current base with ans if bit is set
+                ans = (ans * a) % m;
+            }
+            //drop LSB by right-shifing by 1
+            b >>= 1;
+            //increase base with every iteration
+            a = (a * a) % m;
+        }
+
+        return ans;
+    }
+}