1356. Sort Integers by The Number of 1 Bits | LeetCode Solution

Raunit Verma May 23, 2022

1356. Sort Integers by The Number of 1 Bits | LeetCode Solution

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
{
public:
    static int count_set_bits(int n)
    {
        int ans = 0;
        for (int i = 0; i < 32; i++)
        {
            if (n % 2 == 1)
                ans++;
            n >>= 1;
        }

        return ans;
    }

    static bool comp(int &a, int &b)
    {
        if (count_set_bits(a) < count_set_bits(b))
            return true;
        else if (count_set_bits(a) == count_set_bits(b) and a < b)
            return true;
        else
            return false;
    }

    vector<int> sortByBits(vector<int> &arr)
    {
        sort(arr.begin(), arr.end(), comp);
        return arr;
    }
};