fix(curriculum): Problem 47 - Added optimized solution

curriculum/challenges/english/08-coding-interview-prep/project-euler/problem-47-distinct-primes-factors.english.md

@@ -53,57 +53,54 @@ distinctPrimeFactors(4, 4);
 
 </div>
 
-
-
 </section>
 
 ## Solution
 <section id='solution'>
 
-
 ```js
 function distinctPrimeFactors(targetNumPrimes, targetConsecutive) {
+  function numberOfPrimeFactors(n) {
+    let factors = 0;
 
-  function isPrime(num) {
+    //  Considering 2 as a special case
-    for (let i = 2, s = Math.sqrt(num); i <= s; i++) {
+    let firstFactor = true;
-      if (num % i === 0) {
+    while (n % 2 == 0) {
-        return false;
+      n = n / 2;
+      if (firstFactor) {
+        factors++;
+        firstFactor = false;
+      }
+    }
+    // Adding other factors
+    for (let i = 3; i < Math.sqrt(n); i += 2) {
+      firstFactor = true;
+      while (n % i == 0) {
+        n = n / i;
+        if (firstFactor) {
+          factors++;
+          firstFactor = false;
         }
       }
-    return num !== 1;
     }
 
-  function getPrimeFactors(num) {
+    if (n > 1) { factors++; }
-    const factors = [];
+
-    for (let i = 2; i <= Math.sqrt(num); i++) {
-      if (num % i === 0) {
-        // found a factor
-        if (isPrime(i)) {
-          factors.push(i);
-        }
-        if (isPrime(num / i) && i !== Math.sqrt(num)) {
-          factors.push(num / i);
-        }
-      }
-    }
     return factors;
   }
 
-  function findConsecutiveNumbers() {
   let number = 0;
   let consecutive = 0;
+
   while (consecutive < targetConsecutive) {
     number++;
-      if (getPrimeFactors(number).length >= targetNumPrimes) {
+    if (numberOfPrimeFactors(number) >= targetNumPrimes) {
       consecutive++;
     } else {
       consecutive = 0;
     }
   }
-    return (number - targetConsecutive) + 1;
+  return number - targetConsecutive + 1;
-  }
-
-  return findConsecutiveNumbers();
 }
 ```