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javascriptarrays
ajthinking
ajthinking

Reputation: 4568

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

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]
]

Upvotes: 6

Views: 8648

Answers (4)

Ori Drori
Ori Drori

Reputation: 191986

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);

Upvotes: 15

brk
brk

Reputation: 50291

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)

Upvotes: 1

Ben Aston
Ben Aston

Reputation: 55739

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));

Upvotes: 1

Keith
Keith

Reputation: 24191

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])));

Upvotes: 4

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