Pairs of Songs With Total Durations Divisible by 60 LT1010
In a list of songs, the i-th song has a duration of time[i] seconds.
Return the number of pairs of songs for which their total duration in seconds is divisible by 60. Formally, we want the number of indices i < j with (time[i] + time[j]) % 60 == 0.
Example 1:
Input: [30,20,150,100,40]
Output: 3
Explanation: Three pairs have a total duration divisible by 60:
(time[0] = 30, time[2] = 150): total duration 180
(time[1] = 20, time[3] = 100): total duration 120
(time[1] = 20, time[4] = 40): total duration 60
Example 2:
Input: [60,60,60]
Output: 3
Explanation: All three pairs have a total duration of 120, which is divisible by 60.
Idea 1. Similar to Two Sum LT1, map with modular arithmetic
class Solution {
public int numPairsDivisibleBy60(int[] time) {
Map<Integer, Integer> record = new HashMap<>();
int count = 0;
for(int val: time) {
val = val%60;
count += record.getOrDefault((60 - val)%60, 0);
record.put(val, record.getOrDefault(val, 0) + 1);
}
return count;
}
}
array used as map
class Solution {
public int numPairsDivisibleBy60(int[] time) {
int[] record = new int[60];
int count = 0;
for(int val: time) {
val = val%60;
count += record[(60 - val)%60];
++record[val];
}
return count;
}
}
Idea 1a. count pairs, preprose the array first
class Solution {
public int numPairsDivisibleBy60(int[] time) {
int[] record = new int[60];
for(int val: time) {
val = val%60;
++record[val];
}
int count = 0;
if(record[0] > 0) {
count += record[0] * (record[0]-1)/2;
}
if(record[30] > 0) {
count += record[30] * (record[30]-1)/2;
}
for(int i = 1; i < 30; ++i) {
count += record[i]*record[60-i];
}
return count;
}
}
Note:
1 <= time.length <= 600001 <= time[i] <= 500
Pairs of Songs With Total Durations Divisible by 60 LT1010的更多相关文章
- [Swift]LeetCode1010. 总持续时间可被 60 整除的歌曲 | Pairs of Songs With Total Durations Divisible by 60
In a list of songs, the i-th song has a duration of time[i] seconds. Return the number of pairs of s ...
- 128th LeetCode Weekly Contest Pairs of Songs With Total Durations Divisible by 60
In a list of songs, the i-th song has a duration of time[i] seconds. Return the number of pairs of s ...
- 【leetcode】1013. Pairs of Songs With Total Durations Divisible by 60
题目如下: In a list of songs, the i-th song has a duration of time[i] seconds. Return the number of pair ...
- 【LeetCode】1013. Pairs of Songs With Total Durations Divisible by 60 解题报告(Python)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.c ...
- 1013. Pairs of Songs With Total Durations Divisible by 60总持续时间可被 60 整除的歌曲
网址:https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/submissions/ 参考 ...
- Swift LeetCode 目录 | Catalog
请点击页面左上角 -> Fork me on Github 或直接访问本项目Github地址:LeetCode Solution by Swift 说明:题目中含有$符号则为付费题目. 如 ...
- Weekly Contest 128
1012. Complement of Base 10 Integer Every non-negative integer N has a binary representation. For e ...
- 【LEETCODE】51、数组分类,简单级别,题目:581,830,1010,665
package y2019.Algorithm.array; /** * @ClassName FindUnsortedSubarray * @Description TODO 581. Shorte ...
- Codeforces #364 DIV2
~A题 A. Cards time limit per test 1 second memory limit per test 256 megabytes input standard input ...
随机推荐
- azkaban编译以及安装(调度系统)
编译源码 下载azkaban源码 git clone https://github.com/azkaban/azkaban.git jdk要求是1.8以上版本 export JAVA_HOME=/ ...
- package.json bin
[package.json bin] 1.bin field in your package.json which is a map of command name to local file nam ...
- ncat的使用
由于netcat的缺陷,所以有了升级版ncat,弥补了netcat的一些不足. ncat是nmap工具包的一个工具. 服务器端 ncat -c bash --allow 192.168.1.119 - ...
- Windows消息循环
首先理解一句话:“Windows”向应用程序发送了一条消息.这里是指Windows调用了该程序内部的一个函数. 当UpdateWindow被调用后,新建的窗口在屏幕便完全可见了.此时,Windows会 ...
- ORA-01555 snapshot too old
假设有一张6000万行数据的testdb表,预计testdb全表扫描1次需要2个小时,参考过程如下: 1.在1点钟,用户A发出了select * from testdb;此时不管将来testdb怎么变 ...
- Forms.WebBrowser与Controls.WebBrowser区别
Forms.WebBrowser与Controls.WebBrowser区别 Forms.WebBrowser有ScrollBarsEnabled 属性,即窗口滚动条,可以设置为false即可: Co ...
- idea使用maven打包jar包
1.在pom.xml中加入以下内容: <?xml version="1.0" encoding="UTF-8"?> <project xmln ...
- Linux 向文件末尾追加命令
Linux 向文件末尾追加命令 //echo后边用单引号包围要添加的内容 echo 'add content'>>/home/data/test.sh 注意:>> 是追加 ec ...
- e-olymp Problem11 Big accuracy
传送门:点我 Big accuracy The rational fraction m/n is given. Write it in the decimal notation with k digi ...
- 《学习OpenCV(中文版)》
<模式识别中文版(希)西奥多里蒂斯> <学习OpenCV(中文版)> 矩阵计算 英文版 第四版 Matrix Computations OpenCV 3.x with Pyth ...