Find next array index not from start AND if not found loop from beginning

Question

const arr = [{item:'a', value:false}, {item:'b', value:true}, {item:'c', value:true}, {item:'d', value:false}];
const startIndex = 2;

Given startIndex find next object in array with value=true. Iteration must be forward and, if not found, start from index 0. Should return arr[1];

I've just hit the wall and seems can't find an iteration method:/

let result = {};
for (let i = startIndex + 1; i < arr.length; i++) {
    if (!arr[i] && i < arr.length) {
        result = arr[i];
    }
}
console.log(result);

There is this function called indexOf it takes 2 parameters... look it up — Akxe
– Akxe, Commented Feb 15, 2021 at 22:03
Hint: You aren't checking the value property and not breaking if found — charlietfl
– charlietfl, Commented Feb 15, 2021 at 22:04
How do I start from the beginning when I reach end. Given for loop — Tadas V.
– Tadas V., Commented Feb 15, 2021 at 22:04
Just use 2 loops: the first one from startIndex + 1 forward, if not found start the second one from 0 to startIndex. Why do you want it to be 1 loop? — Yury Tarabanko
– Yury Tarabanko, Commented Feb 15, 2021 at 22:07

Nina Scholz · Accepted Answer · 2021-02-15 22:11:20Z

1

You could take a single loop and take startIndex as offset and shape the index by the remainder with the length.

const
    array = [{ item: 'a', value: false }, { item: 'b', value: true }, { item: 'c', value: true }, { item: 'd', value: false }],
    startIndex = 3;
    
let result;

for (let i = startIndex; i < array.length + startIndex; i++) {
    console.log(i, i % array.length);
    if (array[i % array.length].value) {
        result = array[i % array.length];
        break;
    }
}

console.log(result);

answered Feb 15, 2021 at 22:11

Nina Scholz

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1 Comment

Tadas V. Over a year ago

let i = startIndex + 1; produced desired results.

Samir Ribeiro · Accepted Answer · 2021-02-15 22:17:58Z

1

I'm not sure if I understood. But...

getNext = function(startIndex, arr){
    var i;
    for(i = startIndex; i < arr.length; i++){
        if(arr[i].value){
            return arr[i];
        }
    }

    if(startIndex != 0){
        return getNext(0, arr);
    }
    return null;
}

edited Feb 15, 2021 at 22:17

answered Feb 15, 2021 at 22:08

Samir Ribeiro

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5 Comments

Tadas V. Over a year ago

This works! But you have an error "i > arr.length" should be "i < arr.length". Thank you.

Samir Ribeiro Over a year ago

Yes! That's true XD Edited!

Tadas V. Over a year ago

Great solution. But I would like to avoid recursion if possible.

Samir Ribeiro Over a year ago

Only one level of recursivity. You will not have a stack overflow at all.

Tadas V. Over a year ago

I agree with you. But I prefer to keep recursion only for cases when there is a recursive data structure.

Majed Badawi · Accepted Answer · 2021-02-15 22:51:02Z

1

You can use .splice to create the two halves:

First, search for the element with value=true in the second half, if found, keep in mind to add the length of the first to the index
If not found in the second part, search in the first one

const getElementAtIndexFromStart = (arr=[], startIndex=0) => {
  // get index of element if exists starting from startIndex
  let index = arr.slice(startIndex+1).findIndex(({value}) => value===true);
  if(index!==-1) {
    // if found, add the start index to it to represent the position in in arr
    index = index + startIndex + 1;
  } else {
    // if not found in the second half, search in the first one
    index = arr.slice(0, startIndex+1).findIndex(({value}) => value===true);
  }
  return arr[index];
}

const arr = [{item:'a', value:false}, {item:'b', value:true}, {item:'c', value:true}, {item:'d', value:false}];

console.log('Start: 0 => ', getElementAtIndexFromStart(arr, 0));
console.log('Start: 1 => ', getElementAtIndexFromStart(arr, 1));
console.log('Start: 2 => ', getElementAtIndexFromStart(arr, 2));
console.log('Start: 3 => ', getElementAtIndexFromStart(arr, 3));
console.log('Start: 5 => ', getElementAtIndexFromStart(arr, 5));

edited Feb 15, 2021 at 22:51

answered Feb 15, 2021 at 22:16

Majed Badawi

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7 Comments

Tadas V. Over a year ago

Unless I am misreading something. It should return startIndex=0 -> arr[1]; si=1-> arr[2]; si=2 -> arr[0]; si=3-> arr[0]. Thank you for the effort.

Majed Badawi Over a year ago

@TadasV. so element at startIndex shouldn't be count? (si=1-> arr[2] instead of arr[1] - si=2 -> arr[0] instead of arr[2])

Tadas V. Over a year ago

Should be: f(arr,0)=arr[1], f(arr,1)=arr[2], f(arr,2)=arr[0], f(arr,3)=arr[0]

Majed Badawi Over a year ago

But arr[0] has value=false

Tadas V. Over a year ago

Should be: f(arr,0)=arr[1], f(arr,1)=arr[2], f(arr,2)=arr[1], f(arr,3)=arr[1]

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