Task description

A zero-indexed array A consisting of N different integers is given. The array contains integers in the range [1..(N + 1)], which means that exactly one element is missing.

Your goal is to find that missing element.

Write a function:

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

that, given a zero-indexed array A, returns the value of the missing element.

For example, given array A such that:

A[0] = 2 A[1] = 3 A[2] = 1 A[3] = 5

the function should return 4, as it is the missing element.

Assume that:

  • N is an integer within the range [0..100,000];
  • the elements of A are all distinct;
  • each element of array A is an integer within the range [1..(N + 1)].

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: 4 minutes
Code: 10:43:42 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");
import java.util.Arrays;
class Solution {
public int solution(int[] A) {
// write your code in Java SE 8
int res = 0;
int max = A.length;
if(max==0)
return 1;
Arrays.sort(A);
for(int i=0; i<max; i++){
res = (i+1)^A[i];
if(res != 0)
return (i+1);
}
return max+1;
}
}

Codility----PermMissingElem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. react里执行shouldComponentUpdate时返回false的后果

    大家都知道生命周期shouldComponentUpdate返回false时,不会进行后续的渲染,那这个时候state是什么情况呢.我们看一下demo class Toggle extends Rea ...

  2. Android推送服务——百度云推送

    一.推送服务简介 消息推送,顾名思义,是由一方主动发起,而另一方与发起方以某一种方式建立连接并接收消息.在Android开发中,这里的发起方我们把它叫做推送服务器(Push Server),接收方叫做 ...

  3. 【31.58%】【codeforces 682D】Alyona and Strings

    time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...

  4. 拦截和跟踪HTTP请求的主要方法及实现

    一.HTTP 请求的拦截技术可以广泛地应用在反向代理.拦截 Ajax 通信.网页的在线翻译.网站改版重构等方面.而拦截根据位置可以分为服务器端和客户端两大类,客户端拦截借助 JavaScript 脚本 ...

  5. 开发Android项目中使用androidannotations

    前言 最近接手了一个工程,使用了androidannotations,因为以前使用过butterknife所以有似曾相识的感觉,但是有些用法还是不一样,所以就花时间学习了下androidannotat ...

  6. End-to end provisioning of storage clouds

    Embodiments discussed in this disclosure provide an integrated provisioning framework that automates ...

  7. HTC VIVE 虚拟现实眼镜VR游戏体验

    HTC的VIVE入手一段时间了,体验了几个免费的VR游戏,效果还不错,分享一下. 1. VIVE主要部件 VIVE的主要部件有3个,分别是头盔,两个无线控制手柄和两个定位器. 1.1 头盔 头盔整体照 ...

  8. 半监督学习(semi-supervised learning)

    P(x)P(x,y)}⇒P(y|x) 自 P(x) 生成的无标签样本: 自 P(x,y) 生成的标记样本:

  9. Scatter matrix(散布矩阵)

    n 个 m 维的样本,Xm×n=[x1,x2,-,xn],样本均值定义为: x¯=1n∑i=1nxi 散列矩阵定义为如下的半正定矩阵: S=∑j=1n(xj−x¯)(xj−x¯)T=∑j=1n(xj− ...

  10. Android Widget 小工具(两) 使用configure

    添加Widget在此之前需要做一些处理操作,可以使用 配置活动 在上一篇的实现基础上,加上配置活动(configure=activity).这时加入Widget时.会先打开一个Activity,进行配 ...