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https://github.com/TheAlgorithms/JavaScript.git
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merge: Improved Pow
function (#911)
* feat: add negative power option * docs: add js doc for powOn function * feat: add PowFaster with faster algorithm, complexity O(logN) * chore: rename to exponent * chore: rename fixed * style: formated with standard
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65
Maths/Pow.js
65
Maths/Pow.js
@ -1,11 +1,62 @@
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// Returns the value of x to the power of y
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/**
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* @function powLinear
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const pow = (x, y) => {
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* @description - The powLinear function is a power function with Linear O(n) complexity
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let result = 1
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* @param {number} base
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for (let i = 1; i <= y; i++) {
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* @param {number} exponent
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result *= x
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* @returns {number}
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* @example - powLinear(2, 2) => 4 --> 2 * 2
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* @example - powLinear(3, 3) => 27 --> 3 * 3
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*/
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const powLinear = (base, exponent) => {
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if (exponent < 0) {
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base = 1 / base
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exponent = -exponent
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}
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}
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let result = 1
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while (exponent--) { // Break the execution while the exponent will 0
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result *= base
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}
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return result
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return result
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}
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}
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export { pow }
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/**
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* @function powFaster
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* @description - The powFaster function is a power function with O(logN) complexity
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* @param {number} base
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* @param {number} exponent
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* @returns {number}
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* @example - powFaster(2, 2) => 4 --> 2 * 2
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* @example - powFaster(3, 3) => 27 --> 3 * 3
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*/
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const powFaster = (base, exponent) => {
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if (exponent < 2) { // explanation below - 1
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return base && ([1, base][exponent] || powFaster(1 / base, -exponent))
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}
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if (exponent & 1) { // if the existing exponent is odd
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return base * powFaster(base * base, exponent >> 1) // explanation below - 2
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}
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return powFaster(base * base, exponent / 2)
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}
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/**
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* 1 - Magic of short circuit evaluation (&&, ||)
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* if the base is 0 then it returns 0 cause 0 is falsy
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* if the base is not 0 then it's must be truthy. after that, it will be executed the right portion of the && (AND) operator
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* Now it checks the exponent by the help array index, is it 0 or 1.
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* if the exponent is not 0 or 1 it's definitely less than 0, and a negative number is not a valid index number so it returns "undefined"
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* if the expression is undefined mean -> falsy, the || (OR) operator evaluates the right portion that is a recursive function.
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*/
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/**
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* 2 - Play with right shift bitwise operator (>>)
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* right shift with any odd numbers it returns the floor number instead of float.
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* E.g. if the number is 5, after right shifting with 1 it's will give us 2, not 2.5
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* cause the right shift formula is --> x >> y = |x| / 2^y
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*/
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export { powLinear, powFaster }
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@ -1,15 +1,41 @@
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import { pow } from '../Pow'
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import { powLinear, powFaster } from '../Pow'
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describe('Pow', () => {
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describe('Testing powLinear function', () => {
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it('should return 1 for numbers with exponent 0', () => {
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it('should return 1 for numbers with exponent 0', () => {
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expect(pow(2, 0)).toBe(1)
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expect(powLinear(2, 0)).toBe(1)
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})
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it('should return 0.5 for numbers with exponent -1', () => {
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expect(powLinear(2, -1)).toBe(0.5)
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})
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})
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it('should return 0 for numbers with base 0', () => {
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it('should return 0 for numbers with base 0', () => {
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expect(pow(0, 23)).toBe(0)
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expect(powLinear(0, 23)).toBe(0)
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})
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})
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it('should return the base to the exponent power', () => {
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it('should return the base to the exponent power', () => {
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expect(pow(24, 4)).toBe(331776)
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expect(powLinear(24, 4)).toBe(331776)
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})
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})
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describe('Testing powFaster function', () => {
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it('should return 1 for numbers with exponent 0', () => {
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expect(powFaster(2, 0)).toBe(1)
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})
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it('should return 0.5 for numbers with exponent -1', () => {
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expect(powFaster(2, -1)).toBe(0.5)
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})
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it('should return 0 for numbers with base 0', () => {
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expect(powFaster(0, 23)).toBe(0)
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})
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it('should return the base to the exponent power', () => {
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expect(powFaster(24, 4)).toBe(331776)
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})
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it('should return the result in O(lonN) complexity', () => {
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expect(powFaster(2, 64)).toBe(18446744073709552000) // execution time Math.log2(64) -> 6
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})
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})
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})
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})
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