Given a number N and a power P, the task is to find the exponent of this number raised to the given power, i.e. NP. Examples:
Input: N = 5, P = 2 Output: 25 Input: N = 2, P = 5 Output: 32
Below are the various ways to find NP:
- Method 1: Using Recursion
Java
// Java program to find the power of a number// using Recursionclass GFG { // Function to calculate N raised to the power P static int power(int N, int P) { if (P == 0) return 1; else return N * power(N, P - 1); } // Driver code public static void main(String[] args) { int N = 2; int P = 3; System.out.println(power(N, P)); }} |
Output
8
- Method 2: With the help of Loop
Java
// Java program to find the power of a number// with the help of loopclass GFG { // Function to calculate N raised to the power P static int power(int N, int P) { int pow = 1; for (int i = 1; i <= P; i++) pow *= N; return pow; } // Driver code public static void main(String[] args) { int N = 2; int P = 3; System.out.println(power(N, P)); }} |
- Method 3: Using Math.pow() method
Java
// Java program to find the power of a number// using Math.pow() methodimport java.lang.Math;class GFG { // Function to calculate N raised to the power P static double power(int N, int P) { return Math.pow(N, P); } // Driver code public static void main(String[] args) { int N = 2; int P = 3; System.out.println(power(N, P)); }} |
Output
8.0
- Method 4 : Efficient Approach
In this approach I am going to calculate power of given base using bit manipulation with logarithmic time complexity
Java
/*java program to calculate of power of given base number */public class GFG { public static void main(String[] args) { int base = 2; int power = 3; int ans = 1; while (power > 0){ if ((power&1)==1){ // if == 0 there is no point to calculate ans ans = ans * base; } base = base * base; // keep dividing power by 2 using right shift power = power >> 1; } System.out.println(ans); }} |
Output
8
Time complexity : O(logb)
