# LeetCode - Rearrange Array Elements by Sign

## 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:

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.

Problem statement taken from: https://leetcode.com/problems/rearrange-array-elements-by-sign

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 * 10^5
- nums.length is even
- 1 <= |nums[i]| <= 10^5
- nums consists of equal number of positive and negative integers.

### Explanation

The problem is easy to solve. We can solve it in O(n) time using an additional array.

Let's explore the algorithm directly.

- set n = nums.size()
positiveIndex = 0, negativeIndex = 1

- loop for int i = 0; i < n; i++
- if nums[i] > 0
positiveIndex = positiveIndex + 2
- else
negativeIndex = negativeIndex + 2
- if end
- for end

The time complexity of the above approach is O(n), and the space complexity is O(1).

Let's check our algorithm in C++, Golang, and JavaScript.

#### C++ solution

class Solution {
public:
vector<int> rearrangeArray(vector<int>& nums) {
int n = nums.size();
int positiveIndex = 0, negativeIndex = 1;

for(int i = 0; i < n; i++) {
if(nums[i] > 0) {
positiveIndex += 2;
} else {
negativeIndex += 2;
}
}

}
};

#### Golang solution

func rearrangeArray(nums []int) []int {
n := len(nums)
positiveIndex, negativeIndex := 0, 1

for i := 0; i < n; i++ {
if nums[i] > 0 {
positiveIndex += 2
} else {
negativeIndex += 2
}
}

}

#### JavaScript solution

var rearrangeArray = function(nums) {
let n = nums.length;
let positiveIndex = 0, negativeIndex = 1;

for(let i = 0; i < n; i++) {
if(nums[i] > 0) {
positiveIndex += 2;
} else {
negativeIndex += 2;
}
}

};

#### Dry Run

Let's dry-run our algorithm to see how the solution works.

Input: nums = [3, 1, -2, -5, 2, -4]

Step 1: n = nums.size()
= 6
answer = [0, 0, 0, 0, 0, 0]
positiveIndex = 0
negativeIndex = 1

Step 2: loop for i = 0; i < 6; i++
0 < 6
true

if nums[i] > 0
nums[0] > 0
3 > 0
true

= 3
answer = [3, 0, 0, 0, 0, 0]

positiveIndex = positiveIndex + 2
= 0 + 2
= 2

i++
i = 1

Step 3: loop for i < 6
1 < 6
true

if nums[i] > 0
nums[1] > 0
1 > 0
true

= 1
answer = [3, 0, 1, 0, 0, 0]

positiveIndex = positiveIndex + 2
= 2 + 2
= 4

i++
i = 2

Step 4: loop for i < 6
2 < 6
true

if nums[i] > 0
nums[2] > 0
-2 > 0
false
else
= -2
answer = [3, -2, 1, 0, 0, 0]

negativeIndex = negativeIndex + 2
= 1 + 2
= 3

i++
i = 3

Step 5: loop for i < 6
3 < 6
true

if nums[i] > 0
nums[3] > 0
-5 > 0
false
else
= -5
answer = [3, -2, 1, -5, 0, 0]

negativeIndex = negativeIndex + 2
= 3 + 2
= 5

i++
i = 4

Step 6: loop for i < 6
4 < 6
true

if nums[i] > 0
nums[4] > 0
2 > 0
true

= 1
answer = [3, -2, 1, -5, 2, 0]

positiveIndex = positiveIndex + 2
= 4 + 2
= 6

i++
i = 5

Step 7: loop for i < 6
5 < 6
true

if nums[i] > 0
nums[5] > 0
-4 > 0
false
else
= -4
answer = [3, -2, 1, -5, 2, -4]

negativeIndex = negativeIndex + 2
= 5 + 2
= 7

i++
i = 6

Step 8: loop for i < 6
6 < 6
false