Check consecutive numbers in array javascript

Question

There is a array like this on my hand;

var array = [{value:"0"},{value:"1"},{value:"2"},{value:"3"},{value:"4"}];

I need check, this numbers is going consecutive?

[{value:"0"},{value:"1"},{value:"2"},{value:"3"},{value:"4"}];
TRUE (0,1,2,3,4)

[{value:"0"},{value:"1"},{value:"2"},{value:"3"},{value:"5"}];
FALSE (0,1,2,3,5)

Answer 1

You can use reduce with the initial value as the first element of the array

 const checkIsConsecutive = (array) => Boolean(array.reduce((res, cur) => (res? (Number(res.value) + 1 === Number(cur.value)? cur: false): false))); console.log(checkIsConsecutive([{ value: '0' }, { value: '1' }, { value: '2' }, { value: '3' }, { value: '4' }])); console.log(checkIsConsecutive([{ value: '0' }, { value: '1' }, { value: '2' }, { value: '3' }, { value: '5' }]));

Answer 2

You could check if every item is one more than the previous item. This will stop looping once a non-consecutive values is found

array.every(({ value }, i) => i === 0 || +value === +array[i-1].value + 1)

Here's a snippet:

 function isConsecutive(array) { return array.every(({ value }, i) => i === 0 || +value === +array[i-1].value + 1) } console.log( isConsecutive([{value:"0"},{value:"1"},{value:"2"},{value:"3"},{value:"4"}]), isConsecutive([{value:"0"},{value:"1"},{value:"3"}]) )

Answer 3

The other answer are really well made, and look very pleasing. However as @Yevgen Gorbunkov mentioned, they do not take empty arrays, single item arrays, as well as desc order.

I am quite new to JS so the answer is not optimal and can most likely be improved. But I figured giving it a shot wouldn't hurt.

Attempt 1: jsfiddle

 //var arrr = [{value:"0"}]; var arrr = [{value:"8"},{value:"7"},{value:"6"},{value:"5"},{value:"4"},{value:"3"},{value:"2"},{value:"1"},{value:"0"}]; //var arrr = [{value:"8"},{value:"3"},{value:"6"},{value:"5"},{value:"4"},{value:"3"},{value:"2"},{value:"1"},{value:"0"}]; //var arrr = []; //var arrr = [{value:"0"},{value:"1"},{value:"2"},{value:"3"},{value:"4"},{value:"5"},{value:"6"},{value:"7"},{value:"8"}]; var arr = arrr.map(function (obj) { return Number(obj.value); }); isConsecutive(arr); function isConsecutive(arr) { var consecutive; let previousAsc = arr[0]; let previousDesc = arr[0]; let firstAsc; let firstDesc; for (let i = 1; i < arr.length; i++) { if ((previousDesc - 1);== arr[i]) { if ((previousAsc + 1);== arr[i]) { firstAsc = arr[i]; consecutive = false; break; } else { consecutive = true; previousDesc --; previousAsc ++; } } else { consecutive = true; previousDesc --. previousAsc ++; } } console.log(consecutive) }

Attempt 2: jsfiddle

 //var arrr = [{value:"0"}]; var arrr = [{value:"8"},{value:"7"},{value:"6"},{value:"5"},{value:"4"},{value:"3"},{value:"2"},{value:"1"},{value:"0"}]; //var arrr = [{value:"8"},{value:"3"},{value:"6"},{value:"5"},{value:"4"},{value:"3"},{value:"2"},{value:"1"},{value:"0"}]; //var arrr = []; //var arrr = [{value:"0"},{value:"1"},{value:"2"},{value:"3"},{value:"4"},{value:"5"},{value:"6"},{value:"7"},{value:"8"}]; var arr = arrr.map(function (obj) { return Number(obj.value); }); var arrSort = arrr.map(function (obj) { return Number(obj.value); }); if(arr && arr.length >= 2){ if(JSON.stringify(arrSort.sort()) === JSON.stringify(arr) || JSON.stringify(arrSort.sort().reverse()) === JSON.stringify(arr)){ /* console.log(arrSort.sort().reverse()); console.log(arr); */ console.log("Array is consecutive") } else { /* console.log(arrSort.sort()); console.log(arr); */ console.log("Array is not consecutive") } } else { console.log("array does not exist or does not have enough values to start") }

Answer 4

Here is an auxiliary function that computes the length of the most long subarray of an array.

const maxConsecutive = (arr) => {
    const result = arr.reduce((a, b) => {
        const { count, maxCount, cons } = a;
        if (cons.length === 0)
            return { count: 1, maxCount: Math.max(1, maxCount), cons: cons.concat([b]) };
        else {
            const last = cons.at(-1);
            const isCons = (b - last) === 1;
            return {
                count: isCons ? count + 1 : 0,
                maxCount: isCons ? Math.max(count + 1, maxCount) : maxCount,
                cons: isCons ? cons.concat([b]) : []
            };
        }
    }, { count: 0, maxCount: 0, cons: [] });
    return result.maxCount;
}

Sample outputs:

maxConsecutive([])
> 0
maxConsecutive([0,1,2,3,4])
> 5
maxConsecutive([0,1,2,3,5])
> 4

So now, it's pretty obvious how it can be helpful to check whether the array is consecutive or not: it suffices to maxConsecutive(arr) === arr.length .

Answer 5

Here is another way to to this.

const array = [{value:"0"},{value:"1"},{value:"2"},{value:"3"},{value:"4"}];

function isConsecutive (array) {
    const pred = array.map (i => parseInt(i.value))
    const expect = range(0, 4) // maybe use lodash.
    return pred === expect 
}

I think you do know what numbers are consecutive, so you can create a consecutive array to match your original input.