Gist demonstrating 3 different O(n) Methods For Getting Unique Values Count in JavaScript

CountUniqueValues.js

/**
 * Count Unique Values Problem solved 3 ways
 *
 * Prompt: Implement a function which accepts a sorted array and counts
 * the unique values in the array.
 *
 * Negative numbers are allowed and will only be numbers.
 * The array is always sorted*
 *
 * @param {*} sortedArr
 * @returns
 */

// Using a set O(n) time complexity
const countUniqueValuesWithSet = (sortedArr) => {
  if (!sortedArr.length) return [];

  const charSet = new Set(sortedArr);
  const uniqueArray = Array.from(charSet);

  return uniqueArray.length;
};

// Using sliding pointer + same array -> O(n) t.c.
const countUniqueValues = (sortedArr) => {
  if (!sortedArr.length) return [];

  let i = 0;

  for (let j = 0; j < sortedArr.length; j++) {
    if (sortedArr[i] !== sortedArr[j]) {
      i++;

      sortedArr[i] = sortedArr[j];
    }
  }

  return i + 1;
};

// Using the sliding pointer approach and new array -> O(n) t.c.
const countUniqueValues2 = (sortedArr) => {
  if (!sortedArr.length) return [];
  const store = [];

  for (let i = 0; i < sortedArr.length; i++) {
    let j = i + 1;
    sortedArr[i] === sortedArr[j] ? j++ : store.push(sortedArr[j]);
  }

  return store.length;
};

console.log(countUniqueValues([1, 1, 1, 1, 1, 2]));
console.log(countUniqueValues([1, 2, 3, 4, 4, 4, 7, 7, 12, 12, 13]));
console.log(countUniqueValues([]));
console.log(countUniqueValues([-2, -1, -1, 0, 1]));