Women's Cup stats & predictions

Tips per Seguire le Partite dal Vivo

Ecco alcuni consigli utili per chi vuole seguire le partite dal vivo e vivere appieno l'esperienza sportiva:

• Informarsi sui canali ufficiali di trasmissione delle partite per non perdere nemmeno un minuto dell'azione.
• Cercare streaming online ufficiali o piattaforme affidabili se non puoi essere presente fisicamente allo stadio.
• Avere a portata di mano statistiche aggiornate e analisi pre-partita per comprendere meglio le dinamiche del gioco.
• Pianificare momenti conviviali con amici o familiari per rendere l'esperienza ancora più piacevole.

Possibili Sorprese del Torneo

Ogni torneo può riservare sorprese impreviste. Ecco alcune squadre o giocatrici che potrebbero emergere come vere rivelazioni:

• Squadre Underdog:** Alcune squadre meno blasonate potrebbero sfruttare al meglio l'occasione per dimostrare il proprio valore e sorprendere tutti con prestazioni straordinarie.
• Giovani Talenti:** Attenzione ai giovani talenti che stanno facendo il loro esordio internazionale; potrebbero essere il futuro del calcio femminile!

Risultati Storici della Women's Cup Russia

Ecco un breve excursus sui risultati storici della Women's Cup Russia, che ci aiuta a comprendere meglio il contesto attuale del torneo:

• Nel 2018, la Women's Cup Russia fu vinta dalla squadra locale che stupì tutti con una performance impeccabile lungo tutto il torneo.kubekode/AdventOfCode2020<|file_sep|>/day6/part1.js const fs = require('fs'); const data = fs.readFileSync('data/day6.txt', 'utf-8').split('nn'); const getAnswers = (answers) => { return answers.split('n').reduce((acc, answer) => { return acc.concat([...new Set(answer.split('').filter(a => a !== 'n'))]); }, []); } const count = data.reduce((acc, answers) => { return acc + getAnswers(answers).length; }, 0); console.log(count); <|file_sep|># Advent Of Code My solutions to [Advent Of Code](https://adventofcode.com/) from the year 2020. ## Running To run all the parts of all the days: bash $ node day1/part1.js && node day1/part2.js && node day6/part1.js && node day6/part2.js && node day7/part1.js && node day7/part2.js && node day8/part1.js && node day8/part2.js && node day9/part1.js && node day9/part2.js To run all the parts of one day: bash $ node day8/part1.js && node day8/part2.js <|repo_name|>kubekode/AdventOfCode2020<|file_sep|>/day7/part1.js const fs = require('fs'); const data = fs.readFileSync('data/day7.txt', 'utf-8').split('n'); const bagRules = {}; data.forEach(rule => { const [bagName, bagContains] = rule.split(' bags contain '); bagRules[bagName] = bagContains.split(', ').reduce((acc, item) => { if (item === 'no other bags.') { return acc; } const [amount, color] = item.split(' '); acc[color] = parseInt(amount); return acc; }, {}); }); const containsBagRecursive = (bagColor) => { let containedBagsCount = 0; Object.entries(bagRules[bagColor]).forEach(([color]) => { containedBagsCount += containsBagRecursive(color); if (bagRules[color]) { containedBagsCount++; } }); return containedBagsCount; } const countContainingBagsRecursive = (bagColor) => { let containingBagsCount = 0; Object.entries(bagRules).forEach(([color]) => { const colorBagContents = bagRules[color]; if (colorBagContents[bagColor]) { containingBagsCount++; if (bagRules[color]) { containingBagsCount += countContainingBagsRecursive(color); } } else if (bagRules[color]) { containingBagsCount += countContainingBagsRecursive(color); } }); return containingBagsCount; } console.log(containsBagRecursive('shiny gold')); console.log(countContainingBagsRecursive('shiny gold') - 1); // minus one to not include the shiny gold bag itself. <|repo_name|>kubekode/AdventOfCode2020<|file_sep|>/day9/part1.js const fs = require('fs'); const data = fs.readFileSync('data/day9.txt', 'utf-8').split('n').map(Number); let invalidNumber; for (let i = 25; i <= data.length - 1; i++) { const numbersInLast25Entries = data.slice(i - 25, i); let isSumOfTwoFound; for (let j = i - 25; j <= i - 1; j++) { for (let k = j + 1; k <= i - 1; k++) { if (numbersInLast25Entries[j] + numbersInLast25Entries[k] === data[i]) { isSumOfTwoFound = true; break; } } if (isSumOfTwoFound) break; } if (!isSumOfTwoFound) { invalidNumber = data[i]; break; } } console.log(invalidNumber); <|file_sep|># Day #6 ## Part #1 This one was pretty straight forward: javascript const fs = require('fs'); const data = fs.readFileSync('data/day6.txt', 'utf-8').split('nn'); const getAnswers = (answers) => { return answers.split('n').reduce((acc, answer) => { return acc.concat([...new Set(answer.split('').filter(a => a !== 'n'))]); }, []); } const count = data.reduce((acc, answers) => { return acc + getAnswers(answers).length; }, 0); console.log(count); ## Part #2 The only difference here is that we need to filter out the answers that are not shared by all members of the group: javascript const fs = require('fs'); const data = fs.readFileSync('data/day6.txt', 'utf-8').split('nn'); const getAnswersSharedByAllMembersOfTheGroup = (answers) => { const answersByGroupMember = answers.split('n'); const firstGroupMemberAnswersSet = new Set(answersByGroupMember[0].split('').filter(a => a !== 'n')); return firstGroupMemberAnswersSet.filter(answer => answersByGroupMember.every(groupMemberAnswer => groupMemberAnswer.includes(answer) ) ).length; }; const countSharedAnswersByAllMembersOfGroups = data.reduce((acc, answers) => acc + getAnswersSharedByAllMembersOfTheGroup(answers), 0); console.log(countSharedAnswersByAllMembersOfGroups); <|file_sep|># Day #7 ## Part #1 We need to find out how many different colors of bags can contain at least one shiny gold bag. The best approach I could think of was to use recursion and traverse the tree of bag rules until we find all the bags that can contain the shiny gold bag. Here's how I solved it: javascript const fs = require('fs'); const data = fs.readFileSync('data/day7.txt', 'utf-8').split('n'); const bagRules = {}; data.forEach(rule => { const [bagName, bagContains] = rule.split(' bags contain '); bagRules[bagName] = bagContains.split(', ').reduce((acc, item) => { if (item === 'no other bags.') return acc; const [amount, color] = item.split(' '); acc[color] = parseInt(amount); return acc; }, {}); }); const containsBagRecursive = (bagColor) => { let containedBagsCount = 0; Object.entries(bagRules[bagColor]).forEach( ([color]) => { containedBagsCount += containsBagRecursive(color); if (bagRules[color]) { containedBagsCount++; } } ); return containedBagsCount; }; console.log(containsBagRecursive('shiny gold')); ## Part #2 For this one we need to find out how many individual bags are required inside a single shiny gold bag. Here's how I solved it: javascript const fs = require( 'fs' ); const data = fs.readFileSync( 'data/day7.txt', 'utf-8' ).split( 'n' ); const bagRules = {}; data.forEach( rule => { const [ bagName, bagContains ] = rule.split( ' bags contain ' ); bagRules[ bagName ] = bagContains.split( ', ' ).reduce( ( acc, item ) => {