Round B 2020 Google Kick Start Bike Tour Checkpoints Problem | Java | Know About

Problem

Li has planned a bike tour through the mountains of Switzerland. His tour consists of N checkpoints, numbered from 1 to N in the order he will visit them. The i-th checkpoint has a height of Hi.

A checkpoint is a peak if:

It is not the 1st checkpoint or the N-th checkpoint, and the height of the checkpoint is strictly greater than the checkpoint immediately before it and the checkpoint immediately after it.

Please help Li find out the number of peaks.

Input

The first line of the input gives the number of test cases, T. T test cases follow. Each test case begins with a line containing the integer N. The second line contains N integers. The i-th integer is Hi.

Output

For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the number of peaks in Li's bike tour.

Sample

In sample case #1, the 2nd checkpoint is a peak.
In sample case #2, there are no peaks.
In sample case #3, the 2nd and 4th checkpoint are peaks.
In sample case #4, there are no peaks.

Analysis

For each of the checkpoints, we can determine if it is a peak in O(1) time by comparing its height to the heights of the checkpoints before and after it.

There are N checkpoints, so the total time complexity of this approach is O(N), which is sufficient for both Test Set 1 and Test Set 2.

Sample Code(C++)
int countPeaks(vector<int> checkpoints) { int peaks = 0; for(int i = 1; i < checkpoints.size() - 1; i++) { if(checkpoints[i-1] < checkpoints[i] && checkpoints[i+1] < checkpoints[i]) { peaks++; } } return peaks; }

PROGRAM


import java.io.*; import java.util.*; public class Solution { public static void main(String[] args) { Scanner sc = new Scanner(new BufferedReader(new InputStreamReader(System.in))); int t = sc.nextInt(); for(int i=1; i<=t;i++) { int N,peek=0; N = sc.nextInt(); int a[] = new int[N]; for(int j=0; j<N; j++) { a[j] = sc.nextInt(); } for(int k=1; k<N-1; k++) { if(a[k-1] < a[k] && a[k+1] < a[k]) peek= peek +1; } System.out.printf("Case #%d: %d\n",i,peek); } } }

Here is the link for Google KickStart Round A 2020 Allocation problem using JAVA.