## Description

https://leetcode.com/problems/sort-integers-by-the-number-of-1-bits/

Given an integer array `arr`

. You have to sort the integers in the array in ascending order by the number of **1’s** in their binary representation and in case of two or more integers have the same number of **1’s** you have to sort them in ascending order.

Return *the sorted array*.

**Example 1:**

Input:arr = [0,1,2,3,4,5,6,7,8]Output:[0,1,2,4,8,3,5,6,7]Explantion:[0] is the only integer with 0 bits. [1,2,4,8] all have 1 bit. [3,5,6] have 2 bits. [7] has 3 bits. The sorted array by bits is [0,1,2,4,8,3,5,6,7]

**Example 2:**

Input:arr = [1024,512,256,128,64,32,16,8,4,2,1]Output:[1,2,4,8,16,32,64,128,256,512,1024]Explantion:All integers have 1 bit in the binary representation, you should just sort them in ascending order.

**Example 3:**

Input:arr = [10000,10000]Output:[10000,10000]

**Example 4:**

Input:arr = [2,3,5,7,11,13,17,19]Output:[2,3,5,17,7,11,13,19]

**Example 5:**

Input:arr = [10,100,1000,10000]Output:[10,100,10000,1000]

**Constraints:**

`1 <= arr.length <= 500`

`0 <= arr[i] <= 10^4`

## Explanation

Count the ‘1’ bits of each number and use a heap to sort by ascending order.

## Python Solution

```
class Solution:
def sortByBits(self, arr: List[int]) -> List[int]:
bit_counter = {}
heap = []
for num in arr:
num_binary = bin(num)[2:]
bit_count = num_binary.count('1')
heapq.heappush(heap, (bit_count, num))
results = []
for _ in range(len(heap)):
results.append(heapq.heappop(heap)[1])
return results
```

- Time Complexity: O(Nlog(N)).
- Space Complexity: O(N).