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I have an array of arrays like this, const array = [[1,5],[7,9],[10,14]]; I need to check to see if the arrays are in order.

Basically, for the provided array, I need to compare the first array's second value, (5), with the second array's first value, (7), which I would then use to invalidate a form.

If the arrays or in order, const array = [[1,5],[6,9],[10,14]]; it returns false, it passes validation, if they aren't, validation fails and it returns true.

So far, I've done something like this. 

const arrayIdxZeroValues = [];
const arrayIdxOneValues = []; 
array.forEach((arr,idx) => {
  arrayIdxZeroValues.push(arr[0])
  arrayIdxOneValues.push(arr[1])
})

which separates the values fine. I then thought about using .shift() to remove the first value of arrayIdxZeroValues since it will never need to be compared, but I feel this is hacky and I'm not sure it's the right approach from here. Maybe .reduce() to compare the output arrays, but I haven't been able to get it to work right.

Any guidance would be much appreciated.

Thanks

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2 Answers 2

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Create an array that starts from the 2nd sub-array using Array.slice(). Use Array.some() and check if the 2nd item of the current sub-array is equal or less the 2nd item of the sub-array at the current index i of the original array. In this way you compare the 2nd with the 1st, the 3rd with the 2nd, and so on.

false - means the validation passed true - means validation failed

const isFailed = arr => arr.slice(1)
  .some(([n], i) => n !== arr[i][1] + 1)


console.log(isFailed([[1,5],[6,9],[10,14]])) // false
console.log(isFailed([[1,5],[7,9],[10,14]])) // true

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5 Comments

Thanks Ori, but I don't believe this works given the criteria. The first array, [1,5] would need to then start with a 6, [6,9] to pass validation. Given that array, the third array would need to start with 10, [10,14], and so on for an undetermined number of subarrays
So the distance between last and 1st should be 1 always?
yes. Basically, continuous. The sub array values represent ranges, with a lower and upper value. They can't overlap, for example, [[1,4], [4,7], .........] wouldn't qualify. The second subarray would need to be a 5, [5,7]. If it's [6,7] that wouldn't qualify either. And the number of subarrays has no limit.
Updated the answer. What is that used for? and why do you want false when the criteria is met, and true when it fails?
Thanks again! I use it in a form validation. The way the validation works, is if it return true, it throws a form error. For example, if the input field is required, and the value === '', the logic return true and the form renders the UI message.
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You could check the actual value agains the last value of the previous array plus one and omit the check for the first inner array.

const
    inOrder = array => array.every(([v], i, a) => !i || a[i - 1][1] + 1 === v);

console.log(inOrder([[1, 5], [7, 9], [10, 14]]));
console.log(inOrder([[1, 5], [6, 9], [10, 14]]));

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5 Comments

Really appreciate the help! If you find the time, I'd love an explanation of the implementation.
I'm still getting a true value returned when there are no gaps, however. const array = [[1, 5], [6, 9], [10, 14]]
it looks like true. do you want false? why?
because I need to determine if the subarrays are continuous. In your example array, there is a gap between 5 and 7, so it should return true, as it does. In my example there is no gap, 5 to 6, 9 to 10, and so one. This should return false, meaning there is no gap between the subarrays.
do you need only to check the gaps? or if the value inside of the array are greater than the first value?

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