Task description

A zero-indexed array A consisting of N integers is given. The dominator of array A is the value that occurs in more than half of the elements of A.

For example, consider array A such that

A[0] = 3 A[1] = 4 A[2] = 3 A[3] = 2 A[4] = 3 A[5] = -1 A[6] = 3 A[7] = 3

The dominator of A is 3 because it occurs in 5 out of 8 elements of A (namely in those with indices 0, 2, 4, 6 and 7) and 5 is more than a half of 8.

Write a function

class Solution { public int solution(int[] A); }

that, given a zero-indexed array A consisting of N integers, returns index of any element of array A in which the dominator of A occurs. The function should return −1 if array A does not have a dominator.

Assume that:

  • N is an integer within the range [0..100,000];
  • each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].

For example, given array A such that

A[0] = 3 A[1] = 4 A[2] = 3 A[3] = 2 A[4] = 3 A[5] = -1 A[6] = 3 A[7] = 3

the function may return 0, 2, 4, 6 or 7, as explained above.

Complexity:

  • expected worst-case time complexity is O(N);
  • expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

Solution

 
Programming language used: Java
Total time used: 1 minutes
 
Code: 01:47:41 UTC, java, final, score:  100
// you can also use imports, for example:
// import java.util.*; // you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message"); class Solution {
public int solution(int[] A) {
// write your code in Java SE 8
int size = 0, leader = -1;
for(int i=0; i<A.length;i++) {
if(size == 0) {
leader = A[i];
}
if(leader == A[i]) {
size++;
} else {
size--;
}
}
if(size == 0) return -1;
int index = 0, count = 0;
for(int i=0; i<A.length; i++) {
if(leader == A[i]) {
index = i;
count++;
}
}
if(count*2 <= A.length) return -1;
return index;
}
}
https://codility.com/demo/results/trainingK4VATM-6K8/

Codility---Dominator的更多相关文章

  1. codility上的练习 (1)

    codility上面添加了教程.目前只有lesson 1,讲复杂度的……里面有几个题, 目前感觉题库的题简单. tasks: Frog-Jmp: 一只青蛙,要从X跳到Y或者大于等于Y的地方,每次跳的距 ...

  2. Codility NumberSolitaire Solution

    1.题目: A game for one player is played on a board consisting of N consecutive squares, numbered from ...

  3. codility flags solution

    How to solve this HARD issue 1. Problem: A non-empty zero-indexed array A consisting of N integers i ...

  4. GenomicRangeQuery /codility/ preFix sums

    首先上题目: A DNA sequence can be represented as a string consisting of the letters A, C, G and T, which ...

  5. *[codility]Peaks

    https://codility.com/demo/take-sample-test/peaks http://blog.csdn.net/caopengcs/article/details/1749 ...

  6. *[codility]Country network

    https://codility.com/programmers/challenges/fluorum2014 http://www.51nod.com/onlineJudge/questionCod ...

  7. *[codility]AscendingPaths

    https://codility.com/programmers/challenges/magnesium2014 图形上的DP,先按照路径长度排序,然后依次遍历,状态是使用到当前路径为止的情况:每个 ...

  8. *[codility]MaxDoubleSliceSum

    https://codility.com/demo/take-sample-test/max_double_slice_sum 两个最大子段和相拼接,从前和从后都扫一遍.注意其中一段可以为0.还有最后 ...

  9. *[codility]Fish

    https://codility.com/demo/take-sample-test/fish 一开始习惯性使用单调栈,后来发现一个普通栈就可以了. #include <stack> us ...

  10. *[codility]CartesianSequence

    https://codility.com/programmers/challenges/upsilon2012 求笛卡尔树的高度,可以用单调栈来做. 维持一个单调递减的栈,每次进栈的时候记录下它之后有 ...

随机推荐

  1. Ubuntu server使用命令行上板VPNclient

    Ubuntu server使用命令行上板VPNclient VPN,虚拟专用网络,这个技术还是非常有用的.近期笔者參与的项目中就使用上了VPN,大概情况是这种.有两个开发团队,在异地,代码服务器在深圳 ...

  2. 数据竞赛利器 —— xgboost 学习清单

    1. 入门大全 xgboost 作者给出的一份完备的使用 xgboost 进行数据分析的完整示例代码:A walk through python example for UCI Mushroom da ...

  3. 关于babel和babel-polyfill

    使用babel-cli命令babel xx -d xx把一个js文件转成了ES5的,并在package.json里加了"babel-polyfill": "^6.23.0 ...

  4. HSSFWorkBooK用法 ---Excel表的导出和设置

    public ActionResult excelPrint() { HSSFWorkbook workbook = new HSSFWorkbook();// 创建一个Excel文件 HSSFShe ...

  5. WPF 使用 Pandoc 把 Markdown 转 Docx

    原文:WPF 使用 Pandoc 把 Markdown 转 Docx 本文告诉大家如何通过 WPF 使用 Pandoc 把 Markdown 转 Docx 文件 在之前有文章使用 Pandoc 把 M ...

  6. ubuntu grub 操作

    系统开机时,按住 shift 进入 grub 1. 什么是 Grub GNU GRUB(GRand Unified Bootloader 简称"GRUB")是一个来自GNU项目的多 ...

  7. HTML:描述语义

    一.HTML HTML:Hypertext Markup Launguage,超文本标记语言,是网页的就文件格式,用于描述网页语义. 二.HTML骨架 DTD手册:http://www.w3schoo ...

  8. 使用WPF实现3D场景[二]

    原文:使用WPF实现3D场景[二] 在上一篇的文章里我们知道如何构造一个简单的三维场景,这次的课程我将和大家一起来研究如何用代码,完成对建立好了的三维场景的观察. 首先看一下DEMO的界面:     ...

  9. JavaScript知识树

  10. android 玩愤怒的小鸟等游戏的时候全屏TP失败

    1.tp driver的tpd_down()和tpd_up()函数不需要进行报告id号码.自己主动顶级赛: 2.tpd_up()功能只需要报告BTN_TOUCH和mt_sync信息,其他信息未报告,如 ...