2161-partition-array-according-to-given-pivot.js

/**
 * 2161. Partition Array According to Given Pivot
 * https://leetcode.com/problems/partition-array-according-to-given-pivot/
 * Difficulty: Medium
 *
 * You are given a 0-indexed integer array nums and an integer pivot. Rearrange nums such that
 * the following conditions are satisfied:
 * - Every element less than pivot appears before every element greater than pivot.
 * - Every element equal to pivot appears in between the elements less than and greater than
 *   pivot.
 * - The relative order of the elements less than pivot and the elements greater than pivot
 *   is maintained.
 *   - More formally, consider every pi, pj where pi is the new position of the ith element
 *     and pj is the new position of the jth element. If i < j and both elements are smaller
 *     (or larger) than pivot, then pi < pj.
 *
 * Return nums after the rearrangement.
 */

/**
 * @param {number[]} nums
 * @param {number} pivot
 * @return {number[]}
 */
var pivotArray = function(nums, pivot) {
  return [
    ...nums.filter(x => x < pivot),
    ...nums.filter(x => x === pivot),
    ...nums.filter(x => x > pivot),
  ];
};