check if elements in array are consecutive --- javascript

Question

I have an array as

arr = [1,2,3,4,6,7,8,9]

Now I want to check if the values in the array are consecutive.

Being more specific, I want this

First Check gives first and second element are consecutive and the next element is not consecutive then the algo must return the first element from where the consecutive number started

Like

First Check will give 1
Second Check will give 6
and so on...

Please help Thanks in advance

Is this an assignment? It seems like it. If so, show what you've tried so far and where you are stuck. If not, could you explain the context that requires you to do this? — The Archetypal Paul
– The Archetypal Paul, Commented Nov 26, 2010 at 12:05
I want an algo which will return me the values as I described in the end — Rohit
– Rohit, Commented Nov 26, 2010 at 12:06
@Rohit - What about a standalone number? for example should [1,2,3,4,6,8,9] return 1, 6, 8, or just 1, 8? — Nick Craver
– Nick Craver, Commented Nov 26, 2010 at 12:08
if the array is [1,2,3,4,6,8,9] then it should return 1,6,8 i.e it should return the number from where we started getting the consecutive number — Rohit
– Rohit, Commented Nov 26, 2010 at 12:09

Denis Ivanov · Accepted Answer · 2016-08-24 18:52:27Z

5

/**
 * Given an array of number, group algebraic sequences with d=1
 * [1,2,5,4,8,11,14,13,12] => [[1,2],[4,5],[8],[11,12,13,14]]
 */
import {reduce, last} from 'lodash/fp';

export const groupSequences = (array) => (
  reduce((result, value, index, collection) => {
    if (value - collection[index - 1] === 1) {
      const group = last(result);
      group.push(value);
    } else {
      result.push([value]);
    }
    return result;
  }, [])(array)
);

answered Aug 24, 2016 at 18:52

Denis Ivanov

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Comments

user2983555 · Accepted Answer · 2019-09-18 08:14:42Z

4

 /**
 * Given an array of number, group algebraic sequences with d=1
 * [1,2,3,4,5,6] => true
 * [1,2,4,5,6] => false
 */
 const differenceAry = arr.slice(1).map(function(n, i) { return n - arr[i]; })
 const isDifference= differenceAry.every(value => value == 1)
 console.log(isDifference);

answered Sep 18, 2019 at 8:14

user2983555

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Comments

Marcel Korpel · Accepted Answer · 2010-11-26 12:50:06Z

3

One sidenote is that you want to call it multiple times, so each call should know which array it's working on and what the previous offset in that array was. One thing you can do is to extend the native Array object. [Demo]

Array.prototype.nextCons = (function () {
  var offset = 0; // remember the last offset
  return function () {
    var start = offset, len = this.length;
    for (var i = start + 1; i < len; i++) {
      if (this[i] !== this[i-1] + 1) {
        break;
      }
    }
    offset = i;
    return this[start];
  };
})();

Usage

var arr =  [1,2,3,4,6,8,9];
arr.nextCons(); // 1
arr.nextCons(); // 6
arr.nextCons(); // 8

edited Nov 26, 2010 at 12:50

Marcel Korpel

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answered Nov 26, 2010 at 12:41

gblazex

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5 Comments

The Archetypal Paul Over a year ago

Why not just hand in the last index returned as the place to start this one at?

The Archetypal Paul Over a year ago

I suppose so. In your case,I think it returns increasing values greater than the length of the array if called more times than there are non-consequitive elements, and there's no way of resetting it to the beginning of the array...

gblazex Over a year ago

@Paul - It depends on what the OP expects, but the code is fairly simple to modify.

The Archetypal Paul Over a year ago

How would you modify it to reset it? (just curious)

gblazex Over a year ago

... return function (reset) { if (reset) offset = 0; ... } then both arr.nextCons(true); and arr.nextCons("reset"); will do.

Enrique Briones Arcos · Accepted Answer · 2022-03-18 15:17:22Z

2

Check if all numbers in array are consecutives:

Updated March 2022

 const allConsecutives = (arr) =>{ 
    if(arr.some(n=> typeof n !== "number" || Number.isNaN(n))) return false;
     return arr.every((num, i)=>  arr[i+1]-num === 1 || arr[i+1] === undefined)
  }

edited Mar 18, 2022 at 15:17

answered Oct 12, 2020 at 8:21

Enrique Briones Arcos

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2 Comments

Carl Sala Over a year ago

this returns true when arr = [8,0] also [0,0] so doesn't work perfectly.

Enrique Briones Arcos Over a year ago

You're right @CarlSala I just corrected it, Thank you.

Ali Tarhini · Accepted Answer · 2010-11-26 12:49:17Z

0

pseudo code:

int count = 0 
for i = 0 to array.length - 2 
    if  {array[i + 1] - array[i] = 1 then 
        count+=1 
         return i
    else count=0}

answered Nov 26, 2010 at 12:49

Ali Tarhini

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Comments

user2983555 · Accepted Answer · 2019-09-26 12:21:41Z

0

const array1 = [1,2,3];
const sum = array1.reduce((accumulator, currentValue) =>{
  return accumulator + currentValue;
});
const max = Math.max(...array1);
  maximum = max
  if(sum == maximum * (maximum+1) /2) {
       console.log(true);
  } else {
       console.log(false);
  }

answered Sep 26, 2019 at 12:21

user2983555

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