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
- 给tbody加垂直滚动条的具体思路
[给tbody加垂直滚动条的具体思路] 给tbody加垂直滚动条的思路就是把tbody设置成display:block,然后就对其高度设置一个固定值,overflow设置成auto即可 参考:http ...
- Java反射、动态加载(将java类名、方法、方法参数当做参数传递,执行方法)
需求:将java类名.方法.方法参数当做参数传递,执行方法.可以用java的动态加载实现 反射的过程如下: 第一步:通过反射找到类并创建实例(classname为要实例化的类名,由pack ...
- opsmanage 自动化运维管理平台
关闭防火墙.selinux 更换阿里云 yum源 依赖环境 yum install -y epel-releaseyum install vim net-tools nmon htop rsync t ...
- Centos 7 下 Mysql 5.7 Galera Cluster 集群部署
一.介绍 传统架构的使用,一直被人们所诟病,因为MySQL的主从模式,天生的不能完全保证数据一致,很多大公司会花很大人力物力去解决这个问题,而效果却一般,可以说,只能是通过牺牲性能,来获得数据一致性 ...
- 贪吃蛇 Java实现(一)
贪吃蛇 Java实现 1.面向对象思想 1.创建antition包它是包含sanke Ground Food类 2.创建Controller包controller类 3.创建Game包有game类 ...
- Centos + Maven + Jenkins
下载 JDKwget --no-check-certificate --no-cookie --header "Cookie: oraclelicense=accept-secureback ...
- Householder矩阵,Givens矩阵
householder 矩阵相当于对某一空间中的元素(向量.矩阵)进行镜像变换,但是模值并不发生变化. H=I-2uuT householder矩阵有几个重要的性质: 1 : H-1 = H 2: ...
- TZOJ 4621 Grammar(STL模拟)
描述 Our strings only contain letters(maybe the string contains nothing). Now we define the production ...
- 【环境配置】本地配置sublime text以及和远程linux设置sftp
工具: sublime text 2(mac版) 远程linux(centos 7系) securCRT(for mac) [本地安装并配置securCRT(for mac)] 关于配置: 1.解决终 ...
- gradle项目与maven项目互转
maven to gradle 在maven项目根目录下执行命令: gradle init --type pom 当然你得先下载Gradle,配置完环境变量. gradle to maven grad ...