Decide if a sequence (integer numbers) is monotonically increasing - c++

Question

#include <iostream>
using namespace std;
int main()
{
int n;
bool error;
do
{
cout << "How many numbers would you like to put in this sequence? "<< endl;
cin >> n;
error = cin.fail() || cin.peek() != '\n' || (n < 0);
if (error)
{
cout << "Please input again" << endl;
}
cin.clear();
cin.ignore(999,'\n');
}while(error);
int a[n];
cout << "Please input the numbers in your sequence." << endl;
for(int i = 0; i < n; i++)
{
do
{
cin >> a[i];
error = cin.fail() || cin.peek() != '\n';
if (error)
{
cout << "Please input again" << endl;
}
cin.clear();
cin.ignore(999,'\n');
}while(error);
}
for(int i=0; i<n; i++)
{
if (a[i]<a[i+1])
{
cout << "The sequence is monotonically increasing." << endl;
}
else
{
cout << "The sequence is not monotonically increasing." << endl;
}
}
}
Hello guys, I want to check if this sequence (integer numbers) is monotonically increasing or not.
And I do not know which way I should use, so I just choose to use the 'For' loop, which it just compare two numbers each time, which its not comparing all numbers together.
Could you please help me?
Thanks ::>

Answer 1

Your initial attempt using the for loop is on the right track. The key here to fix the message being printed multiple times is to move the print statement (cout << ...) outside the loop. Now, you may ask, "How do I do that?" The proper way to approach it is to negate the condition: how do you know that a sequence is not monotonically increasing? (Based on your code, I will assume you are looking for a strictly increasing sequence.)
You know a sequence is not monotonically increasing if the current term in the sequence is greater than or equal to the next term in the sequence. Once you find one such pair of terms, you immediately know the sequence does not satisfy that property and can exit the loop early. Thus, our tactic to solving the problem is to first assume the condition is true, check if the condition is ever not true for any two elements, and then to check the condition again and print the result.
Thus, your final for loop should look like this:
bool is_increasing = true; // Begin by assuming the condition is true.
for (int i = 0; i < n - 1; ++i) { // Note the condition. More info below.
if (a[i] >= a[i + 1]) { // Here we check if the condition is NOT met.
is_increasing = false;
break; // This exits the loop. We don't have to keep going because
// of the way monotonically increasing is defined.
}
}
// Now we check the condition and present the result.
if (is_increasing) {
cout << "The sequence is monotonically increasing." << endl;
}
else {
cout << "The sequence is not monotonically increasing." << endl;
}
There is another issue with your code. You want to be very careful about not attempting to read past the end of an array. If we trace over the for loop you wrote, what happens when i = n - 1? Clearly, n - 1 < n, so the body of the for loop runs. Then we run into the if statement: if a[i] < a[i + 1]. Okay, the first part a[i] is good because that is really asking for a[n - 1] which is legal, but what about a[i + 1]? Whoops, if n is supposed to be size of the array, then a[n] does not exist! (Remember, arrays are zero-based!) Off-by-one errors are ubiquitous in programming, but if you take the time to trace through the code, you can catch most of them. (Notice how I in the above fixed the issue by changing the loop condition.)
Edit:
As pointed out by paddy in the comments, you do not want to use an array here because that is nonstandard c++. Instead, you want to use the std::vector (or just vector if you are using namespace std) class. Vectors are very similar to arrays, but they can grow and shrink as needed at runtime. The syntax for using them is also a little different. An example of how to use vector follows:
std::vector<int> a {}; // This constructs an empty vector of ints.
a.push_back(1); // This puts 1 at the end of the vector. In this case,
// you can access this number at location 0.
a.push_back(2); // This puts 2 at the end.
cout << print(a.at(0)) << endl; // This will print 1.
cout << print(a.at(1)) << endl; // This will print 2.
You can also use [] like with arrays to access existing elements of the vector, but using .at may be preferable because you get bounds checking.