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feat: add solutions for lc No.1009,1010 (#5073)
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Lines changed: 134 additions & 287 deletions

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‎solution/1000-1099/1009.Complement of Base 10 Integer/README.md‎

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@@ -179,6 +179,24 @@ impl Solution {
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}
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```
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#### C#
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```cs
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public class Solution {
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public int BitwiseComplement(int n) {
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if (n == 0) {
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return 1;
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}
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int ans = 0, i = 0;
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while (n != 0) {
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ans |= ((n & 1) ^ 1) << (i++);
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n >>= 1;
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}
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return ans;
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/1000-1099/1009.Complement of Base 10 Integer/README_EN.md‎

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}
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```
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#### C#
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```cs
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public class Solution {
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public int BitwiseComplement(int n) {
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if (n == 0) {
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return 1;
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}
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int ans = 0, i = 0;
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while (n != 0) {
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ans |= ((n & 1) ^ 1) << (i++);
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n >>= 1;
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}
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return ans;
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/1000-1099/1009.Complement of Base 10 Integer/Solution.cs‎

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public class Solution {
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public int BitwiseComplement(int n) {
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if (n == 0) {
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return 1;
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}
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int ans = 0, i = 0;
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while (n != 0) {
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ans |= ((n & 1) ^ 1) << (i++);
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n >>= 1;
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}
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return ans;
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}
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}

‎solution/1000-1099/1010.Pairs of Songs With Total Durations Divisible by 60/README.md‎

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#### Python3
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```python
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class Solution:
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def numPairsDivisibleBy60(self, time: List[int]) -> int:
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cnt = Counter(t % 60 for t in time)
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ans = sum(cnt[x] * cnt[60 - x] for x in range(1, 30))
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ans += cnt[0] * (cnt[0] - 1) // 2
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ans += cnt[30] * (cnt[30] - 1) // 2
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return ans
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```
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#### Java
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```java
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class Solution {
91-
public int numPairsDivisibleBy60(int[] time) {
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int[] cnt = new int[60];
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for (int t : time) {
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++cnt[t % 60];
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}
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int ans = 0;
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for (int x = 1; x < 30; ++x) {
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ans += cnt[x] * cnt[60 - x];
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}
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ans += (long) cnt[0] * (cnt[0] - 1) / 2;
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ans += (long) cnt[30] * (cnt[30] - 1) / 2;
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return ans;
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}
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}
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```
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#### C++
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```cpp
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class Solution {
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public:
112-
int numPairsDivisibleBy60(vector<int>& time) {
113-
int cnt[60]{};
114-
for (int& t : time) {
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++cnt[t % 60];
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}
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int ans = 0;
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for (int x = 1; x < 30; ++x) {
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ans += cnt[x] * cnt[60 - x];
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}
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ans += 1LL * cnt[0] * (cnt[0] - 1) / 2;
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ans += 1LL * cnt[30] * (cnt[30] - 1) / 2;
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return ans;
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}
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};
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```
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#### Go
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```go
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func numPairsDivisibleBy60(time []int) (ans int) {
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cnt := [60]int{}
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for _, t := range time {
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cnt[t%60]++
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}
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for x := 1; x < 30; x++ {
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ans += cnt[x] * cnt[60-x]
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}
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ans += cnt[0] * (cnt[0] - 1) / 2
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ans += cnt[30] * (cnt[30] - 1) / 2
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return
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}
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```
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#### TypeScript
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```ts
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function numPairsDivisibleBy60(time: number[]): number {
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const cnt: number[] = new Array(60).fill(0);
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for (const t of time) {
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++cnt[t % 60];
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}
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let ans = 0;
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for (let x = 1; x < 30; ++x) {
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ans += cnt[x] * cnt[60 - x];
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}
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ans += (cnt[0] * (cnt[0] - 1)) / 2;
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ans += (cnt[30] * (cnt[30] - 1)) / 2;
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return ans;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- solution:start -->
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### 方法二
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def numPairsDivisibleBy60(self, time: List[int]) -> int:
@@ -253,6 +155,24 @@ function numPairsDivisibleBy60(time: number[]): number {
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn num_pairs_divisible_by60(time: Vec<i32>) -> i32 {
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let mut cnt = [0i32; 60];
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let mut ans: i32 = 0;
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for mut x in time {
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x %= 60;
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let y = (60 - x) % 60;
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ans += cnt[y as usize];
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cnt[x as usize] += 1;
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}
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ans
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/1000-1099/1010.Pairs of Songs With Total Durations Divisible by 60/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Math + Counting
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def numPairsDivisibleBy60(self, time: List[int]) -> int:
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cnt = Counter(t % 60 for t in time)
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ans = sum(cnt[x] * cnt[60 - x] for x in range(1, 30))
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ans += cnt[0] * (cnt[0] - 1) // 2
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ans += cnt[30] * (cnt[30] - 1) // 2
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return ans
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```
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If the sum of a pair $(a, b)$ is divisible by $60$, i.e., $(a + b) \bmod 60 = 0$, then $(a \bmod 60 + b \bmod 60) \bmod 60 = 0$. Let $x = a \bmod 60$ and $y = b \bmod 60$, then $(x + y) \bmod 60 = 0$, which means $y = (60 - x) \bmod 60$.
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#### Java
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```java
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class Solution {
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public int numPairsDivisibleBy60(int[] time) {
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int[] cnt = new int[60];
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for (int t : time) {
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++cnt[t % 60];
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}
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int ans = 0;
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for (int x = 1; x < 30; ++x) {
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ans += cnt[x] * cnt[60 - x];
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}
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ans += (long) cnt[0] * (cnt[0] - 1) / 2;
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ans += (long) cnt[30] * (cnt[30] - 1) / 2;
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return ans;
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}
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}
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```
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Therefore, we can iterate over the song list and use an array $cnt$ of length $60$ to record the number of occurrences of each remainder $x$. For the current $x$, if there exists a remainder $y = (60 - x) \bmod 60$ in array $cnt$, we add $cnt[y]$ to the answer. Then we increment the count of $x$ in array $cnt$ by $1$. We continue iterating until the entire song list has been traversed.
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#### C++
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After the iteration, we get the number of song pairs that satisfy the condition.
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```cpp
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class Solution {
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public:
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int numPairsDivisibleBy60(vector<int>& time) {
103-
int cnt[60]{};
104-
for (int& t : time) {
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++cnt[t % 60];
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}
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int ans = 0;
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for (int x = 1; x < 30; ++x) {
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ans += cnt[x] * cnt[60 - x];
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}
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ans += 1LL * cnt[0] * (cnt[0] - 1) / 2;
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ans += 1LL * cnt[30] * (cnt[30] - 1) / 2;
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return ans;
114-
}
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};
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```
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#### Go
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```go
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func numPairsDivisibleBy60(time []int) (ans int) {
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cnt := [60]int{}
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for _, t := range time {
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cnt[t%60]++
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}
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for x := 1; x < 30; x++ {
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ans += cnt[x] * cnt[60-x]
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}
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ans += cnt[0] * (cnt[0] - 1) / 2
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ans += cnt[30] * (cnt[30] - 1) / 2
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return
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}
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```
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#### TypeScript
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```ts
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function numPairsDivisibleBy60(time: number[]): number {
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const cnt: number[] = new Array(60).fill(0);
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for (const t of time) {
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++cnt[t % 60];
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}
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let ans = 0;
144-
for (let x = 1; x < 30; ++x) {
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ans += cnt[x] * cnt[60 - x];
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}
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ans += (cnt[0] * (cnt[0] - 1)) / 2;
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ans += (cnt[30] * (cnt[30] - 1)) / 2;
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return ans;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- solution:start -->
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### Solution 2
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The time complexity is $O(n)$ and the space complexity is $O(C)$, where $n$ is the length of the song list and $C$ is the number of possible remainders, here $C = 60$.
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<!-- tabs:start -->
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn num_pairs_divisible_by60(time: Vec<i32>) -> i32 {
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let mut cnt = [0i32; 60];
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let mut ans: i32 = 0;
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for mut x in time {
164+
x %= 60;
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let y = (60 - x) % 60;
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ans += cnt[y as usize];
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cnt[x as usize] += 1;
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}
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ans
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/1000-1099/1010.Pairs of Songs With Total Durations Divisible by 60/Solution.cpp‎

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public:
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int numPairsDivisibleBy60(vector<int>& time) {
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int cnt[60]{};
5-
for (int& t : time) {
6-
++cnt[t % 60];
7-
}
85
int ans = 0;
9-
for (int x = 1; x < 30; ++x) {
10-
ans += cnt[x] * cnt[60 - x];
6+
for (int x : time) {
7+
x %= 60;
8+
int y = (60 - x) % 60;
9+
ans += cnt[y];
10+
++cnt[x];
1111
}
12-
ans += 1LL * cnt[0] * (cnt[0] - 1) / 2;
13-
ans += 1LL * cnt[30] * (cnt[30] - 1) / 2;
1412
return ans;
1513
}
16-
};
14+
};

‎solution/1000-1099/1010.Pairs of Songs With Total Durations Divisible by 60/Solution.go‎

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func numPairsDivisibleBy60(time []int) (ans int) {
22
cnt := [60]int{}
3-
for _, t := range time {
4-
cnt[t%60]++
3+
for _, x := range time {
4+
x %= 60
5+
y := (60 - x) % 60
6+
ans += cnt[y]
7+
cnt[x]++
58
}
6-
for x := 1; x < 30; x++ {
7-
ans += cnt[x] * cnt[60-x]
8-
}
9-
ans += cnt[0] * (cnt[0] - 1) / 2
10-
ans += cnt[30] * (cnt[30] - 1) / 2
119
return
12-
}
10+
}

‎solution/1000-1099/1010.Pairs of Songs With Total Durations Divisible by 60/Solution.java‎

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class Solution {
22
public int numPairsDivisibleBy60(int[] time) {
33
int[] cnt = new int[60];
4-
for (int t : time) {
5-
++cnt[t % 60];
6-
}
74
int ans = 0;
8-
for (int x = 1; x < 30; ++x) {
9-
ans += cnt[x] * cnt[60 - x];
5+
for (int x : time) {
6+
x %= 60;
7+
int y = (60 - x) % 60;
8+
ans += cnt[y];
9+
++cnt[x];
1010
}
11-
ans += (long) cnt[0] * (cnt[0] - 1) / 2;
12-
ans += (long) cnt[30] * (cnt[30] - 1) / 2;
1311
return ans;
1412
}
15-
}
13+
}

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