chore(i18n,curriculum): update translations (#44573)

curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-50-consecutive-prime-sum.md

@@ -54,42 +54,70 @@ consecutivePrimeSum(1000000);
 # --solutions--
 
 ```js
+// Initalize prime number list with sieve
+const NUM_PRIMES = 1000000;
+const PRIMES = [2];
+const PRIME_SIEVE = Array(Math.floor((NUM_PRIMES-1)/2)).fill(true);
+(function initPrimes(num) {
+  const upper = Math.floor((num - 1) / 2);
+  const sqrtUpper = Math.floor((Math.sqrt(num) - 1) / 2);
+  for (let i = 0; i <= sqrtUpper; i++) {
+    if (PRIME_SIEVE[i]) {
+      // Mark value in PRIMES array
+      const prime = 2 * i + 3;
+      PRIMES.push(prime);
+      // Mark all multiples of this number as false (not prime)
+      const primeSqaredIndex = 2 * i ** 2 + 6 * i + 3;
+      for (let j = primeSqaredIndex; j < upper; j += prime)
+        PRIME_SIEVE[j] = false;
+    }
+  }
+  for (let i = sqrtUpper + 1; i < upper; i++) {
+    if (PRIME_SIEVE[i])
+      PRIMES.push(2 * i + 3);
+  }
+})(NUM_PRIMES);
+
+function isPrime(num) {
+  if (num === 2)
+    return true;
+  else if (num % 2 === 0)
+    return false
+  else
+    return PRIME_SIEVE[(num - 3) / 2];
+}
+
 function consecutivePrimeSum(limit) {
-  function isPrime(num) {
+  // Initalize for longest sum < 100
-    if (num < 2) {
+  let bestPrime = 41;
-      return false;
+  let bestI = 0;
-    } else if (num === 2) {
+  let bestJ = 5;
-      return true;
-    }
-    const sqrtOfNum = Math.floor(num ** 0.5);
-    for (let i = 2; i <= sqrtOfNum + 1; i++) {
-      if (num % i === 0) {
-        return false;
-      }
-    }
-    return true;
-  }
-  function getPrimes(limit) {
-    const primes = [];
-    for (let i = 0; i <= limit; i++) {
-      if (isPrime(i)) primes.push(i);
-    }
-    return primes;
-  }
 
-  const primes = getPrimes(limit);
+  // Find longest sum < limit
-  let primeSum = [...primes];
+  let sumOfCurrRange = 41;
-  primeSum.reduce((acc, n, i) => {
+  let i = 0, j = 5;
-    primeSum[i] += acc;
+  // -- Loop while current some starting at i is < limit
-    return acc += n;
+  while (sumOfCurrRange < limit) {
-  }, 0);
+    let currSum = sumOfCurrRange;
-
+    // -- Loop while pushing j towards end of PRIMES list
-  for (let j = primeSum.length - 1; j >= 0; j--) {
+    //      keeping sum under limit
-    for (let i = 0; i < j; i++) {
+    while (currSum < limit) {
-      const sum = primeSum[j] - primeSum[i];
+      if (isPrime(currSum)) {
-      if (sum > limit) break;
+        bestPrime = sumOfCurrRange = currSum;
-      if (isPrime(sum) && primes.indexOf(sum) > -1) return sum;
+        bestI = i;
+        bestJ = j;
       }
+      // -- Increment inner loop
+      j++;
+      currSum += PRIMES[j];
     }
+    // -- Increment outer loop
+    i++;
+    j = i + (bestJ - bestI);
+    sumOfCurrRange -= PRIMES[i - 1];
+    sumOfCurrRange += PRIMES[j];
+  }
+  // Return
+  return bestPrime;
 }
 ```

curriculum/challenges/ukrainian/10-coding-interview-prep/project-euler/problem-50-consecutive-prime-sum.md

@@ -54,42 +54,70 @@ consecutivePrimeSum(1000000);
 # --solutions--
 
 ```js
+// Initalize prime number list with sieve
+const NUM_PRIMES = 1000000;
+const PRIMES = [2];
+const PRIME_SIEVE = Array(Math.floor((NUM_PRIMES-1)/2)).fill(true);
+(function initPrimes(num) {
+  const upper = Math.floor((num - 1) / 2);
+  const sqrtUpper = Math.floor((Math.sqrt(num) - 1) / 2);
+  for (let i = 0; i <= sqrtUpper; i++) {
+    if (PRIME_SIEVE[i]) {
+      // Mark value in PRIMES array
+      const prime = 2 * i + 3;
+      PRIMES.push(prime);
+      // Mark all multiples of this number as false (not prime)
+      const primeSqaredIndex = 2 * i ** 2 + 6 * i + 3;
+      for (let j = primeSqaredIndex; j < upper; j += prime)
+        PRIME_SIEVE[j] = false;
+    }
+  }
+  for (let i = sqrtUpper + 1; i < upper; i++) {
+    if (PRIME_SIEVE[i])
+      PRIMES.push(2 * i + 3);
+  }
+})(NUM_PRIMES);
+
+function isPrime(num) {
+  if (num === 2)
+    return true;
+  else if (num % 2 === 0)
+    return false
+  else
+    return PRIME_SIEVE[(num - 3) / 2];
+}
+
 function consecutivePrimeSum(limit) {
-  function isPrime(num) {
+  // Initalize for longest sum < 100
-    if (num < 2) {
+  let bestPrime = 41;
-      return false;
+  let bestI = 0;
-    } else if (num === 2) {
+  let bestJ = 5;
-      return true;
-    }
-    const sqrtOfNum = Math.floor(num ** 0.5);
-    for (let i = 2; i <= sqrtOfNum + 1; i++) {
-      if (num % i === 0) {
-        return false;
-      }
-    }
-    return true;
-  }
-  function getPrimes(limit) {
-    const primes = [];
-    for (let i = 0; i <= limit; i++) {
-      if (isPrime(i)) primes.push(i);
-    }
-    return primes;
-  }
 
-  const primes = getPrimes(limit);
+  // Find longest sum < limit
-  let primeSum = [...primes];
+  let sumOfCurrRange = 41;
-  primeSum.reduce((acc, n, i) => {
+  let i = 0, j = 5;
-    primeSum[i] += acc;
+  // -- Loop while current some starting at i is < limit
-    return acc += n;
+  while (sumOfCurrRange < limit) {
-  }, 0);
+    let currSum = sumOfCurrRange;
-
+    // -- Loop while pushing j towards end of PRIMES list
-  for (let j = primeSum.length - 1; j >= 0; j--) {
+    //      keeping sum under limit
-    for (let i = 0; i < j; i++) {
+    while (currSum < limit) {
-      const sum = primeSum[j] - primeSum[i];
+      if (isPrime(currSum)) {
-      if (sum > limit) break;
+        bestPrime = sumOfCurrRange = currSum;
-      if (isPrime(sum) && primes.indexOf(sum) > -1) return sum;
+        bestI = i;
+        bestJ = j;
       }
+      // -- Increment inner loop
+      j++;
+      currSum += PRIMES[j];
     }
+    // -- Increment outer loop
+    i++;
+    j = i + (bestJ - bestI);
+    sumOfCurrRange -= PRIMES[i - 1];
+    sumOfCurrRange += PRIMES[j];
+  }
+  // Return
+  return bestPrime;
 }
 ```