Business
Jobs
  • About Us
  • Solutions
    • Job Postings
      Post your job and receive qualified candidates in 48h.
    • Candidate Assessments
      500+ technical and psychological tests, plus anti-fraud.
    • Headhunting
      Tailor-made executive search from start to finish.
    • Payroll + EOR
      Payroll dispersal and EOR across 15+ LATAM countries.
  • Pricing
  • Jobs

0

130
Views
Max consecutive items of numbers in array with +1 difference

I am trying to return the max number of consecutive numbers or same numbers which difference is no more than + 1.

Example*

const array = [1,2,3,5,5,6,6,6,6,7,8]
solution= 556666

const array2 = [2,2,3,4,4,5,5]
solution= 4455

I am a new coder and seems like there should be a simpler way to solve this but I am stuck at this point.

function getMaximumNumberItems(arr) {
    let initial = {'0': 0, '1':0, '2':0, '3':0, '4':0, '5':0, '6':0, '7':0, '8':0, '9':0 }
    let counts = {}
    arr.forEach((element) => {
       counts[element] = (counts[element] || 0) + 1
    })
    const numbers = {...initial,...counts}
    const arrValues = Object.values(numbers)
    let sum = []
    for (let i = 0; i < arrValues.length; i++) {
      arrValues[i] === arrValues[arrValues.length - 1] ? null : sum.push(arrValues[i] + arrValues[i + 1])
    }
    console.log(sum)
    let maxIndex = sum.indexOf(Math.max(...sum))
  }

What I have done is set a count for each number which is numbers, then I add each element with next element to see the max number of elements that are consecutive and add them to an array which is sum. The index of this maximum number should also be the index of the first element that should be returned from counts.

Mi idea was to return key from object and access how many times this number has appeared and add it some way and then use next element of numbers the same way to get the solution.

Obviously I see that this is the worst way of doing it, so would appreciate any help.

Thanks!

about 4 years ago · Juan Pablo Isaza
1 answers
Answer question

0

You can do it in O(n) time. Take number, match all numbers after that, which difference with it is 1 or smaller. If you find number, which difference is bigger, than 1, then you need to remember the current result and repeat the previous steps, fixing the number of the element from which the best sequence begins and its length.

function maxSubsequence(array) {
    let ind = 0;
    let bestInd = 0;
    let cnt = 1;
    let maxCnt = 0;

    for (let i = 1; i < array.length; i++) {
        if (Math.abs(array[ind] - array[i]) <= 1) {
            cnt++;
        } else {
            if(cnt > maxCnt) {
                bestInd = ind;
                maxCnt = cnt;
            }
            cnt = 1;
            ind = i;
        }
    }

    if (cnt > maxCnt) {
        bestInd = ind;
        maxCnt = cnt;
    }

    return array.slice(bestInd, bestInd + maxCnt);
}

Output:

maxSubsequence(array)
[5, 5, 6, 6, 6, 6]

maxSubsequence(array2)
[4, 4, 5, 5]
about 4 years ago · Juan Pablo Isaza Report
Answer question
Find remote jobs

Discover the new way to find a job!

Top jobs
Top job categories
Business
Post vacancy Pricing Sales
Legal
Terms and conditions Privacy policy
© 2026 PeakU Inc. All Rights Reserved.
Andres GPT
Show me some job opportunities
There's an error!