Task description

A non-empty zero-indexed array A consisting of N integers is given. The consecutive elements of array A represent consecutive cars on a road.

Array A contains only 0s and/or 1s:

  • 0 represents a car traveling east,
  • 1 represents a car traveling west.

The goal is to count passing cars. We say that a pair of cars (P, Q), where 0 ≤ P < Q < N, is passing when P is traveling to the east and Q is traveling to the west.

For example, consider array A such that:

A[0] = 0 A[1] = 1 A[2] = 0 A[3] = 1 A[4] = 1

We have five pairs of passing cars: (0, 1), (0, 3), (0, 4), (2, 3), (2, 4).

Write a function:

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

that, given a non-empty zero-indexed array A of N integers, returns the number of pairs of passing cars.

The function should return −1 if the number of pairs of passing cars exceeds 1,000,000,000.

For example, given:

A[0] = 0 A[1] = 1 A[2] = 0 A[3] = 1 A[4] = 1

the function should return 5, as explained above.

Assume that:

  • N is an integer within the range [1..100,000];
  • each element of array A is an integer that can have one of the following values: 0, 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
Code: 02:08:07 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 zeroCnt=0, oneCnt =0;
for(int i=0; i<A.length; i++) {
if(A[i] == 0) {
zeroCnt += 1;
} else if(A[i] == 1) {
oneCnt += zeroCnt;
}
if(oneCnt > 1000000000)
return -1;
}
return oneCnt;
}
}
https://codility.com/demo/results/training8U5YGT-RJR/

Codility----PassingCars的更多相关文章

  1. Codility NumberSolitaire Solution

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

  2. codility flags solution

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

  3. 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 ...

  4. *[codility]Peaks

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

  5. *[codility]Country network

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

  6. *[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 ...

  7. *[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 ...

  8. *[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 ...

  9. *[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 ...

  10. [codility]CountDiv

    https://codility.com/demo/take-sample-test/count_div 此题比较简单,是在O(1)时间里求区间[A,B]里面能被K整除的数字,那么就计算一下就能得到. ...

随机推荐

  1. 给定正整数n,计算出n个元素的集合{1,2,....,n}能够划分为多少个不同的非空集合

    给定正整数n,计算出n个元素的集合{1,2,....,n}能够划分为多少个不同的非空集合 附源码: #include<iostream> using namespace std; int ...

  2. WPF去除边框的方法

    原文:WPF去除边框的方法 版权声明:本文为博主原创文章,未经博主允许不得转载. https://blog.csdn.net/yangsen600/article/details/81978125 W ...

  3. 初探js

    第一章   1.JS的位置 1-1.行间 1-2.内嵌 1-3.外联 2.JS的标签位置 页面中的代码在一般情况下会按从上到下的顺序,从左往右的顺序执行. 因此当JS放在了元素上面的时候,就不能正常执 ...

  4. Java--分布式系统高并发解决方案

    对于我们开发的网站,如果网站的访问量非常大的话,那么我们就需要考虑相关的并发访问问题了.而并发问题是绝大部分的程序员头疼的问题, 但话又说回来了,既然逃避不掉,那我们就坦然面对吧~今天就让我们一起来研 ...

  5. 手把手教你启用Win10的Linux子系统(超详细)

    原文:手把手教你启用Win10的Linux子系统(超详细) 版权声明:转载请保留出处,谢谢! https://blog.csdn.net/zhangdongren/article/details/82 ...

  6. Opencv中SVM样本训练、归类流程及实现

    支持向量机(SVM)中最核心的是什么?个人理解就是前4个字--"支持向量",一旦在两类或多累样本集中定位到某些特定的点作为支持向量,就可以依据这些支持向量计算出来分类超平面,再依据 ...

  7. 跟我学ASP.NET MVC之九:SportsStrore产品管理

    摘要: 在这篇文章中,我将继续完成SportsStore应用程序,让站点管理者可以管理产品列表.我将添加创建.修改和删除产品功能. 本篇文章将分模块的方式,逐个介绍SportsStore站点管理功能的 ...

  8. C. Adidas vs Adivon

    C. Adidas vs Adivon Time Limit: 1000ms Case Time Limit: 1000ms Memory Limit: 65536KB 64-bit integer ...

  9. Analysis of variance(ANOVA)

    方差分析,也称为"变异数分析",用于两个及两个以上样本均值(group means)差别的显著性检验.在 ANOVA 的环境下,一个观测得到的方差视为是由不同方差的源组合而成.

  10. js到字符串数组,实现阵列成一个字符串

    数组字符串(阵列元件与字符串连接) var a, b; a = new Array(0,1,2,3,4); b = a.join("-");   字符串转数组(根据一个字符串被分成 ...