Question

In a series of numbers, a "mini peak" is a set of 3 numbers where the middle number is strictly greater than its adjacent left and right neighbors. For example, the array:

[ 4, 6, 2, 1 ]

contains 1 mini peak, with 6 in the middle ([ 4, 6, 2]).

Write full and complete C++ program to open a text file named numbers.txt in the working folder, which contains groups of numbers, one per line, with each group starting with "how many" are in the set:

<how_many_numbers>
<number 1>
<number 2>
...
<number n>

For each group of numbers, output the number of mini peaks in that group.  Do not count boundary numbers (at the beginning and end of the array) since they only have one left/right neighbor.

For example, if the file contains these lines:

5
5
4
1
3
2
3
1
1
2

Your program would first read the number 5, indicating there are 5 numbers in the series. Then it would read the next 5 numbers, count the number of mini peaks, then output the result. Then it would read the number 3, indicating there are 3 numbers in the next set, read 3 numbers, and so on.

Requirements:

• Continue reading sets of numbers until EOF.
• Use dynamic memory to allocate enough space for the input. There should be no hard-coded limits on the size of the input.
• You can assume each series has at least 3 numbers.
• Free (delete) any memory that you allocate dynamically.
• Use only the coding structures taught in class.

The output on the above example would be:

Read 5 numbers.  Mini peaks: 1
Read 2 numbers. Mini peaks: 0

Answer 1

If you have any doubts, please give me comment...

#include<iostream>

#include<fstream>

using namespace std;

int main(){

ifstream in;

in.open("numbers.txt");

if(in.fail()){

cout<<"Unable to open file"<<endl;

return 0;

}

int n;

int *arr;

while(!in.eof()){

in>>n;

arr = new int[n];

for(int i=0; i<n; i++)

in>>arr[i];

int peaks = 0;

for(int i=1; i<n-1; i++){

if(arr[i-1]>arr[i] && arr[i]<arr[i+1])

peaks++;

}

delete arr;

cout<<"Read "<<n<<" numbers. Mini peaks: "<<peaks<<endl;

}

in.close();

return 0;

}