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. Dll注入技术之消息钩子

    转自:黑客反病毒 DLL注入技术之消息钩子注入 消息钩子注入原理是利用Windows 系统中SetWindowsHookEx()这个API,他可以拦截目标进程的消息到指定的DLL中导出的函数,利用这个 ...

  2. Swift下CoreData的使用

    我之前的随笔中有写过一些iOS持久化存储的方法,包含了sqlite.解归档.沙盒存放等等.这些方式中,能够大规模存储,并保持性能的只有使用sqlite了.而这里将记录下Cocoa自身继承的数据库的存储 ...

  3. C#实现拼图游戏

    C#实现<拼图游戏> (下) 原理篇   前言:在 http://www.cnblogs.com/labixiaohei/p/6698887.html 程序设计 之 C#实现<拼图游 ...

  4. 【26.87%】【codeforces 712D】Memory and Scores

    time limit per test2 seconds memory limit per test512 megabytes inputstandard input outputstandard o ...

  5. Parallel.For

    Parallel.For 你可能忽视的一个非常实用的重载方法    说起Parallel.For大家都不会陌生,很简单,不就是一个提供并行功能的for循环吗? 或许大家平时使用到的差不多就是其中最简单 ...

  6. 使用JScript编译指定目录下所有工程

    作者:朱金灿 来源:http://blog.csdn.net/clever101 我遇到这样一个问题:在一个插件工程目录下的插件工程越来越多,因此通过建一个解决方案然后把新增加的工程逐个添加进解决方案 ...

  7. CUDA查询和选取设备信息

    CUDA查询设备信息 CUDA C中的cudaGetDeviceProperties函数可以很方便的获取到设备的信息,函数原型是: cudaError_t CUDARTAPI cudaGetDevic ...

  8. Go程序开发---Go环境配置:CentOS6.5+Go1.8标准包安装

    1.Go安装 1.1Go的三种安装方式 Go有多种安装方式,可以选择自己习惯的方式进行,这里介绍三种安装方式: 1)Go源码安装 2)Go标准包安装 3)第三方工具安装 这里主要介绍下Go标准包在Ce ...

  9. WPF 透明掩码 OpactiyMask

    原文:WPF 透明掩码 OpactiyMask 在WPF中提供了Opacity属性使得元素的所有内容都是透明的.而OpacityMask属性可以使元素的特定区域变成透明. OpacityMask属性接 ...

  10. C#中将字符串转换成Md5值的方法

    原文:C#中将字符串转换成Md5值的方法 版权声明:有问题可联系博主QQ:15577969,大家一起相互交流和学习. https://blog.csdn.net/qq15577969/article/ ...