Rearrange Array Elements by Sign - LeetCode Daily Challenge

Problem Statement

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:

1. Every consecutive pair of integers have opposite signs.
2. For all integers with the same sign, the order in which they were present in nums is preserved.
3. 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].

Try here before watching the video.

Video Explanation

Java Code

class Solution {
    public int[] rearrangeArray(int[] nums) {
        int[] answer = new int[nums.length];
        int positiveIndex = 0, negativeIndex = 1;
        for(int i = 0; i < nums.length; i++) {
            if(nums[i] > 0) {
                answer[positiveIndex] = nums[i];
                positiveIndex += 2;
            } else {
                answer[negativeIndex] = nums[i];
                negativeIndex += 2;
            }
        }
        return answer;
    }
}

C++ Code

class Solution {
public:
    vector<int> rearrangeArray(vector<int>& nums) {
        vector<int> answer(nums.size());
        int positiveIndex = 0, negativeIndex = 1;
        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] > 0) {
                answer[positiveIndex] = nums[i];
                positiveIndex += 2;
            } else {
                answer[negativeIndex] = nums[i];
                negativeIndex += 2;
            }
        }
        return answer;
    }
};

Python Code

class Solution:
    def rearrangeArray(self, nums: List[int]) -> List[int]:
        answer = [0] * len(nums)
        positive_index, negative_index = 0, 1
        for num in nums:
            if num > 0:
                answer[positive_index] = num
                positive_index += 2
            else:
                answer[negative_index] = num
                negative_index += 2
        return answer

Javascript Code

var rearrangeArray = function(nums) {
    const answer = new Array(nums.length).fill(0);
    let positiveIndex = 0, negativeIndex = 1;
    for (let i = 0; i < nums.length; i++) {
        if (nums[i] > 0) {
            answer[positiveIndex] = nums[i];
            positiveIndex += 2;
        } else {
            answer[negativeIndex] = nums[i];
            negativeIndex += 2;
        }
    }
    return answer;
};

Go Code

func rearrangeArray(nums []int) []int {
    answer := make([]int, len(nums))
    positiveIndex, negativeIndex := 0, 1
    for _, num := range nums {
        if num > 0 {
            answer[positiveIndex] = num
            positiveIndex += 2
        } else {
            answer[negativeIndex] = num
            negativeIndex += 2
        }
    }
    return answer
}

Complexity Analysis

Time Complexity: O(N)
Space Complexity: O(1), other than the answer array, we are not taking any extra space.