2019/6/24

Find The Parity Outlier

[2, 4, 0, 100, 4, 11, 2602, 36]
Should return: 11 (the only odd number)

[160, 3, 1719, 19, 11, 13, -21]
Should return: 160 (the only even number)

// mine
function findOutlier(int){
  const arr = int.map(v => v%2===0)
  let index = 0
  arr.map(v => {
    if(arr.indexOf(v) === arr.lastIndexOf(v)){
      index = arr.indexOf(v)   
    }
  })
  return int[index]
}

// the best
function findOutlier(int){
  var even = int.filter(a=>a%2==0);
  var odd = int.filter(a=>a%2!==0);
  return even.length==1? even[0] : odd[0];
}

Format a string of names like ‘Bart, Lisa & Maggie’

list([ {name: ‘Bart’}, {name: ‘Lisa’}, {name: ‘Maggie’} ])
// returns ‘Bart, Lisa & Maggie’

list([ {name: ‘Bart’}, {name: ‘Lisa’} ])
// returns ‘Bart & Lisa’

list([ {name: ‘Bart’} ])
// returns ‘Bart’

list([])
// returns ‘’

// mine
function list(names){
  const arr = names.map(v => v.name)
  let str = arr.join(', ')
  if(arr.length === 0 || arr.length === 1) return str
  let index = str.lastIndexOf(',')
  const newArr = str.split('')
  newArr.splice(index, 1, ' &')
  return newArr.join('')
}

// the best
function list(names){
  return names.reduce(function(prev, current, index, array){
    if (index === 0){
      return current.name;
    }
    else if (index === array.length - 1){
      return prev + ' & ' + current.name;
    } 
    else {
      return prev + ', ' + current.name;
    }
  }, '');
 }

Jaden Casing Strings

Not Jaden-Cased: “How can mirrors be real if our eyes aren’t real”
Jaden-Cased: “How Can Mirrors Be Real If Our Eyes Aren’t Real”

// mine
String.prototype.toJadenCase = function () {
  return this.replace(/(^|\s)\S/g, v => v.toUpperCase())
};

// the best
String.prototype.toJadenCase = function () { 
  return this.split(" ").map(function(word){
    return word.charAt(0).toUpperCase() + word.slice(1);
  }).join(" ");
}

2019/6/25

Sum of the first nth term of Series

SeriesSum(1) => 1 = “1.00”
SeriesSum(2) => 1 + 1/4 = “1.25”
SeriesSum(5) => 1 + 1/4 + 1/7 + 1/10 + 1/13 = “1.57”

// mine
function SeriesSum(n) {
  return [...Array(n).keys()].reduce((acc, cur) => acc + 1 / (1 + cur * 3), 0).toFixed(2)
}

// the best
function SeriesSum(n, s = 0) {
    return n ? SeriesSum(n - 1, s + 1 / (3 * n - 2)) : s.toFixed(2)
}

Exes and Ohs(统计相同数量的o和x)

// mine
function XO(str) {
    const newStr = str.toLowerCase();
    const obj = {};
    for(let i=0;i<newStr.length;i++){
      obj[newStr[i]] ? obj[newStr[i]]+=1 : obj[newStr[i]]=0
    } 
    return (obj.x===obj.o)
}

// the best
function XO(str) {
  let x = str.match(/x/gi);
  let o = str.match(/o/gi);
  return (x && x.length) === (o && o.length);
}

Equal Sides Of An Array

找到一个索引N，其中N左边的整数和等于右边的整数和，如果没有请返回 -1

Test.assertEquals(findEvenIndex([1,2,3,4,3,2,1]),3, "The array was: [1,2,3,4,3,2,1] \n");
Test.assertEquals(findEvenIndex([1,100,50,-51,1,1]),1, "The array was: [1,100,50,-51,1,1] \n");
Test.assertEquals(findEvenIndex([1,2,3,4,5,6]),-1, "The array was: [1,2,3,4,5,6] \n");
Test.assertEquals(findEvenIndex([20,10,30,10,10,15,35]),3, "The array was: [20,10,30,10,10,15,35] \n");

// mine
function findEvenIndex(arr)
{
  const bool = arr.map((v,i) => {
    const left = arr.slice(0,i)
    const right = arr.slice(i+1,arr.length)
    return (left.length === 0 ? 0 : left.reduce((a,b)=>a+b)) === (right.length === 0 ? 0 : right.reduce((a,b)=>a+b))
  })
  return bool.indexOf(true)
}

// the best
function findEvenIndex(arr)
{
  for(var i=1; i<arr.length-1; i++) {
    if(arr.slice(0, i).reduce((a, b) =>  a+b) === arr.slice(i+1).reduce((a, b) =>  a+b)) {
      return i;
    }
  }
  return -1;
}

Does my number look big in this?

自恋数是一个数字，它是自己数字的总和，每个数字都增加到给定基数中的数字位数。
1^3 + 5^3 + 3^3 = 1 + 125 + 27 = 153
1^4 + 6^4 + 3^4 + 4^4 = 1 + 1296 + 81 + 256 = 1634

// mine
function narcissistic(value) {
  return value.toString().split('').reduce((acc, cur, index, arr) => acc + Math.pow(cur, arr.length), 0) === value
}

// the best
function narcissistic( value ) {
  return ('' + value).split('').reduce(function(p, c){
    return p + Math.pow(c, ('' + value).length)
    }, 0) == value;
}

2019/6/26

IQ Test

iqTest(“2 4 7 8 10”) => 3 // Third number is odd, while the rest of the numbers are even
iqTest(“1 2 1 1”) => 2 // Second number is even, while the rest of the numbers are odd

// mine
function iqTest(numbers){
  const arr = numbers.split(' ').map(v => v%2 === 0)
  for(let i=0;i<arr.length;i++){
    if(arr.indexOf(arr[i]) === arr.lastIndexOf(arr[i])){
      return i+1
    }  
  }
}

// the best
function iqTest(numbers){
  numbers = numbers.split(" ").map(function(el){return parseInt(el)});
  
  var odd = numbers.filter(function(el){ return el % 2 === 1});
  var even = numbers.filter(function(el){ return el % 2 === 0});
  
  return odd.length < even.length ? (numbers.indexOf(odd[0]) + 1) : (numbers.indexOf(even[0]) + 1);
}

2019/6/27

Find the missing letter

[‘a’,’b’,’c’,’d’,’f’] -> ‘e’
[‘O’,’Q’,’R’,’S’] -> ‘P’
(Use the English alphabet with 26 letters!)

// mine
function findMissingLetter(array)
{ 
  const arr = 'abcdefghijklmnopqrstuvwxyz'
  const ARR = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
  for (let i=0;i<array.length;i++){
    if(arr.indexOf(array[i]) >= 0) {
      if(arr.indexOf(array[i]) !== arr.indexOf(array[i+1])-1){
        return arr[arr.indexOf(array[i])+1]
      }
    } else {
      if(ARR.indexOf(array[i]) !== ARR.indexOf(array[i+1])-1){
        return ARR[ARR.indexOf(array[i])+1]
      }
    }
    
  }
}

// the best
function findMissingLetter(array)
{
   var i=array[0].charCodeAt();
   array.map(x=>  x.charCodeAt()==i?i++:i);
   return String.fromCharCode(i);
}

Roman Numerals Encoder

solution(1000); // should return ‘M’
把数字转化为罗马数字
1
2
3
4
5
6
7
8
9
10
// help
Symbol    Value
I          1
V          5
X          10
L          50
C          100
D          500
M          1,000
```   
// the best
function solution(number){
// convert the number to a roman numeral
var roman = {M:1000,CM:900, D:500,CD:400,C:100,XC:90,L:50,XL:40,X:10,IX:9,V:5,IV:4,I:1 }

var ans = ‘’;
while(number>0){
for(a in roman){
if(roman[a]<=number){ ans += a; number-=roman[a]; break;}

}
return ans;
}

- # Stop gninnipS My sdroW!
> spinWords( "Hey fellow warriors" ) => returns "Hey wollef sroirraw"   
> spinWords( "This is a test") => returns "This is a test"  
> spinWords( "This is another test" )=> returns "This is rehtona test"
> 翻转长度大于等于5的单词

// mine
function spinWords(str){
return str.split(‘ ‘).map(v => v.length>=5 ? v.split(‘’).reverse().join(‘’) : v ).join(‘ ‘)
}

```
// the best
function spinWords(string){
  return string.replace(/\w{5,}/g, function(w) { return w.split('').reverse().join('') })
}

2019/6/28

Persistent Bugger

persistence(39) === 3 // because 3*9 = 27, 2*7 = 14, 1*4=4
                      // and 4 has only one digit

persistence(999) === 4 // because 9*9*9 = 729, 7*2*9 = 126,
                       // 1*2*6 = 12, and finally 1*2 = 2

persistence(4) === 0 // because 4 is already a one-digit number

// mine 
function persistence(num) {
  let index = 0
  let result = num
  while (result.toString().length !== 1) {
    console.log(result)
    result = result.toString().split('').reduce((acc, cur) => acc*cur, 1)
    index += 1
  }
  return index
}

// the best
const persistence = num => {
  return `${num}`.length > 1 
    ? 1 + persistence(`${num}`.split('').reduce((a, b) => a * +b)) 
    : 0;
}

Is a number prime

is_prime(1) /* false /
is_prime(2) / true /
is_prime(-1) / false */
是否为质数

// mine
function isPrime(num) {
  let result = 0
  for (let i=1;i<=num;i++){
    num % i === 0 ? result += 1 : null
  }
  return result === 2
}
// function isPrime(num) {
//   return [...Array(num).keys()].reduce((acc, cur, index) => num%(index+1)===0?acc+=1:acc,0) === 2
// } // 此方法会出现数组最大长度BUG

// the best
function isPrime(num) {
  for (var i = 2; i < num; i++) if (num % i == 0) return false;
  return num >= 2; 
}

Where my anagrams at?

'abba' & 'baab' == true

'abba' & 'bbaa' == true

'abba' & 'abbba' == false

'abba' & 'abca' == false

>anagrams('abba', ['aabb', 'abcd', 'bbaa', 'dada']) => ['aabb', 'bbaa']

>anagrams('racer', ['crazer', 'carer', 'racar', 'caers', 'racer']) => ['carer', 'racer']

>anagrams('laser', ['lazing', 'lazy',  'lacer']) => []

// mine
function anagrams(word, words) {
  return words.filter(v => v.split('').sort().join('') === word.split('').sort().join(''))
}

// the best
String.prototype.sort = function() {
  return this.split("").sort().join("");
};

function anagrams(word, words) {
  return words.filter(function(x) {
      return x.sort() === word.sort();
  });
}

2019/7/9

Permutations

您必须创建输入字符串的所有排列并删除重复项（如果存在）。
1
2
3
permutations('a'); // ['a']
permutations('ab'); // ['ab', 'ba']
permutations('aabb'); // ['aabb', 'abab', 'abba', 'baab', 'baba', 'bbaa']

// mine  
function permutations(string) {
  let array = string.split('')
  let arr = []
  for (let i = 0; i < 10000; i++) {
    let str = array.sort(() => {
      return Math.random() - 0.5
    }).join('')
    arr.push(str)
  }
  return Array.from(new Set(arr))
}

// the best
function permutations(str) {
 return (str.length <= 1) ? [str] :
         Array.from(new Set(
           str.split('')
              .map((char, i) => permutations(str.substr(0, i) + str.substr(i + 1)).map(p => char + p))
              .reduce((r, x) => r.concat(x), [])
         ));
}

2019/7/14

First non-repeating character

‘a’ => ‘a’
‘stress’ => ‘t’
‘Moonmen’ => ‘e’

// mine
function firstNonRepeatingLetter(s) {
  const result = []
  s.toLowerCase().split('').filter((v,i,arr) => {
    if( arr.indexOf(v) === arr.lastIndexOf(v) ) { 
      console.log(v, i, s.split('')[i])
      result.push(s.split('')[i])
    }
  })
  return result[0] || ''
}

// the best
function firstNonRepeatingLetter(s) {
  var t=s.toLowerCase();
  for (var x=0;x<t.length;x++)
    if(t.indexOf(t[x]) === t.lastIndexOf(t[x]))
      return s[x];
  return "";
}

codewars 闯关纪录

codewars 闯关纪录

2019/6/24

Find The Parity Outlier

Format a string of names like ‘Bart, Lisa & Maggie’

Jaden Casing Strings

2019/6/25

Sum of the first nth term of Series

Exes and Ohs(统计相同数量的o和x)

Equal Sides Of An Array

Does my number look big in this?

2019/6/26

IQ Test

2019/6/27

Find the missing letter

Roman Numerals Encoder

2019/6/28

Persistent Bugger

Is a number prime

Where my anagrams at?

2019/7/9

Permutations

2019/7/14

First non-repeating character