Sequence of equal numbers in array

I want to push the first sequence to currentSequence, then store it in maxSequence, then push the next sequence in currentSequence, check if its length is greater, if it isn't proceed further, if it is store it in maxSequence. I'd like to keep the original order of the array.

 let arr = ['2','1','1','2','3','3','2','2','2','1'].map(Number)

let maxSequence = []

for(let i = 0; i< arr.length; i++){
     let currentSequence = []
  if( arr[i] === arr[i+1] ){ 
    currentSequence.push(arr[i]);
  //currentSequence = arr.toSpliced(i, 1);
  }
  if(currentSequence.length > maxSequence.length){
      maxSequence = currentSequence  
  }
  }
   console.log(maxSequence.length) // expected output: 3
   console.log(maxSequence) // expected output: [2, 2, 2]
 

This can be achieved like so:

<!-- begin snippet: js hide: false console: true babel: false -->

<!-- language: lang-js -->

let arr = ['2', '1', '1', '2', '3', '3', '2', '2', '2', '1'].map(Number)

let maxSequence = [];
let currentSequence = [];
let currentValue = -1;

for (let i = 0; i < arr.length; i++) {
  if (arr[i] === currentValue) {
    currentSequence.push(arr[i]);
  } else {
    currentValue = arr[i];
    currentSequence = [currentValue];
  }
  if (currentSequence.length > maxSequence.length) {
    maxSequence = currentSequence
  }
}
console.log(maxSequence.length) // expected output: 3
console.log(maxSequence) // expected output: [2, 2, 2]

<!-- end snippet -->

One way of doing what you need is:

<!-- begin snippet: js hide: false console: true babel: false -->

<!-- language: lang-js -->

let arr = ["2", "1", "1", "2", "3", "3", "2", "2", "2", "1"].map(Number);

let maxSequence = [];
let currentSequence = [arr[0]];

for (let i = 1; i < arr.length; i++) {
  if (arr[i] !== arr[i - 1]) { // if current number does not match the one before it
    if (currentSequence.length > maxSequence.length) { // it means currentSequence is complete and we need to check if it's larger than current maxSequence or not
      maxSequence = currentSequence;
    }
    currentSequence = [arr[i]]; // then need to reset the currentSequence array
    continue;
  }
  currentSequence.push(arr[i]); // else if numbers match, we push current number to the currentSequence
}
console.log(maxSequence.length); // expected output: 3
console.log(maxSequence); // expected output: [2, 2, 2]

<!-- end snippet -->

Can be done without an intermediate array.

<!-- begin snippet: js hide: false console: true babel: false -->

<!-- language: lang-js -->

let arr = ['2','1','1','2','3','3','2','2','2','1'].map(Number);
let maxSequence;

(() => {
    let maxRunVal = arr[0], maxRun = 1;
    for (let i = 1, run = 1; i < arr.length; i++) {
        if (arr[i] == arr[i-1]) {
            run++;
            if (run > maxRun) {
                maxRun = run;
                maxRunVal = arr[i];
            }
        }
        else {
            run = 1;
        }
    }
    maxSequence = new Array(maxRun);
    maxSequence.fill(maxRunVal);
})();

console.log(maxSequence.length);
console.log(maxSequence);

<!-- end snippet -->

For fun: you can also use a regular expression to filter the groups combined with [Array.] filter and reduce (see MDN)

It may be a problem if the input array contains multiple sequences of the same length. The function in the snippet therefore returns all sequences of consecutive equal values with a found maximum length.

<!-- begin snippet: js hide: false console: true babel: false -->

<!-- language: lang-js -->

const arr = ['2','1','1','2','3','3','2','2','2','1','3','3','3'];
const maxSequences = longestSequencesOfEqualValues(arr);

console.log(`In [${arr}] the max sequence length is ${
  maxSequences[0].length ?? 0}`);
console.log(`found sequence(s):`);
maxSequences.forEach( seq => 
 console.log( `\u2713 [${seq}]`) 
);

function longestSequencesOfEqualValues(arr = []) {
  const groups = arr.join(``).match(/(.)*/g);
  const max = groups
    ?.filter(v => v.length > 1)
    .reduce( (a, v) => v.length > a ? v.length : a, 0 );
  return groups?.filter(el => el.length === max, 0)
    .reduce( (acc, v) => ([...acc, [...v] ]), []) ?? [];
}

<!-- end snippet -->