Question

1356. Sort Integers by The Number of 1 Bits

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Description

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

Solutions

Solution 1: Custom Sorting

We sort the array $arr$ according to the requirements of the problem, that is, sort in ascending order according to the number of $1$s in the binary representation. If there are multiple numbers with the same number of $1$s in the binary representation, they must be sorted in ascending order by numerical value.

The time complexity is $O(n \log n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $arr$.

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

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;
  }
}

class Solution {
public:
  vector<int> sortByBits(vector<int>& arr) {
    for (int& v : arr) {
      v += __builtin_popcount(v) * 100000;
    }
    sort(arr.begin(), arr.end());
    for (int& v : arr) {
      v %= 100000;
    }
    return arr;
  }
};

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
  }
}

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int countOnes(int n) {
  int res = 0;
  while (n) {
    n &= n - 1;
    res++;
  }
  return res;
}

int cmp(const void* _a, const void* _b) {
  int a = *(int*) _a;
  int b = *(int*) _b;
  int res = countOnes(a) - countOnes(b);
  if (res == 0) {
    return a - b;
  }
  return res;
}

int* sortByBits(int* arr, int arrSize, int* returnSize) {
  qsort(arr, arrSize, sizeof(int), cmp);
  *returnSize = arrSize;
  return arr;
}

Solution 2

class Solution {
  public int[] sortByBits(int[] arr) {
    int n = arr.length;
    Integer[] t = new Integer[n];
    for (int i = 0; i < n; ++i) {
      t[i] = arr[i];
    }
    Arrays.sort(t, (a, b) -> {
      int x = Integer.bitCount(a), y = Integer.bitCount(b);
      return x == y ? a - b : x - y;
    });
    for (int i = 0; i < n; ++i) {
      arr[i] = t[i];
    }
    return arr;
  }
}

class Solution {
public:
  vector<int> sortByBits(vector<int>& arr) {
    sort(arr.begin(), arr.end(), [&](auto& a, auto& b) -> bool {
      int x = __builtin_popcount(a), y = __builtin_popcount(b);
      return x < y || (x == y && a < b);
    });
    return arr;
  }
};

func sortByBits(arr []int) []int {
  sort.Slice(arr, func(i, j int) bool {
    a, b := bits.OnesCount(uint(arr[i])), bits.OnesCount(uint(arr[j]))
    return a < b || (a == b && arr[i] < arr[j])
  })
  return arr
}