Leetcode problem 98 - Partition Array According to Given Pivot

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. For elements less than pivot, if i < j and nums[i] < pivot and nums[j] < pivot, then pi < pj. Similarly for elements greater than pivot, if i < j and nums[i] > pivot and nums[j] > pivot, then pi < pj. Return nums after the rearrangement.

Example 1:

Input: nums = [9,12,5,10,14,3,10], pivot = 10 Output: [9,5,3,10,10,12,14] Explanation: The elements 9, 5, and 3 are less than the pivot so they are on the left side of the array. The elements 12 and 14 are greater than the pivot so they are on the right side of the array. The relative ordering of the elements less than and greater than pivot is also maintained. [9, 5, 3] and [12, 14] are the respective orderings. Example 2:

Input: nums = [-3,4,3,2], pivot = 2 Output: [-3,2,4,3] Explanation: The element -3 is less than the pivot so it is on the left side of the array. The elements 4 and 3 are greater than the pivot so they are on the right side of the array. The relative ordering of the elements less than and greater than pivot is also maintained. [-3] and [4, 3] are the respective orderings.

Constraints:

1 <= nums.length <= 105 -106 <= nums[i] <= 106 pivot equals to an element of nums.

/**
 * @param {number[]} nums
 * @param {number} pivot
 * @return {number[]}
 */
var pivotArray = function (nums, pivot) {
  // Initialize arrays to store elements smaller, equal, and greater than the pivot
  let smaller = [];
  let greater = [];
  let center = [];

  // Iterate through each element in the array 'nums'
  for (let i = 0; i < nums.length; i++) {
    // Check if the current element is smaller than the pivot
    if (nums[i] < pivot) {
      // If smaller, add it to the 'smaller' array
      smaller.push(nums[i]);
    }
    // Check if the current element is greater than the pivot
    else if (nums[i] > pivot) {
      // If greater, add it to the 'greater' array
      greater.push(nums[i]);
    }
    // If the current element is equal to the pivot
    else {
      // Add it to the 'center' array
      center.push(nums[i]);
    }
  }

  // Concatenate the 'smaller', 'center', and 'greater' arrays and return the result
  return [...smaller, ...center, ...greater];
};

The code iterates through each element in the array nums exactly once. The time complexity of the iteration is O(n), where n is the length of the array.

The space complexity is also O(n) because the function creates three arrays of size n.