1356. Sort Integers by The Number of 1 Bits

Easy

---

You are given an integer array arr. 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 array after sorting it.

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.

Constraints:

• 1 <= arr.length <= 500

• 0 <= arr[i] <= 104

class Solution:
    def sortByBits(self, arr: List[int]) -> List[int]:
        # Pass through array and get number of 1's bits in each number
        # Store this in dict and sort using `value`
        d = []
        for num in arr:
            d.append([num, self.countBits(num)])
        def sort(item1, item2):
            if item1[1] < item2[1]:
                return -1
            elif item1[1] > item2[1]:
                return 1
            else:
                if item1[0] < item2[0]:
                    return -1
                else:
                    return 1
        d = sorted(d, key=functools.cmp_to_key(sort))
        return map(lambda item: item[0],d )
        
    def countBits(self, num: int) -> int:
        count = 0
        while num:
            count += 1 & num
            num = num >> 1
        return count