Leetcode problem 99 - Rearrange Array Elements by Sign

You are given a 0-indexed integer array nums of even length consisting of an equal number of positive and negative integers.

You should rearrange the elements of nums such that the modified array follows the given conditions:

Every consecutive pair of integers have opposite signs. For all integers with the same sign, the order in which they were present in nums is preserved. The rearranged array begins with a positive integer. Return the modified array after rearranging the elements to satisfy the aforementioned conditions.

Example 1:

Input: nums = [3,1,-2,-5,2,-4] Output: [3,-2,1,-5,2,-4] Explanation: The positive integers in nums are [3,1,2]. The negative integers are [-2,-5,-4]. The only possible way to rearrange them such that they satisfy all conditions is [3,-2,1,-5,2,-4]. Other ways such as [1,-2,2,-5,3,-4], [3,1,2,-2,-5,-4], [-2,3,-5,1,-4,2] are incorrect because they do not satisfy one or more conditions.
Example 2:

Input: nums = [-1,1] Output: [1,-1] Explanation: 1 is the only positive integer and -1 the only negative integer in nums. So nums is rearranged to [1,-1].

Constraints:

2 <= nums.length <= 2 * 105 nums.length is even 1 <= |nums[i]| <= 105 nums consists of equal number of positive and negative integers.

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var rearrangeArray = function (nums) {
  // Declare an empty array called result to store the rearranged elements.
  let result = [];

  // Declare two variables, posIndex and negIndex, to keep track of the current positions for positive and negative numbers in the result array.
  let posIndex = 0,
    negIndex = 1;

  // Iterate through each element in the input array nums.
  for (let i = 0; i < nums.length; i++) {
    // Check if the current element is positive.
    if (nums[i] > 0) {
      // Assign the positive element to the position specified by posIndex in the result array.
      result[posIndex] = nums[i];
      // Increment posIndex by 2 to skip the next position for positive numbers.
      posIndex += 2;
    } else {
      // Assign the negative or zero element to the position specified by negIndex in the result array.
      result[negIndex] = nums[i];
      // Increment negIndex by 2 to skip the next position for negative numbers.
      negIndex += 2;
    }
  }
  // Return the rearranged array.
  return result;
};

Time Complexity: O(n), where n is the length of the input array nums. Space Complexity: O(n), where n is the length of the input array nums.