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feat: add solutions to lc problem: No.1010
No.1010.Pairs of Songs With Total Durations Divisible by 60
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6 files changed

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

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：计数 + 枚举**
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我们可以用一个哈希表或数组 `cnt` 记录当前已经遍历过的歌曲的持续时间 `time[i]` 的数量。
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枚举当前遍历到的歌曲的持续时间 `time[i]`，假设其为 `t`，那么我们只需要枚举 `60` 的倍数 `s`，并统计 `cnt[s - t]` 即可。然后将 `cnt[t]` 的值加 `1`。
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时间复杂度 $O(n \times C)$，空间复杂度 $O(M)$。其中 $n$ 为数组 `time` 的长度，而 $C$ 和 $M$ 分别为数组 `time` 中的最大值以及 `60` 的倍数的个数。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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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 = defaultdict(int)
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ans = 0
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for t in time:
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s = 60
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for _ in range(17):
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ans += cnt[s - t]
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s += 60
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cnt[t] += 1
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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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[501];
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int ans = 0;
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for (int t : time) {
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int s = 60;
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for (int i = 0; i < 17; ++i) {
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if (s - t >= 0 && s - t < cnt.length) {
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ans += cnt[s - t];
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}
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s += 60;
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}
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cnt[t]++;
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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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### **C++**
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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) {
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int cnt[501]{};
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int ans = 0;
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for (int& t : time) {
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int s = 60;
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for (int i = 0; i < 17; ++i) {
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if (s - t >= 0 && s - t < 501) {
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ans += cnt[s - t];
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}
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s += 60;
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}
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cnt[t]++;
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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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### **Go**
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```go
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func numPairsDivisibleBy60(time []int) (ans int) {
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cnt := [501]int{}
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for _, t := range time {
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s := 60
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for i := 0; i < 17; i++ {
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if s-t >= 0 && s-t < 501 {
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ans += cnt[s-t]
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}
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s += 60
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}
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cnt[t]++
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}
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return
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}
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```
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### **...**

‎solution/1000-1099/1010.Pairs of Songs With Total Durations Divisible by 60/README_EN.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 = defaultdict(int)
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ans = 0
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for t in time:
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s = 60
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for _ in range(17):
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ans += cnt[s - t]
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s += 60
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cnt[t] += 1
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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 {
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public int numPairsDivisibleBy60(int[] time) {
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int[] cnt = new int[501];
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int ans = 0;
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for (int t : time) {
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int s = 60;
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for (int i = 0; i < 17; ++i) {
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if (s - t >= 0 && s - t < cnt.length) {
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ans += cnt[s - t];
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}
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s += 60;
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}
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cnt[t]++;
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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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### **C++**
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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) {
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int cnt[501]{};
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int ans = 0;
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for (int& t : time) {
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int s = 60;
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for (int i = 0; i < 17; ++i) {
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if (s - t >= 0 && s - t < 501) {
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ans += cnt[s - t];
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}
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s += 60;
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}
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cnt[t]++;
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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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### **Go**
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```go
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func numPairsDivisibleBy60(time []int) (ans int) {
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cnt := [501]int{}
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for _, t := range time {
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s := 60
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for i := 0; i < 17; i++ {
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if s-t >= 0 && s-t < 501 {
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ans += cnt[s-t]
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}
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s += 60
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}
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cnt[t]++
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}
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return
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}
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```
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### **...**

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

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class Solution {
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public:
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int numPairsDivisibleBy60(vector<int>& time) {
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int cnt[501]{};
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int ans = 0;
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for (int& t : time) {
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int s = 60;
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for (int i = 0; i < 17; ++i) {
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if (s - t >= 0 && s - t < 501) {
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ans += cnt[s - t];
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}
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s += 60;
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}
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cnt[t]++;
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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/Solution.go

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func numPairsDivisibleBy60(time []int) (ans int) {
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cnt := [501]int{}
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for _, t := range time {
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s := 60
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for i := 0; i < 17; i++ {
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if s-t >= 0 && s-t < 501 {
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ans += cnt[s-t]
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}
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s += 60
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}
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cnt[t]++
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}
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return
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}

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

Lines changed: 9 additions & 19 deletions
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class Solution {
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public int numPairsDivisibleBy60(int[] time) {
3-
Arrays.sort(time);
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int[] cnt = new int[501];
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int ans = 0;
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for (int i = 0; i < time.length - 1; i++) {
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int num = (time[i] + time[time.length - 1]) / 60;
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while (num > 0) {
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int key = num * 60;
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int index = Arrays.binarySearch(time, i + 1, time.length, key - time[i]);
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if (index >= 0) {
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int temp = index;
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ans++;
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while (++temp < time.length && time[temp] == time[index]) {
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ans++;
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}
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temp = index;
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while (--temp > i && time[temp] == time[index]) {
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ans++;
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}
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for (int t : time) {
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int s = 60;
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for (int i = 0; i < 17; ++i) {
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if (s - t >= 0 && s - t < cnt.length) {
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ans += cnt[s - t];
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}
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num--;
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s += 60;
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}
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cnt[t]++;
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}
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return ans;
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}
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}
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}

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

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class Solution:
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def numPairsDivisibleBy60(self, time: List[int]) -> int:
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cnt = defaultdict(int)
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ans = 0
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for t in time:
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s = 60
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for _ in range(17):
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ans += cnt[s - t]
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s += 60
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cnt[t] += 1
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return ans

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