Returning index of odd one out in array

Question

I need to return the index of the odd word out in an array of words(strings).

For example...

['junper', 'jumper'] should return 0 
['Jumper'] should return 0 
['coast', 'coast', 'coal', 'coast'] should return 2
['sofa', 'sofa', 'sofa', 'soft'] should return 3

I already have a worked solution but I am aiming to make this cleaner.

I have been thinking about whats the best way to go about returning the index of the word which occurs only once...

This is my aforementioned working solution:

function findWrongWord(householditems) {
    if (householditems.length < 3) return 0;

    function findOccurencesLength (array, toEqual) {
        return array.filter(x => x === toEqual).length
    }

    let copyOfInitialArray = [...householditems];
    let [, mostOccuredItem = ''] = copyOfInitialArray.sort((a, b) => 
        findOccurencesLength(copyOfInitialArray, a) - 
        findOccurencesLength(copyOfInitialArray, b));

    return householditems.findIndex(x => x === 
        mostOccuredItem.split('').reverse().join(''));
}

hi maheer, not length, it can be any length but essentially returning the index of the word which occurs only once :) — JL91
– JL91, Commented May 14, 2019 at 16:03
@CL91 You mean the first element which is not repeated in array? — Maheer Ali
– Maheer Ali, Commented May 14, 2019 at 16:04

Shidersz · Accepted Answer · 2020-10-18 12:48:00Z

If you want to find one word that only appears once on the array, maybe you can use a combination of Array.findIndex() and Array.filter(), like this:

const tests = [
  ['junper', 'jumper'],
  ['Jumper'],
  ['coast', 'coast', 'coal', 'coast'],
  ['sofa', 'sofa', 'sofa', 'soft']
];

const getIndex = (arr) =>
{
    return arr.findIndex(word => arr.filter(x => x === word).length === 1);
}

tests.forEach(
    test => console.log(`${test} -> ${getIndex(test)}`)
);

.as-console {background-color:black !important; color:lime;}
.as-console-wrapper {max-height:100% !important; top:0;}

Another approach is to first use Array.reduce() to generate a Map() with the frequencies of each word, and then find the first word with frequency 1.

const tests = [
  ['junper', 'jumper'],
  ['Jumper'],
  ['coast', 'coast', 'coal', 'coast'],
  ['sofa', 'sofa', 'sofa', 'soft']
];

const getIndex = (arr) =>
{
    let freq = arr.reduce((acc, word) =>
    {        
        acc.set(word, acc.has(word) ? acc.get(word) + 1 : 1);
        return acc;
    }, new Map())

    return arr.findIndex(word => freq.get(word) === 1);
}

tests.forEach(
    test => console.log(`${test} -> ${getIndex(test)}`)
);

.as-console {background-color:black !important; color:lime;}
.as-console-wrapper {max-height:100% !important; top:0;}

Maheer Ali · Accepted Answer · 2019-05-14 16:17:24Z

You can compare indexOf and lastIndexOf to check if element is repeated.

const nonRepeated = arr => arr.findIndex(x => arr.indexOf(x) === arr.lastIndexOf(x))

console.log(nonRepeated(['junper', 'jumper']))
console.log(nonRepeated(['Jumper'] ))
console.log(nonRepeated(['coast', 'coast', 'coal', 'coast']))
console.log(nonRepeated(['sofa', 'sofa', 'sofa', 'soft']))

Another way is to use an object get count of all the elements and then return the first of those whose count is 1.

const nonRepeated = arr => {
  let countObj = arr.reduce((ac,a,i) => {
    if(!ac[a]) ac[a] = {count:0,index:i};
    ac[a].count++;
    return ac;
  },{})
  return Object.values(countObj).find(x => x.count === 1).index;

}
console.log(nonRepeated(['junper', 'jumper']))
console.log(nonRepeated(['Jumper'] ))
console.log(nonRepeated(['coast', 'coast', 'coal', 'coast']))
console.log(nonRepeated(['sofa', 'sofa', 'sofa', 'soft']))

Kamil Kiełczewski · Accepted Answer · 2019-05-14 16:14:03Z

0

try (where h={})

a.indexOf(a.map(e=>(h[e]=++h[e]|0,e)).find(w=>!h[w]));

let test = [
  ['junper', 'jumper'],
  ['Jumper'],
  ['coast', 'coast', 'coal', 'coast'],
  ['sofa', 'sofa', 'sofa', 'soft'],
];

function findWrongWord(a) {
  let h={};
  return a.indexOf(a.map(w=>(h[w]=++h[w]|0,w)).find(w=>!h[w]));
}

test.map(t=> console.log(JSON.stringify(t),findWrongWord(t)));

answered May 14, 2019 at 16:14

Kamil Kiełczewski

93.7k34 gold badges401 silver badges375 bronze badges

Comments

Keith · Accepted Answer · 2019-05-14 16:24:35Z

0

One idea is to use a map to maintain a list of index position and count, Next sort, and then unshift the first item.

Below is a working snippet.

const oddOne = arr => 
  arr.map((m,i,a) => (
    {i, c: a.filter(f => f === m).length})
  ).sort((a,b) => a.c - b.c).shift().i;
  
const tests = [
  ['junper', 'jumper'],
  ['Jumper'],
  ['coast', 'coast', 'coal', 'coast'],
  ['sofa', 'sofa', 'sofa', 'soft']
];

for (const test of tests) {
  console.log(JSON.stringify(test), "=", oddOne(test));
}

answered May 14, 2019 at 16:24

Keith

24.4k2 gold badges32 silver badges48 bronze badges

Comments

marc_s · Accepted Answer · 2019-08-24 14:23:22Z

First I'll make an assumption, you already know that there is only one "odd" word in the array. My javascript is not good so I'll put it down in words. Its not the most beautiful solution but is logical

Create an array of integers with the same length as the array you want to test. Each element in this array will represent the number of times the corresponding word in the array you are testing appears.
```
Array appearances = new Array(testArray.length);
// instantiate all element to zero
```

For each element in the test array, test if it is equal to the element at position 1, 2, 3,...last position, increment each by one where they are equal

for (int i = 0; i < testArray.length; i++) {
    for(testWord in testArray) {
        if(word == testWord) {
           appearances[i] = appearances[i]++;
        }

        // we don't really need to know the exact number of times it appears, just knowing it appears more than once is enough
        if(appearances[i] ==2) {
            break;
        }
    }
}

Now we find 1 in the appearances array I think array.find() is the way, it index will also be the index of the word not similar to others.

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