javascriptarrayssequencepushsubsequence

Find consecutive numbers when decreased by 1 in JavaScript

I am trying to find consecutive numbers from an array of numbers. This is my code:

let array = [4,3,5,4,3];
let new_arr = [];
for (let i=0; i<array.length; i++) {
    let arr = [array[i]];
    new_arr.push(arr);
    if (array[i] - array[i+1] == 1) {
        new_arr.push([array[i], array[i+1]]);
    }
}
console.log(new_arr);

How do I get more than 2 elements in an array? In my example, [5,4,3] should also be in the new_arr. But I am only able to get 2 element arrays. What am I missing?

Solution

  • You need a nested loop. One possibility is, for each index after i, in an inner loop, check if the following number is 1 less than the current. If so, create and push a new array - otherwise, break out and go to the outer loop's next iteration.

    let inputArray = [4, 3, 5, 4, 3];
const allSequences = [];
for (let i = 0; i < inputArray.length; i++) {
  let oneSequence = [inputArray[i]];
  allSequences.push(oneSequence);
  for (let j = i + 1; j < inputArray.length; j++) {
    if (inputArray[j] === oneSequence[oneSequence.length - 1] - 1) {
      oneSequence = [...oneSequence, inputArray[j]];
      allSequences.push(oneSequence);
    } else {
      break;
    }
  }
}
console.log(allSequences);