LeetCode 1356. Sort Integers by The Number of 1 Bits

Question

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

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]

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.

Solution

It’s a customized sorting problem. But we need to convert int to Integer, so that we can sort them.

Code

// time: O(nlogn), space: O(n)
public int count(int num) {
        int res = 0;
        for (int i = 0; i < 32; i++) {
            res += num & 1;
            num >>= 1;
        }
        return res;
    }
    public int[] sortByBits(int[] arr) {

        if (arr == null || arr.length == 0) return new int[]{};
        int n = arr.length;
        Integer[] nums = new Integer[n];
        for (int i = 0; i < n; i++) nums[i] = arr[i];

        Comparator comp = new myComp();
        Arrays.sort(nums, comp);

        for (int i = 0; i < arr.length; i++) arr[i] = nums[i];
        return arr;
    }
    class myComp implements Comparator<Integer> {
        @Override
        public int compare(Integer o1, Integer o2) {
            int c1 = count(o1);
            int c2 = count(o2);
            return c1 != c2 ? c1 - c2 : o1 - o2;
        }
    }