Javascript group the numbers from an array with series of consecutive numbers

Question

Given a sorted array of ints for example

a = [0,1,2,5,6,9];

I would like to identify the ranges like

[
    [0,1,2],
    [5,6],
    [9]
]

So far I have tried a double/triple loop but it nests to really nasty code. Maybe this problem can be solved using recursion or other smart tricks?

Additional example:

input

b = [0,1,5,6,7,9];

output

[
    [0,1],
    [5,6,7],
    [9]
]

so, the first array should contain 3, the next 2, and the last one should have only 1 element. right? — Sal-laS, Dec 20 '17 at 13:12
For this example yes, but generally no. It should segment out int ranges — ajthinking, Dec 20 '17 at 13:13
@Salman no, it creates an array if the numbers are in sequance, aka 1,2,3 - 7,8,9 — George, Dec 20 '17 at 13:14
So you want to split a sorted array in other arrays of consecutive numbers? — Federico klez Culloca, Dec 20 '17 at 13:14
@Anders What is the logical base for the output? you are randomly changing the sequence. — Nitin Dhomse, Dec 20 '17 at 13:16
@NitinDhomse - read the question properly - there is nothing random happening at all — Jaromanda X, Dec 20 '17 at 13:19
@JaromandaX ohh got it, it is the consecutive numbers grouping. — Nitin Dhomse, Dec 20 '17 at 13:20

score 15 · Accepted Answer · edited Aug 20 '19 at 18:47

15

Iterate with Array#reduce, and whenever the last number is not equal to the new number - 1, add another sub array. Add the current number to the last sub array:

const a = [0,1,2,5,6,9];

const result = a.reduce((r, n) => {
  const lastSubArray = r[r.length - 1];
  
  if(!lastSubArray || lastSubArray[lastSubArray.length - 1] !== n - 1) {
    r.push([]);
  } 
  
  r[r.length - 1].push(n);
  
  return r;  
}, []);

console.log(result);

edited Aug 20 '19 at 18:47

Peter Brown

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answered Dec 20 '17 at 13:15

Ori Drori

183,571
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1

I love you, thanks for the good solution! – Charis Moutafidis Feb 09 '21 at 11:35
3

Happily married with children, but you're welcome :) – Ori Drori Feb 09 '21 at 12:29

Keith · Answer 2 · 2017-12-20T13:58:45.770

Array.reduce like @OriDrori has done works well.

Another idea is just using a simple for loop, and slice.

function groupArray(a) {
  const ret = [];
  if (!a.length) return ret;
  let ixf = 0;
  for (let ixc = 1; ixc < a.length; ixc += 1) {
    if (a[ixc] !== a[ixc-1] + 1) {
      ret.push(a.slice(ixf, ixc));
      ixf = ixc;  
    }
  }
  ret.push(a.slice(ixf, a.length));
  return ret;
}

console.log(JSON.stringify(groupArray([0,1,2,5,6,9])));

Ben Aston · Answer 3 · 2017-12-20T14:09:29.833

1

This is what I came up with:

const a = [0,1,2,5,6,9];

const solution = (arr, expected = 0, group = []) => 
    arr.reduce((acc, c, i) => 
        ((c === expected 
            ? (group.push(c), expected++)
            : (acc.push(group), group = [c], expected=++c)), 
          (i === arr.length-1 && acc.push(group)), 
          acc), []);

console.log(solution(a));

edited Dec 20 '17 at 14:09

answered Dec 20 '17 at 14:04

Ben Aston

53,718
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score 1 · Answer 4 · answered Apr 25 '19 at 18:32

You can use a normal for loop and check the difference between current and previous number

var a = [0, 1, 2, 5, 6, 7, 9];
// this variable will contain arrays
let finalArray = [];
// Create a recursive function
function checkPrevNextNumRec(array) {
  let tempArr = [];
  // if the array contaon only 1 element then push it in finalArray and
  // return it
  if (array.length === 1) {
    finalArray.push(array);
    return
  }
  // otherside check the difference between current & previous number
  for (let i = 1; i < array.length; i++) {
    if (array[i] - array[i - 1] === 1) {
      // if current & previous number is 1,0 respectively 
      // then 0 will be pushed
      tempArr.push(array[i - 1]);
    } else {
      // if current & previous number is 5,2 respectively 
      // then 2 will be pushed
      tempArr.push(array[i - 1])
      // create a an array and recall the same function
      // example from [0, 1, 2, 5, 6, 9] after removing 0,1,2 it 
      // will create a new array [5,6,9]
      let newArr = array.splice(i);
      finalArray.push(tempArr);
      checkPrevNextNumRec(newArr)
    }
    // for last element if it is not consecutive of
    // previous number 
    if (i === array.length - 1) {
      tempArr.push(array[i]);
      finalArray.push(tempArr)
    }
  }

}
checkPrevNextNumRec(a)
console.log(finalArray)

Javascript group the numbers from an array with series of consecutive numbers

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