# sort-an-array


Try it on <a href='https://leetcode.com/problems/sort-an-array'>leetcode</a>

## Description
<div class="description">
<div><p>Given an array of integers <code>nums</code>, sort the array in ascending order and return it.</p>

<p>You must solve the problem <strong>without using any built-in</strong> functions in <code>O(nlog(n))</code> time complexity and with the smallest space complexity possible.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre><strong>Input:</strong> nums = [5,2,3,1]
<strong>Output:</strong> [1,2,3,5]
<strong>Explanation:</strong> After sorting the array, the positions of some numbers are not changed (for example, 2 and 3), while the positions of other numbers are changed (for example, 1 and 5).
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre><strong>Input:</strong> nums = [5,1,1,2,0,0]
<strong>Output:</strong> [0,0,1,1,2,5]
<strong>Explanation:</strong> Note that the values of nums are not necessairly unique.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>-5 * 10<sup>4</sup> &lt;= nums[i] &lt;= 5 * 10<sup>4</sup></code></li>
</ul>
</div>
</div>

## Solution(Python)
```Python
class Solution:
    def sortArray(self, nums: List[int]) -> List[int]:
        return self.countingSort(nums)

    # merge two sorted arrays
    # O(n) time complexity
    # O(n) space complexity

    def mergeSort(self, nums: List[int]) -> List[int]:
        if len(nums) <= 1:
            return nums
        mid = len(nums) // 2
        left = self.mergeSort(nums[:mid])
        right = self.mergeSort(nums[mid:])
        return self.merge(left, right)

    # Heap sort
    # O(nlogn) time complexity
    # O(n) space complexity

    def heapSort(self, nums: List[int]) -> List[int]:
        heapq.heapify(nums)
        return [heapq.heappop(nums) for _ in range(len(nums))]

    # Counting Sort
    # Time Complexity: O(n + k)
    # SpaceComplexity : O(k)
    def countingSort(self, nums: List[int]) -> List[int]:
        min_num, max_num = min(nums), max(nums)
        count = [0] * (max_num - min_num + 1)
        for num in nums:
            count[num - min_num] += 1
        result = []
        for i, c in enumerate(count):
            result.extend([i + min_num] * c)
        return result

    def merge(self, left: List[int], right: List[int]) -> List[int]:
        result = []
        i = j = 0
        while i < len(left) and j < len(right):
            if left[i] < right[j]:
                result.append(left[i])
                i += 1
            else:
                result.append(right[j])
                j += 1
        result.extend(left[i:])
        result.extend(right[j:])
        return result

```