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

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camperbot
2021-12-23 23:35:44 +05:30
committed by GitHub
parent cb201a8e8b
commit 7c6c8bfd35
2 changed files with 118 additions and 62 deletions
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90
curriculum/challenges/portuguese/10-coding-interview-prep/project-euler/problem-50-consecutive-prime-sum.md
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@ -54,42 +54,70 @@ consecutivePrimeSum(1000000);
# --solutions-- # --solutions--
```js ```js
function consecutivePrimeSum(limit) { // Initalize prime number list with sieve
function isPrime(num) { const NUM_PRIMES = 1000000;
if (num < 2) { const PRIMES = [2];
return false; const PRIME_SIEVE = Array(Math.floor((NUM_PRIMES-1)/2)).fill(true);
} else if (num === 2) { (function initPrimes(num) {
return true; const upper = Math.floor((num - 1) / 2);
} const sqrtUpper = Math.floor((Math.sqrt(num) - 1) / 2);
const sqrtOfNum = Math.floor(num ** 0.5); for (let i = 0; i <= sqrtUpper; i++) {
for (let i = 2; i <= sqrtOfNum + 1; i++) { if (PRIME_SIEVE[i]) {
if (num % i === 0) { // Mark value in PRIMES array
return false; 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; return true;
} else if (num % 2 === 0)
function getPrimes(limit) { return false
const primes = []; else
for (let i = 0; i <= limit; i++) { return PRIME_SIEVE[(num - 3) / 2];
if (isPrime(i)) primes.push(i); }
}
return primes;
}
const primes = getPrimes(limit); function consecutivePrimeSum(limit) {
let primeSum = [...primes]; // Initalize for longest sum < 100
primeSum.reduce((acc, n, i) => { let bestPrime = 41;
primeSum[i] += acc; let bestI = 0;
return acc += n; let bestJ = 5;
}, 0);
for (let j = primeSum.length - 1; j >= 0; j--) { // Find longest sum < limit
for (let i = 0; i < j; i++) { let sumOfCurrRange = 41;
const sum = primeSum[j] - primeSum[i]; let i = 0, j = 5;
if (sum > limit) break; // -- Loop while current some starting at i is < limit
if (isPrime(sum) && primes.indexOf(sum) > -1) return sum; while (sumOfCurrRange < limit) {
let currSum = sumOfCurrRange;
// -- Loop while pushing j towards end of PRIMES list
// keeping sum under limit
while (currSum < limit) {
if (isPrime(currSum)) {
bestPrime = sumOfCurrRange = currSum;
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;
} }
``` ```

90
curriculum/challenges/ukrainian/10-coding-interview-prep/project-euler/problem-50-consecutive-prime-sum.md
View File

@ -54,42 +54,70 @@ consecutivePrimeSum(1000000);
# --solutions-- # --solutions--
```js ```js
function consecutivePrimeSum(limit) { // Initalize prime number list with sieve
function isPrime(num) { const NUM_PRIMES = 1000000;
if (num < 2) { const PRIMES = [2];
return false; const PRIME_SIEVE = Array(Math.floor((NUM_PRIMES-1)/2)).fill(true);
} else if (num === 2) { (function initPrimes(num) {
return true; const upper = Math.floor((num - 1) / 2);
} const sqrtUpper = Math.floor((Math.sqrt(num) - 1) / 2);
const sqrtOfNum = Math.floor(num ** 0.5); for (let i = 0; i <= sqrtUpper; i++) {
for (let i = 2; i <= sqrtOfNum + 1; i++) { if (PRIME_SIEVE[i]) {
if (num % i === 0) { // Mark value in PRIMES array
return false; 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; return true;
} else if (num % 2 === 0)
function getPrimes(limit) { return false
const primes = []; else
for (let i = 0; i <= limit; i++) { return PRIME_SIEVE[(num - 3) / 2];
if (isPrime(i)) primes.push(i); }
}
return primes;
}
const primes = getPrimes(limit); function consecutivePrimeSum(limit) {
let primeSum = [...primes]; // Initalize for longest sum < 100
primeSum.reduce((acc, n, i) => { let bestPrime = 41;
primeSum[i] += acc; let bestI = 0;
return acc += n; let bestJ = 5;
}, 0);
for (let j = primeSum.length - 1; j >= 0; j--) { // Find longest sum < limit
for (let i = 0; i < j; i++) { let sumOfCurrRange = 41;
const sum = primeSum[j] - primeSum[i]; let i = 0, j = 5;
if (sum > limit) break; // -- Loop while current some starting at i is < limit
if (isPrime(sum) && primes.indexOf(sum) > -1) return sum; while (sumOfCurrRange < limit) {
let currSum = sumOfCurrRange;
// -- Loop while pushing j towards end of PRIMES list
// keeping sum under limit
while (currSum < limit) {
if (isPrime(currSum)) {
bestPrime = sumOfCurrRange = currSum;
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;
} }
``` ```