fix(curriculum): Problem 35 - update solution

curriculum/challenges/english/08-coding-interview-prep/project-euler/problem-35-circular-primes.english.md

@@ -9,6 +9,9 @@ title: 'Problem 35: Circular primes'
 The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.
 There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.
 How many circular primes are there below n, whereas 100 <= n <= 1000000?
+  
+<br><strong>Note:</strong><br>
+Circular primes individual rotation can exceeed `n`.
 </section>
 
 ## Instructions
@@ -63,37 +66,57 @@ circularPrimes(1000000);
 
 
 ```js
-const circularPrimes = (n) => {
+function rotate(n) {
-  const primeCheck = (num) => {
+  if (n.length == 1) return n;
-    if (num === 1) {
+  return n.slice(1) + n[0];
-      return false;
+}
+
+function circularPrimes(n) {
+  // Nearest n < 10^k
+  const bound = 10 ** Math.ceil(Math.log10(n));
+  const primes = [0, 0, 2];
+  let count = 0;
+
+  // Making primes array
+  for (let i = 4; i <= bound; i += 2) {
+    primes.push(i - 1);
+    primes.push(0);
   }
-    for (let i = 2; i <= Math.floor(Math.sqrt(num)); i++) {
+
-      if (num % i === 0) {
+  // Getting upperbound
-        return false;
+  const upperBound = Math.ceil(Math.sqrt(bound));
+
+  // Setting other non-prime numbers to 0
+  for (let i = 3; i < upperBound; i += 2) {
+    if (primes[i]) {
+      for (let j = i * i; j < bound; j += i) {
+        primes[j] = 0;
       }
     }
-    return true;
+  }
-  };
+
-  let count = 1;
+  // Iterating through the array
-  for (let i = 1; i < n; i += 2) {
+  for (let i = 2; i < n; i++) {
-    if (primeCheck(i)) {
+    if (primes[i]) {
-      let flag = true;
+      let curr = String(primes[i]);
-      let circularNum = i.toString();
+      let tmp = 1; // tmp variable to hold the no of rotations
-      for (let j = 1; j < i.toString().length; j++) {
+      for (let x = rotate(curr); x != curr; x = rotate(x)) {
-        circularNum = circularNum.substring(1) + circularNum.substring(0, 1);
+        if (x > n && primes[x]) {
-        if (primeCheck(Number(circularNum)) === false) {
+          continue;
-          flag = false;
+        }
+        else if (!primes[x]) {
+          // If the rotated value is 0 then its not a ciruclar prime, break the loop
+          tmp = 0;
           break;
         }
+        tmp++;
+        primes[x] = 0;
       }
-      if (flag) {
+      count += tmp;
-        count++;
-      }
     }
   }
   return count;
-};
+}
 ```
 
 </section>