Odd One Out

The tricky part about this challenge is that we need to figure out which specific element is different from the others. But there can be only one element that is a different length, the rest need to be the same length as one another. In this case, using sort doesn’t quite get us to a conscise solution.

Here, I made an array of the lengths of all the elements. Then I initiated an object called counts to hold all the lengths and how many times they occur. Then, I used Object.entries to make an array from the counts object, then compared the length of the array with all elements minus one to the array with element with only one occurence removed. If they were equal, this indicates that all elements are the same length except for one. and therefore true.

Instructions

Write a function that returns true if exactly one word in the array differs in length from the rest. Return false in all other cases. Examples

oddOneOut([“silly”, “mom”, “let”, “the”]) ➞ true

oddOneOut([“swanky”, “rhino”, “moment”]) ➞ true

oddOneOut([“the”, “them”, “theme”]) ➞ false

oddOneOut([“very”, “to”, “an”, “some”]) ➞ false

Notes

The length of the array will always have at least three or more words.

My solution

function oddOneOut(arr) {
  const counts = {}
  arr.map(x=> x.length).map(e => counts[e] = counts[e] ? counts[e] + 1 : 1 )
  return Object.entries(counts).filter(x=> x[1] !== 1).length === Object.entries(counts).length-1
}

oddOneOut(["silly", "mom", "let", "the"]) //  true
oddOneOut(["swanky", "rhino", "moment"]) // true
oddOneOut(["the", "them", "theme"]) // false
oddOneOut(["very", "to", "an", "some"]) // false

Other solutions

function oddOneOut(arr) {
	return arr
		.map(word => word.length)
		.filter((len,_,a) => a.indexOf(len) === a.lastIndexOf(len))
		.length === 1
}
// cefalexin

function oddOneOut(arr) {
	let arrLen = arr.map( e => e.length ? e.length : 0 );
	let result = 0;
	
	new Set(arrLen).forEach( s => {
		if (arrLen.indexOf(s) === arrLen.lastIndexOf(s)) {
			result++;
		}
	} )
	
	return result === 1;
}
// Ân