JavaScript: Return 1 if the array is sorted in ascending order, -1 if it is sorted in descending order or 0 if it is not sorted

JavaScript fundamental (ES6 Syntax): Exercise-188 with Solution

Write a JavaScript program that returns 1 if the array is sorted in ascending order. It returns -1 if it is sorted in descending order or 0 if it is not sorted.

• Calculate the ordering direction for the first pair of adjacent array elements.
• Return 0 if the given array is empty, only has one element or the direction changes for any pair of adjacent array elements.
• Use Math.sign() to covert the final value of direction to -1 (descending order) or 1 (ascending order).

Sample Solution:

JavaScript Code:

// Define a function 'isSorted' that checks if the array 'arr' is sorted in either ascending or descending order
const isSorted = arr => {
  // Determine the initial direction of sorting by comparing the first two elements
  let direction = -(arr[0] - arr[1]);

  // Iterate over each element in the array
  for (let [i, val] of arr.entries()) {
    // If the direction is not set (0), calculate the direction based on the previous elements
    direction = !direction ? -(arr[i - 1] - arr[i]) : direction;

    // If it's the last element in the array, return 0 if the array is sorted, otherwise return the direction
    if (i === arr.length - 1) return !direction ? 0 : direction;
    // If the current element and the next one violate the sorting order, return 0
    else if ((val - arr[i + 1]) * direction > 0) return 0;
  }
};

// Test cases to check if arrays are sorted
console.log(isSorted([0, 1, 2, 2])); // 1 (ascending order)
console.log(isSorted([4, 3, 2])); // -1 (descending order)
console.log(isSorted([4, 3, 5])); // 0 (not sorted)

Output:

1
-1
0

Visual Presentation:

Flowchart:

Live Demo:

See the Pen javascript-basic-exercise-188-1 by w3resource (@w3resource) on CodePen.

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