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problems/countNumbers.js
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| function countNumbers(arr) { | ||
| let count = {} | ||
| for (let i = 0; i < arr.length; i++) { |
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Did it! seems like a longer code. Thank you
problems/isOdd.js
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| function isOdd() { | ||
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| function isOdd(n) { | ||
| if (n % 2 === 1) { |
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can just return the condition n % 2 === 1
problems/numberOfDigits.js
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| function numberOfDigits(n) { | ||
| let num = n | ||
| return Math.floor(num.toString().length) |
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No need for Math.floor here. A string will always be an integer length and never a decimal
| function removeNumbersAtOddIndices() { | ||
| function removeNumbersAtOddIndices(arr) { | ||
| let output = []; | ||
| for (i = 0; i < arr.length; i += 2) { |
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Try using filter here. But I do love your i += 2
| let arr = str.split("") | ||
| let out = []; | ||
| for (let i = 0; i < arr.length; i++) { | ||
| if (arr[i] !== "o" && arr[i] !== "e" && arr[i] !== "u" && arr[i] !== "i" && arr[i] !== "A" && arr[i] !== "E" && arr[i] !== "I" && arr[i] !== "U" && arr[i] !== "O" && arr[i] !== "a") { |
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check out the includes method "aeiouAEIOU".includes(arr[i])
| function removeVowels(str) { | ||
| let arr = str.split("") | ||
| let out = []; | ||
| for (let i = 0; i < arr.length; i++) { |
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try using a for of loop when going through strings and not needing the index.
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