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Formatted question description: https://leetcode.ca/all/1356.html

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

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

Related Topics:
Sort, Bit Manipulation

## Solution 1.

• class Solution {
public int[] sortByBits(int[] arr) {
int length = arr.length;
int[][] numsBits = new int[length][2];
for (int i = 0; i < length; i++) {
numsBits[i][0] = arr[i];
numsBits[i][1] = Integer.bitCount(arr[i]);
}
Arrays.sort(numsBits, new Comparator<int[]>() {
public int compare(int[] numBit1, int[] numBit2) {
if (numBit1[1] != numBit2[1])
return numBit1[1] - numBit2[1];
else
return numBit1[0] - numBit2[0];
}
});
int[] sorted = new int[length];
for (int i = 0; i < length; i++)
sorted[i] = numsBits[i][0];
return sorted;
}
}

############

class Solution {
public int[] sortByBits(int[] arr) {
int n = arr.length;
for (int i = 0; i < n; ++i) {
arr[i] += Integer.bitCount(arr[i]) * 100000;
}
Arrays.sort(arr);
for (int i = 0; i < n; ++i) {
arr[i] %= 100000;
}
return arr;
}
}

• // OJ: https://leetcode.com/problems/sort-integers-by-the-number-of-1-bits/
// Time: O(NlogN)
// Space: O(N)
class Solution {
int cnt(int n) {
int c = 0;
for (; n; n >>= 1) {
if (n & 1) ++c;
}
return c;
}
public:
vector<int> sortByBits(vector<int>& arr) {
int N = arr.size();
vector<pair<int, int>> v(N);
for (int i = 0; i < N; ++i) v[i] = make_pair(cnt(arr[i]), arr[i]);
sort(v.begin(), v.end());
for (int i = 0; i < N; ++i) arr[i] = v[i].second;
return arr;
}
};

• class Solution:
def sortByBits(self, arr: List[int]) -> List[int]:
return sorted(arr, key=lambda x: (x.bit_count(), x))


• func sortByBits(arr []int) []int {
for i, v := range arr {
arr[i] += bits.OnesCount(uint(v)) * 100000
}
sort.Ints(arr)
for i := range arr {
arr[i] %= 100000
}
return arr
}

• function sortByBits(arr: number[]): number[] {
const countOnes = (n: number) => {
let res = 0;
while (n) {
n &= n - 1;
res++;
}
return res;
};
return arr.sort((a, b) => countOnes(a) - countOnes(b) || a - b);
}


• impl Solution {
pub fn sort_by_bits(mut arr: Vec<i32>) -> Vec<i32> {
arr.sort_by(|a, b| {
let res = a.count_ones().cmp(&b.count_ones());
if res == std::cmp::Ordering::Equal {
return a.cmp(&b);
}
res
});
arr
}
}