Check if an array is descending, ascending or not sorted?

Question

I\'m just beginning with programming using javascript and I need to practice some questions to get EXP with the logic of code build. I got this question for homework but I c

Answer 1

This is a question that requires some sort of loop, with several if statements because there are several cases you need to tackle:

1. Array is empty or has only one element.
2. All items in the array are equal
3. Array is ascending - delta between 2 elements > 0, but some deltas may be 0
4. Array is descending - delta between 2 elements < 0, but some deltas may be 0
5. Not sorted - some deltas are > 0, and some are < 0

Depending on how the sorted is defined in the question, cases 1 & 2 might be regarded as unsorted as well.

function findSortOrder(arr) {
  if(arr.length < 2) { // case 1
    return 'not enough items'; // can also be 'unsorted'
  }
  
  var ascending = null;
  var nextArr = arr.slice(1); // create an array that starts from the 2nd element of the original array

  for(var i = 0; i < nextArr.length; i++) {
    if (nextArr[i] === arr[i]) { // neutral - do nothing
    } else if(ascending === null) { // define the the direction by the 1st delta encountered
      ascending = nextArr[i] > arr[i];
    } else if (ascending !== nextArr[i] > arr[i]) { // case 5
      return 'unsorted';
    }
  }
  
  if(ascending === null) { // case 2
    return 'all items are equal'; // can also be 'unsorted'
  }
  
  return ascending ? 'ascending' : 'descending'; // cases 3 & 4
}

console.log(findSortOrder([1])); // case 1

console.log(findSortOrder([1, 1, 1, 1])); // case 2

console.log(findSortOrder([1, 1, 2, 3, 7, 7])); // case 3

console.log(findSortOrder([7, 2, 2, 1])); // case 4

console.log(findSortOrder([7, 2, 1, 3, 2, 1])); // case 5