#### Fast Inverse Sqrt

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```/** Author : Siddhant Swarup Mallick
* Github : https://github.com/siddhant2002
*/

/** Program description - To find out the inverse square root of the given number*/

/** Wikipedia Link - https://en.wikipedia.org/wiki/Fast_inverse_square_root */

package com.thealgorithms.maths;

public class FastInverseSqrt {

public static boolean inverseSqrt(float number) {
float x = number;
float xhalf = 0.5f * x;
int i = Float.floatToIntBits(x);
i = 0x5f3759df - (i >> 1);
x = Float.intBitsToFloat(i);
x = x * (1.5f - xhalf * x * x);
return x == (float) ((float) 1 / (float) Math.sqrt(number));
}

/**
* Returns the inverse square root of the given number upto 6 - 8 decimal places.
* calculates the inverse square root of the given number and returns true if calculated answer matches with given answer else returns false
*/

public static boolean inverseSqrt(double number) {
double x = number;
double xhalf = 0.5d * x;
long i = Double.doubleToLongBits(x);
i = 0x5fe6ec85e7de30daL - (i >> 1);
x = Double.longBitsToDouble(i);
for (int it = 0; it < 4; it++) {
x = x * (1.5d - xhalf * x * x);
}
x *= number;
return x == 1 / Math.sqrt(number);
}
/**
* Returns the inverse square root of the given number upto 14 - 16 decimal places.
* calculates the inverse square root of the given number and returns true if calculated answer matches with given answer else returns false
*/
}
/**
* OUTPUT :
* Input - number = 4522
* Output: it calculates the inverse squareroot of a number and returns true with it matches the given answer else returns false.
* 1st approach Time Complexity : O(1)
* Auxiliary Space Complexity : O(1)
* Input - number = 4522
* Output: it calculates the inverse squareroot of a number and returns true with it matches the given answer else returns false.
* 2nd approach Time Complexity : O(1)
* Auxiliary Space Complexity : O(1)
*/
```  