1356 - Sort Integers by The Number of 1 Bits (Easy)

Problem Link

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

Problem Statement

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] <= 10^4

Approach 1: Sorting + __builtin_popcount / Bit Count

C++

Written by @wingkwong

class Solution {
public:
    vector<int> sortByBits(vector<int>& arr) {
        sort(arr.begin(), arr.end(), [&](int a, int b){
            // __builtin_popcount(x) returns the number of 1-bits set in an int x.
            int x = __builtin_popcount(a), y = __builtin_popcount(b);
            return x == y ? 
                // in case of two or more integers have the same number of 1's you have to sort them in ascending order
                a < b : 
                // else sort the integers in the array in ascending order by the number of 1's in their binary representation 
                x < y;
        });
        return arr;
    }
};

Java

Written by @vigneshshiv

class Solution {
    public int[] sortByBits(int[] arr) {
        return Arrays.stream(arr)
                .boxed()
                .sorted(Comparator.comparing(Integer::bitCount).thenComparing(Integer::intValue))
                .mapToInt(num -> num)
                .toArray();
    }
}

Approach 2: Sorting with Math Logic

Since the input values can range from 0 to 10000, we can add to each element on the array the value of the bit count (number of 1s) multiplied by 10001.

10001 is not a magic number, it is merely the max possible number the array may have + 1, which ensures that the bit count has the maximum priority

Java

Written by @vigneshshiv

class Solution {
    public int[] sortByBits(int[] arr) {
        for (int i = 0; i < arr.length; i++) {
            arr[i] += (Integer.bitCount(arr[i]) * 10001);
        }
        Arrays.sort(arr);
        for (int i = 0; i < arr.length; i++) {
            arr[i] %= 10001;
        }
        return arr;
    }
}