Your code has a bug because it will accept decreasing sequences as true as well, for example, increasingSequence([4,3,2,1]); will be true.

Having said that, you could do the tests a whole lot simpler by using a search:

  const increasingSequence = (arr) => !arr.some((val, idx) => val != arr[0] + idx);

How does that work? Well, each element in the array is expected to be the value of the first item, plus the index from that first item, so, if the first value is 4 then the 3rd value will be expected to be 6 (4 + 2).

The above code simply looks for any value that does not meet those expectations.

If there is a value that does not meet the expectations, the some returns true, so we negate that, and return false.

Since the some(...) method is a short-circuiting function so it will terminate when the first mismatch is encountered (just like your logic).

Also, note that the short-circuit does not impact the time-complexity of the function, it is still \$O(N)\$ because the performance will scale linearly in relation to the number of elements.

You can see this running on Ideone.

Your code has a bug because it will accept decreasing sequences as true as well, for example, increasingSequence([4,3,2,1]); will be true.

Having said that, you could do the tests a whole lot simpler by using a search:

  const increasingSequence = (arr) => !arr.some((val, idx) => val != arr[0] + idx);

How does that work? Well, each element in the array is expected to be the value of the first item, plus the index address of the current item, so, if the first value is 4 then the 3rd value will be expected to be 6 (4 + 2).

The above code simply looks for any value that does not meet those expectations.

If there is a value that does not meet the expectations, the some returns true, so we negate that, and return false.

Since the some(...) method is a short-circuiting function so it will terminate when the first mismatch is encountered (just like your logic).

Also, note that the short-circuit does not impact the time-complexity of the function, it is still \$O(N)\$ because the performance will scale linearly in relation to the number of elements.

You can see this running on Ideone.

Your code has a bug because it will accept decreasing sequences as true as well, for example, increasingSequence([4,3,2,1]); will be true.

Having said that, you could do the tests a whole lot simpler by using a search:

  const increasingSequence = (arr) => !arr.some((val, idx) => val != arr[0] + idx);

How does that work? Well, each element in the array is expected to be the value of the first item, plus the index from that first item, so, if the first value is 4 then the 3rd value will be expected to be 6 (4 + 2).

The above code simply looks for any value that does not meet those expectations.

If there is a value that does not meet the expectations, the some returns true, so we negate that, and return false.

Since the some(...) method is a short-circuiting function so it will terminate when the first mismatch is encountered (just like your logic).

Also, note that the short-circuit does not impact the time-complexity of the function, it is still \$O(N)\$ because the performance will scale linearly in relation to the number of elements.

Your code has a bug because it will accept decreasing sequences as true as well, for example, increasingSequence([4,3,2,1]); will be true.

Having said that, you could do the tests a whole lot simpler by using a search:

  const increasingSequence = (arr) => !arr.some((val, idx) => val != arr[0] + idx);

How does that work? Well, each element in the array is expected to be the value of the first item, plus the index from that first item, so, if the first value is 4 then the 3rd value will be expected to be 6 (4 + 2).

The above code simply looks for any value that does not meet those expectations.

If there is a value that does not meet the expectations, the some returns true, so we negate that, and return false.

Since the some(...) method is a short-circuiting function so it will terminate when the first mismatch is encountered (just like your logic).

Also, note that the short-circuit does not impact the time-complexity of the function, it is still \$O(N)\$ because the performance will scale linearly in relation to the number of elements.

You can see this running on Ideone.