How to refactor this simple code to avoid code duplication? - c++

Question

I am solving the following simple problem(on one of OnlineJugde sites which is in Russian, so I won't give a link here:). It is easier to state the problem via an example than definition.
Input:
10 // this is N, the number of the integers to follow
1 1 1 2 2 3 3 1 4 4
Output:
3 times 1.
2 times 2.
2 times 3.
1 times 1.
2 times 4.
Constraints:
All the numbers in the input(including N) are positive integer less than 10000.
Here is the code I got Accepted with:
#include <iostream>
using namespace std;
int main()
{
int n;
cin >> n;
int prevNumber = -1;
int currentCount = 0;
int currentNumber;
while(n --> 0) // do n times
{
cin >> currentNumber;
if(currentNumber != prevNumber)
{
if(currentCount != 0) //we don't print this first time
{
cout << currentCount << " times " << prevNumber << "." << endl;
}
prevNumber = currentNumber;
currentCount = 1;
}
else //if(currentNumber == prevNumber)
{
++currentCount;
}
}
cout << currentCount << " times " << prevNumber << "." << endl;
}
Now here's my problem. A little voice inside me keeps telling me that I am doing this line two times:
cout << currentCount << " times " << prevNumber << "." << endl;
I told that voice inside me that it might be possible to avoid printing separately in the end. It told me that there would then be perhaps way too many if's and else's for such a simple problem. Now, I don't want to make the code shorter. Nor do I want do minimize the number of if's and else's. But I do want to get rid of the special printing in the end of the loop without making the code more complicated.
I really believe this simple problem can be solved with simpler code than mine is. Hope I was clear and the question won't be deemed as not constructive :)
Thanks in advance.

Answer 1

i came up with this. no code duplication, but slightly less readable. Using vector just for convenience of testing
EDIT my answer assumes you know the numbers ahead of time and not processing them on the fly
vector<int> numbers;
numbers.push_back(1);
numbers.push_back(1);
numbers.push_back(1);
numbers.push_back(2);
numbers.push_back(2);
numbers.push_back(3);
numbers.push_back(3);
numbers.push_back(1);
numbers.push_back(4);
numbers.push_back(4);
for (int i=0; i<numbers.size(); i++)
{
int count = 1;
for (int j=i+1; j<numbers.size() && numbers[i] == numbers[j]; i++, j++)
{
count++;
}
cout << count << " times " << numbers[i] << "." << endl;
}

Answer 2

My version: reading the first value as a special case instead.
#include <iostream>
int main()
{
int n;
std::cin >> n;
int value;
std::cin >> value;
--n;
while (n >= 0) {
int count = 1;
int previous = value;
while (n --> 0 && std::cin >> value && value == previous) {
++count;
}
std::cout << count << " times " << previous << ".\n";
}
}

Answer 3

Run your loop one longer (>= 0 instead of > 0), and in the last round, instead of reading currentNumber from cin, do currentNumber = lastNumber + 1 (so that it's guaranteed to differ).

Answer 4

slightly more CREATIVE answer, this one does not make assumption about input being all known before the start of the loop. This prints the total every time, but makes use of \r carriage return but not line feed. A new line is inserted when a different number is detected.
int prev_number = -1;
int current_number;
int count = 0;
for (int i=0; i<numbers.size(); i++)
{
current_number = numbers[i];
if (current_number != prev_number)
{
count = 0;
cout << endl;
}
count++;
prev_number = current_number;
cout << count << " times " << numbers[i] << "." << "\r";
}
only problem is that the cursor is left on the last line. you may need to append cout << endl;

Answer 5

I think this will work:
#include <iostream>
using namespace std;
int main()
{
int n;
cin >> n;
int prevNumber = -1;
int currentCount = 0;
int currentNumber;
int i = 0;
while(i <= n)
{
if(i != n) cin >> currentNumber;
if(currentNumber != prevNumber || i == n)
{
if(currentCount != 0)
{
cout << currentCount << " times " << prevNumber << "." << endl;
}
prevNumber = currentNumber;
currentCount = 1;
}
else
{
++currentCount;
}
i++;
}
}
I would use a for loop, but I wanted to stay as close to the original as possible.