codewars.js

//////////////convert sum to binarys
const addBinary = function(a,b) {
return (a+b).toString(2);
}

///////////////find Square number
const isSquare = function(n){
	//math.pow "1/2" oblicz pierwiastek kwadratowy, "2" podniosłoby do potęgi 2
  return Number.isInteger(Math.pow(n, 1/2));
};
console.log("isSquare(3) - " + isSquare(3)); // => false
console.log("isSquare(25) - " + isSquare(25)); // => true

//////////////Capitalize string
String.prototype.toJadenCase = function () {
  let arr = this.split(" ");
  let arrUpper = arr.map(function(x) {
  	//slice dodaje resztę słowa do dużej litery
    return x.charAt(0).toUpperCase() + x.slice(1);
  });
  let str = "";
  for (let i = 0; i < arrUpper.length; i++) {
    i === 0 ? str = str + arrUpper[i] : str = str + " " + arrUpper[i];
  }
  return str;
};
console.log("Jaden string: " + "How can mirrors be real if our eyes aren't real".toJadenCase())

///////////check, if it is possible to construct triangle from given number
const isTriangle = function(a,b,c) {
  if (a+b>c&&a+c>b&&b+c>a) {
    return true;
  }else{return false;}
};
console.log("isTriangle 9,2,4 - " + isTriangle(9,2,4));
console.log("isTriangle 9,6,4 - " + isTriangle(9,6,4));

///////////Help with IQ test - find out which one of the given numbers differs from the others starting from one
const iqTest = function(numbers){
  let arr = numbers.split(" ");
  let odd = arr.filter(x => x % 2 !== 0);
  let even = arr.filter(x => x % 2 == 0);
  return even.length > odd.length ? arr.indexOf(odd[0]) + 1 : arr.indexOf(even[0]) + 1;
};

//longer solution:
const iqTest2 = function(numbers){
  let odd = [];
  let even = [];
  let arr = numbers.split(" ");
  for (let i = 0; i<arr.length; i++){
    arr[i] % 2 === 0 ? even.push(arr[i]) : odd.push(arr[i]);
  }
  return even.length > odd.length ? arr.findIndex(x=>x===odd[0]) + 1 : arr.findIndex(x=>x===even[0]) + 1;
};

console.log('iqTest("2 4 7 8 10") - ' + iqTest("2 4 7 8 10"));// => 3 // Third number is odd, while the rest of the numbers are even
console.log('iqTest("1 2 1 1") - ' + iqTest("1 2 1 1"));// => 2 // Second number is even, while the rest of the numbers are odd


/////////similar to IQtest, but return integer
const findOutlier = function(integers){
  let odd = integers.filter(x => x % 2 !== 0);
  let even = integers.filter(x => x % 2 == 0);
  return even.length > odd.length ? odd[0] : even[0];
};

/////////Which are in??/////Given two arrays of strings a1 and a2 return a sorted array r in lexicographical order of the strings of a1 which are substrings of strings of a2.
function inArray(array1,array2){
  let r = [];
  for (let i = 0; i < array1.length; i++) {
      for (let j = 0; j < array2.length; j++) {
      	//String.prototype.includes() compare if given string contains substring, String.prototype.match() also works
        if (array2[j].includes(array1[i]) == true) {
          r.push(array1[i]);
          break;
        }
    }
  }
  return r.sort();
}
////////shorter solution
function inArray2(array1,array2){
  return array1
    .filter(a1 => array2.find(a2 => a2.includes(a1)))
    .sort()
}

console.log(inArray2(["ups", "mad", "ba"], ["upsy", "ball", "madness"]));
///////// find number which is equal to its digpow like 89 = 8^1 + 9^2
function sumDigPow(a, b) {
return Array(b-a).fill() //fill array with given range using fill() and map()
  .map((_, i) => i + a)
  .filter(function (x) {
   	return x > 9 ? 
   	x.toString()
   	.split("")
   	.reduce((a, b, i) => +a + Math.pow(b, i+1)) == x //+a convert it into number 
   	: Math.pow(x, 1) == x;
  });
}

console.log(sumDigPow(1, 200));
//////////My favorite answer
// function sumDigPow2(a, b) {
//   var ans = [];
//   while(a <= b){
//     if(a.toString().split('').reduce((x,y,i)=>x + +y ** (i + 1),0) == a)
//       ans.push(a);
//     a++;
//   }
//   return ans;
// }