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feat: add solutions to lc problem: No.0477 #2760

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May 8, 2024
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feat: add solutions to lc problem: No.0477 #2760

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88 changes: 56 additions & 32 deletions solution/0400-0499/0477.Total Hamming Distance/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -45,35 +45,33 @@ HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2

### 方法一：位运算

我们在 $[0, 31]$ 的范围内枚举每一位，对于当前枚举的位 $i$，我们统计所有数字中的第 $i$ 位为 $1$ 的个数 $a$，那么这些数字中的第 $i$ 位为 $0$ 的个数就是 $b = n - a$，其中 $n$ 是数组的长度。这样的话，在第 $i$ 位上的汉明距离之和就是 $a \times b$，我们把所有的位的汉明距离相加即为答案。

时间复杂度 $O(n \times \log M)$，其中 $n$ 和 $M$ 分别是数组的长度和数组中的元素的最大值。空间复杂度 $O(1)$。

<!-- tabs:start -->

```python
class Solution:
def totalHammingDistance(self, nums: List[int]) -> int:
ans = 0
for i in range(31):
a = b = 0
for v in nums:
t = (v >> i) & 1
if t:
a += 1
else:
b += 1
ans, n = 0, len(nums)
for i in range(32):
a = sum(x >> i & 1 for x in nums)
b = n - a
ans += a * b
return ans
```

```java
class Solution {
public int totalHammingDistance(int[] nums) {
int ans = 0;
for (int i = 0; i < 31; ++i) {
int a = 0, b = 0;
for (int v : nums) {
int t = (v >> i) & 1;
a += t;
b += t ^ 1;
int ans = 0, n = nums.length;
for (int i = 0; i < 32; ++i) {
int a = 0;
for (int x : nums) {
a += (x >> i & 1);
}
int b = n - a;
ans += a * b;
}
return ans;
Expand All @@ -85,14 +83,13 @@ class Solution {
class Solution {
public:
int totalHammingDistance(vector<int>& nums) {
int ans = 0;
for (int i = 0; i < 31; ++i) {
int a = 0, b = 0;
for (int& v : nums) {
int t = (v >> i) & 1;
a += t;
b += t ^ 1;
int ans = 0, n = nums.size();
for (int i = 0; i < 32; ++i) {
int a = 0;
for (int x : nums) {
a += x >> i & 1;
}
int b = n - a;
ans += a * b;
}
return ans;
Expand All @@ -101,18 +98,45 @@ public:
```

```go
func totalHammingDistance(nums []int) int {
ans := 0
for i := 0; i < 31; i++ {
a, b := 0, 0
for _, v := range nums {
t := (v >> i) & 1
a += t
b += t ^ 1
func totalHammingDistance(nums []int) (ans int) {
for i := 0; i < 32; i++ {
a := 0
for _, x := range nums {
a += x >> i & 1
}
b := len(nums) - a
ans += a * b
}
return ans
return
}
```

```ts
function totalHammingDistance(nums: number[]): number {
let ans = 0;
for (let i = 0; i < 32; ++i) {
const a = nums.filter(x => (x >> i) & 1).length;
const b = nums.length - a;
ans += a * b;
}
return ans;
}
```

```rust
impl Solution {
pub fn total_hamming_distance(nums: Vec<i32>) -> i32 {
let mut ans = 0;
for i in 0..32 {
let mut a = 0;
for &x in nums.iter() {
a += (x >> i) & 1;
}
let b = (nums.len() as i32) - a;
ans += a * b;
}
ans
}
}
```

Expand Down
90 changes: 57 additions & 33 deletions solution/0400-0499/0477.Total Hamming Distance/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -40,37 +40,35 @@ HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2

## Solutions

### Solution 1
### Solution 1: Bit Manipulation

We enumerate each bit in the range $[0, 31]$. For the current enumerated bit $i$, we count the number of numbers where the $i$-th bit is $1$, denoted as $a$. Therefore, the number of numbers where the $i$-th bit is $0$ is $b = n - a$, where $n$ is the length of the array. In this way, the sum of the Hamming distance on the $i$-th bit is $a \times b$. We add the Hamming distances of all bits to get the answer.

The time complexity is $O(n \times \log M)$, where $n$ and $M$ are the length of the array and the maximum value in the array, respectively. The space complexity is $O(1)$.

<!-- tabs:start -->

```python
class Solution:
def totalHammingDistance(self, nums: List[int]) -> int:
ans = 0
for i in range(31):
a = b = 0
for v in nums:
t = (v >> i) & 1
if t:
a += 1
else:
b += 1
ans, n = 0, len(nums)
for i in range(32):
a = sum(x >> i & 1 for x in nums)
b = n - a
ans += a * b
return ans
```

```java
class Solution {
public int totalHammingDistance(int[] nums) {
int ans = 0;
for (int i = 0; i < 31; ++i) {
int a = 0, b = 0;
for (int v : nums) {
int t = (v >> i) & 1;
a += t;
b += t ^ 1;
int ans = 0, n = nums.length;
for (int i = 0; i < 32; ++i) {
int a = 0;
for (int x : nums) {
a += (x >> i & 1);
}
int b = n - a;
ans += a * b;
}
return ans;
Expand All @@ -82,14 +80,13 @@ class Solution {
class Solution {
public:
int totalHammingDistance(vector<int>& nums) {
int ans = 0;
for (int i = 0; i < 31; ++i) {
int a = 0, b = 0;
for (int& v : nums) {
int t = (v >> i) & 1;
a += t;
b += t ^ 1;
int ans = 0, n = nums.size();
for (int i = 0; i < 32; ++i) {
int a = 0;
for (int x : nums) {
a += x >> i & 1;
}
int b = n - a;
ans += a * b;
}
return ans;
Expand All @@ -98,18 +95,45 @@ public:
```

```go
func totalHammingDistance(nums []int) int {
ans := 0
for i := 0; i < 31; i++ {
a, b := 0, 0
for _, v := range nums {
t := (v >> i) & 1
a += t
b += t ^ 1
func totalHammingDistance(nums []int) (ans int) {
for i := 0; i < 32; i++ {
a := 0
for _, x := range nums {
a += x >> i & 1
}
b := len(nums) - a
ans += a * b
}
return ans
return
}
```

```ts
function totalHammingDistance(nums: number[]): number {
let ans = 0;
for (let i = 0; i < 32; ++i) {
const a = nums.filter(x => (x >> i) & 1).length;
const b = nums.length - a;
ans += a * b;
}
return ans;
}
```

```rust
impl Solution {
pub fn total_hamming_distance(nums: Vec<i32>) -> i32 {
let mut ans = 0;
for i in 0..32 {
let mut a = 0;
for &x in nums.iter() {
a += (x >> i) & 1;
}
let b = (nums.len() as i32) - a;
ans += a * b;
}
ans
}
}
```

Expand Down
13 changes: 6 additions & 7 deletions solution/0400-0499/0477.Total Hamming Distance/Solution.cpp
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Open in desktop
Original file line number Diff line number Diff line change
@@ -1,14 +1,13 @@
class Solution {
public:
int totalHammingDistance(vector<int>& nums) {
int ans = 0;
for (int i = 0; i < 31; ++i) {
int a = 0, b = 0;
for (int& v : nums) {
int t = (v >> i) & 1;
a += t;
b += t ^ 1;
int ans = 0, n = nums.size();
for (int i = 0; i < 32; ++i) {
int a = 0;
for (int x : nums) {
a += x >> i & 1;
}
int b = n - a;
ans += a * b;
}
return ans;
Expand Down
16 changes: 7 additions & 9 deletions solution/0400-0499/0477.Total Hamming Distance/Solution.go
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Open in desktop
Original file line number Diff line number Diff line change
@@ -1,13 +1,11 @@
func totalHammingDistance(nums []int) int {
ans := 0
for i := 0; i < 31; i++ {
a, b := 0, 0
for _, v := range nums {
t := (v >> i) & 1
a += t
b += t ^ 1
func totalHammingDistance(nums []int) (ans int) {
for i := 0; i < 32; i++ {
a := 0
for _, x := range nums {
a += x >> i & 1
}
b := len(nums) - a
ans += a * b
}
return ans
return
}
13 changes: 6 additions & 7 deletions solution/0400-0499/0477.Total Hamming Distance/Solution.java
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,13 +1,12 @@
class Solution {
public int totalHammingDistance(int[] nums) {
int ans = 0;
for (int i = 0; i < 31; ++i) {
int a = 0, b = 0;
for (int v : nums) {
int t = (v >> i) & 1;
a += t;
b += t ^ 1;
int ans = 0, n = nums.length;
for (int i = 0; i < 32; ++i) {
int a = 0;
for (int x : nums) {
a += (x >> i & 1);
}
int b = n - a;
ans += a * b;
}
return ans;
Expand Down
13 changes: 4 additions & 9 deletions solution/0400-0499/0477.Total Hamming Distance/Solution.py
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,13 +1,8 @@
class Solution:
def totalHammingDistance(self, nums: List[int]) -> int:
ans = 0
for i in range(31):
a = b = 0
for v in nums:
t = (v >> i) & 1
if t:
a += 1
else:
b += 1
ans, n = 0, len(nums)
for i in range(32):
a = sum(x >> i & 1 for x in nums)
b = n - a
ans += a * b
return ans
14 changes: 14 additions & 0 deletions solution/0400-0499/0477.Total Hamming Distance/Solution.rs
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Original file line number Diff line number Diff line change
@@ -0,0 +1,14 @@
impl Solution {
pub fn total_hamming_distance(nums: Vec<i32>) -> i32 {
let mut ans = 0;
for i in 0..32 {
let mut a = 0;
for &x in nums.iter() {
a += (x >> i) & 1;
}
let b = (nums.len() as i32) - a;
ans += a * b;
}
ans
}
}
9 changes: 9 additions & 0 deletions solution/0400-0499/0477.Total Hamming Distance/Solution.ts
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Original file line number Diff line number Diff line change
@@ -0,0 +1,9 @@
function totalHammingDistance(nums: number[]): number {
let ans = 0;
for (let i = 0; i < 32; ++i) {
const a = nums.filter(x => (x >> i) & 1).length;
const b = nums.length - a;
ans += a * b;
}
return ans;
}
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