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:

1. must have only one set of consecutive 1
2. 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

Answer 1

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));