Resolve duplicate entries for sieve of eratosthenes (#1770)

* remove intarr test

* Remove main file oops

* FIXES: #1666 , remove references to SieveOfEratosthenesIntArray

* Finally fix the requirements, passes vitest

* Updated Documentation in README.md

* FIXES: #1666 and conform to alg comment standards

---------

Co-authored-by: SpiderMath <SpiderMath@users.noreply.github.com>

Author: SpiderMath

DIRECTORY.md

@@ -254,7 +254,6 @@
   * [RowEchelon](Maths/RowEchelon.js)
   * [ShorsAlgorithm](Maths/ShorsAlgorithm.js)
   * [SieveOfEratosthenes](Maths/SieveOfEratosthenes.js)
-  * [SieveOfEratosthenesIntArray](Maths/SieveOfEratosthenesIntArray.js)
   * [Signum](Maths/Signum.js)
   * [SimpsonIntegration](Maths/SimpsonIntegration.js)
   * [Softmax](Maths/Softmax.js)

Maths/SieveOfEratosthenes.js

@@ -1,25 +1,26 @@
-const sieveOfEratosthenes = (n) => {
-  /*
-   * Calculates prime numbers till a number n
-   * :param n: Number up to which to calculate primes
-   * :return: A boolean list containing only primes
-   */
-  const primes = new Array(n + 1)
-  primes.fill(true) // set all as true initially
-  primes[0] = primes[1] = false // Handling case for 0 and 1
-  const sqrtn = Math.ceil(Math.sqrt(n))
-  for (let i = 2; i <= sqrtn; i++) {
-    if (primes[i]) {
-      for (let j = i * i; j <= n; j += i) {
-        /*
-        Optimization.
-        Let j start from i * i, not 2 * i, because smaller multiples of i have been marked false.
+/**
+ * @function sieveOfEratosthenes
+ * @description Function to get all the prime numbers below a given number using sieve of eratosthenes algorithm
+ * @param {Number} max The limit below which all the primes are required to be
+ * @returns {Number[]} An array of all the prime numbers below max
+ * @see [Sieve of Eratosthenes](https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes)
+ * @example
+ * sieveOfEratosthenes(1) // ====> []
+ * @example
+ * sieveOfEratosthenes(20) // ====> [2, 3, 5, 7, 11, 13, 17, 19]
+ *
+ */
+function sieveOfEratosthenes(max) {
+  const sieve = []
+  const primes = []
 
-        For example, let i = 4.
-        We do not have to check from 8(4 * 2) to 12(4 * 3)
-        because they have been already marked false when i=2 and i=3.
-        */
-        primes[j] = false
+  for (let i = 2; i <= max; ++i) {
+    if (!sieve[i]) {
+      // If i has not been marked then it is prime
+      primes.push(i)
+      for (let j = i << 1; j <= max; j += i) {
+        // Mark all multiples of i as non-prime
+        sieve[j] = true
       }
     }
   }

Maths/SieveOfEratosthenesIntArray.js

@@ -1,14 +1,29 @@
 import { sieveOfEratosthenes } from '../SieveOfEratosthenes'
-import { PrimeCheck } from '../PrimeCheck'
-
-describe('should return an array of prime booleans', () => {
-  it('should have each element in the array as a prime boolean', () => {
-    const n = 30
-    const primes = sieveOfEratosthenes(n)
-    primes.forEach((primeBool, index) => {
-      if (primeBool) {
-        expect(PrimeCheck(index)).toBeTruthy()
-      }
-    })
+
+describe('sieveOfEratosthenes', () => {
+  test('returns an empty array for max < 2', () => {
+    expect(sieveOfEratosthenes(1)).toEqual([])
+  })
+
+  test('returns [2] for max = 2', () => {
+    expect(sieveOfEratosthenes(2)).toEqual([2])
+  })
+
+  test('returns [2, 3] for max = 3', () => {
+    expect(sieveOfEratosthenes(3)).toEqual([2, 3])
+  })
+
+  test('returns [2, 3, 5, 7] for max = 10', () => {
+    expect(sieveOfEratosthenes(10)).toEqual([2, 3, 5, 7])
+  })
+
+  test('returns [2, 3, 5, 7, 11, 13, 17, 19] for max = 20', () => {
+    expect(sieveOfEratosthenes(20)).toEqual([2, 3, 5, 7, 11, 13, 17, 19])
+  })
+
+  test('returns [2, 3, 5, 7, 11, 13, 17, 19, 23, 29] for max = 30', () => {
+    expect(sieveOfEratosthenes(30)).toEqual([
+      2, 3, 5, 7, 11, 13, 17, 19, 23, 29
+    ])
   })
 })

Maths/test/SieveOfEratosthenes.test.js

@@ -9,7 +9,7 @@
  *
  * @author ddaniel27
  */
-import { sieveOfEratosthenes } from '../Maths/SieveOfEratosthenesIntArray'
+import { sieveOfEratosthenes } from '../Maths/SieveOfEratosthenes'
 
 function problem35(n) {
   if (n < 2) {