Check for a single consecutive character in an array

Question

I have an array of numbers which can be a value of either 1 or 0. I need to create a function that detects if there is one instance where there is a consecutive 1 and no other 1 exist outside of that instance, returns true else false

So to sum it up here's a clearer view of the constraints to return true:

must have only one set of consecutive 1
there must be no other 1 outside of that one instance of consecutive 1

Test cases:

[0, 0, 1, 1, 0, 0] true

[1, 0, 1, 0, 0, 0] false

[1, 0, 1, 1, 0, 0] false

[1, 1, 0, 1, 1, 0] false

[0, 1, 1, 1, 1, 0] true

[0, 0, 1, 1, 1, 1] true

That's a fine problem statement. What is your intuition on how to solve it? — Matt Ellen
– Matt Ellen, Commented Nov 10, 2022 at 12:22
I'm thinking of reducing the array, check if it has found a 1, once it does, it will flag that it started to detect the 1. Once that flag is detected, it will continue to check for more 1s. If there are more than one 1 already and it has detected a 0 and has detected another 1, then it returns false, otherwise true. My problem is with that in mind [0, 0, 1, 1] would be incorrect since it hasn't detected a 0 yet. — Lelouch
– Lelouch, Commented Nov 10, 2022 at 12:29

Trevor Dixon · Accepted Answer · 2022-11-21 13:27:46Z

3

function hasOneRun(arr) {
  let switches = 0;
  for (let i = 0; i < arr.length; i++) {
    switches += arr[i] ^ arr[i-1];
    if (switches > 2) return false;
  }
  return switches > 0;
}

Counts how many times it switches from 0 to 1 or 1 to 0. arr[i] ^ arr[i-1] is 1 only if the value changes from the previous value. If switches is 0 at the end, there were only 0s, so false. If it's greater than 2, then it switched to 1, then to 0, then back to 1, so there were too many runs.

Here's a fun one-liner :D

Math.ceil(arr.reduce((switches, val, i) => switches + (val ^ arr[i-1]))/2) === 1;

Edit: Some other ideas

const startOne = (el, i, arr) => el == 1 && arr[i-1] != 1;
const is1 = x => x === 1;

arr.filter(startOne).length == 1;
arr.findIndex(startOne) === arr.findLastIndex(startOne);
arr.slice(arr.findIndex(is1), arr.findLastIndex(is1)+1).every(is1)


/^0*1+0*$/.test(arr.join('')); // zero or more zeroes, one or more ones

Or if you treat it as a sequence of bits…

function hasOneRun(x) {
  while (!(x&1)) x >>= 1; // shift off trailing zeroes
  return (x&(x+1)) === 0; // see if one less than power of 2 (all 1s)
}
console.log(hasOneRun(0b001111000));
console.log(hasOneRun(0b11000));
console.log(hasOneRun(0b1111));
console.log(hasOneRun(0b1110000111000));

edited Nov 21, 2022 at 13:27

answered Nov 10, 2022 at 13:06

Trevor Dixon

24.6k13 gold badges77 silver badges115 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

Lelouch Over a year ago

for some reason this works, can you please expound what ^ does? I have no idea what it is

Trevor Dixon Over a year ago

Exclusive or (xor). 0 ^ 0 == 0, 0 ^ 1 == 1, 1 ^ 0 == 1, 1 ^ 1 == 0. developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

Trevor Dixon Over a year ago

You could also write if (arr[i] != (arr[i-1] ?? 0)) ++switches;

Lelouch Over a year ago

thanks! that is so much easier to digest.

Andrew Parks Over a year ago

Your one-liner fails for the test cases [0, 0, 1, 1, 0, 0] and [0, 1, 1, 1, 1, 0]

|

Andrew Parks · Accepted Answer · 2022-11-10 13:48:37Z

1

You can also solve this by treating it as a string, and splitting it up where there are any sequences of zeroes:

const testCases = [
  [[0, 0, 1, 1, 0, 0], true],
  [[1, 0, 1, 0, 0, 0], false],
  [[1, 0, 1, 1, 0, 0], false],
  [[1, 1, 0, 1, 1, 0], false],
  [[0, 1, 1, 1, 1, 0], true],
  [[0, 0, 1, 1, 1, 1], true]
];
const f = a => a.join('').split(/0+/).filter(i=>i).length===1;
testCases.forEach(t=>console.log(f(t[0])===t[1] ? 'pass' : 'fail'));

answered Nov 10, 2022 at 13:48

Andrew Parks

8,1672 gold badges22 silver badges35 bronze badges

3 Comments

Lelouch Over a year ago

sweet, this is also a nice way to solve the problem, thanks you are brilliant.

Trevor Dixon Over a year ago

If it's a string, regex does nicely. /^0*1+0*$/.test(arr.join(''))

Andrew Parks Over a year ago

@TrevorDixon very nice!

Collectives™ on Stack Overflow

Check for a single consecutive character in an array

2 Answers 2

6 Comments

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

6 Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related