How do I parse arrays in C++ for duplicates - c++

Question

Please help me use C++ to parse an array and only display numbers that are unique. I have written a program which answers most of the question below.
Question:
Use a one-demensional array to solve the following problem. Read in 20 numbers, each of which is between 10 and 100, inclusive. As each number is read, validate and store it in the array only if it isn't a duplicate of a number already read. After reading all the values, display only the unique values that the user entered. Provide for the "worst case" in which all 20 number are different. Use the smallest possible array to solve this problem.
What I've done:
My program creates a 20 item array. Prompts user for data, validates it and displays it. I have tried several way to only display unique data, however I have not accomplished what the question asks.
Sample User Input:
11, 12, 12, 12, 13, 14, 15, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27
My Program Output:
{ 11, 12, 12, 12, 13, 14, 15, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27 }
Correct Program Output:
{ 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27 }
Any advice where to go with this. See below for code. Thank you for your help!
#include <iostream>
#include <array> // needed for c++11 arrays
using namespace std;
int main()
{
const int arraySize = 20; // limit array length to 20
array <int, arraySize userArray> = {};
//Populate array with user input and validate it to be between 10 and 100
for (int i = 0; i < userArray.size(); i++)
{
cout << "Enter a number between 10 and 100" << endl;
cin >> userArray[i]; // get user input assign it to proper array subscript
while(userArray[i] > 100 || userArray[i] < 10)//validate user input to be between 10 and 100
{
cout << "Number needs to be between 10 and 100. Enter a new number" << endl;
cin >> userArray[i]; //reassign the proper array subscript if needed
}
}
cout << endl;
//display the information to look like an array
//result looks like [ v, w, x, y, z ]
cout << "[ ";
//display array values
for (int i = 0; i < userArray.size() - 1; i++)
{
cout << userArray[i] << ", ";
}
//properly display last array item
cout << userArray[(userArray.size() - 1)] << " ]" << endl;
return 0; }

Answer 1

If you can use std::vector, then you can use following solution.
template <typename Type>
std::vector<Type> unique_entries (std::vector<Type> vec) {
for (auto iter = vec.begin (); iter != vec.end (); ++iter) {
auto f = std::find_if (iter+1, vec.end (), [&] (const Type& val) {
return *iter == val; // (X)
});
if (f != vec.end ()) {
vec.erase (std::remove (iter+1, vec.end (), *iter), vec.end ());
}
}
return vec;
}
template <typename T>
void show (std::vector<T> vec) {
for (const auto& v : vec) {
std::cout << v << " ";
}
std::cout << std::endl;
}
And example would be like that:
std::vector<int> vec {11, 12, 12, 12, 13, 14, 15, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27};
std::vector<short> vec2 {1, 1, 1, 2, 3, 2, 1, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 10};
std::vector<std::string> vec3 {"a", "a", "a", "aa", "b", "b", "bb", "c"};
show (vec);
show (unique_entries (vec));
show (vec2);
show (unique_entries (vec2));
show (vec3);
show (unique_entries (vec3));
And the output:
11 12 12 12 13 14 15 15 16 17 18 19 20 21 22 23 24 25 26 27
11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
1 1 1 2 3 2 1 2 1 2 3 2 1 2 3 2 1 2 10
1 2 3 10
a a a aa b b bb c
a aa b bb c
Look at the line (X), basically you can use this unique_entries for all types which offer operator==, but you have to keep in mind that this will probably fail if you use this function for floating point types.

Answer 2

Hope this is what you are looking for........... Basically I use an another array to store the elements already entered by the user. So when the user enters a number it checks whether it is already in the array of 'entered elements' or not.
#include<iostream>
#define MAX 20
using namespace std;
int main()
{
int list[MAX],visitedElements[MAX];
int noElements,element,currentIndex;
bool flag=false;
cout<<"Enter total no. of elements(MAX=20):"<<endl;
cin>>noElements;
for(int i=0; i<noElements; i++)
{
visitedElements[i] = -1;
}
for(int i=0;i<noElements;i++)
{
currentIndex=i;
cout<<"Enter Element:"<<endl;
cin>>element;
for(i=0;i<currentIndex;i++)
{
if(visitedElements[i] == element)
{
flag==true;
break;
}
}
if(flag == false)
{
list[i]=element;
visitedElements[i] = element;
}
flag = false;
}
cout<<"Elements in list are"<<endl;
for(int i=0 ; i<noElements ;i++)
cout<<list[i]<<endl;
return 0;
}

Answer 3

#include <iostream>
using namespace std;
int main(){
int tab[10];
int i=0,j;
int number=0;
int counter=1;
bool flag=true;
int tries=0;
while(tries<10){
do{
cin>>number;
tries++;
if((number>100)||(number<10)){
cout<<"Wrong number, has to be between 10 and 100"<<endl;
}
}
while((number>100)||(number<10));
flag=true;
if(i==0){
tab[i]=number;
}
else{
for(j=0; j<counter; j++){
if(tab[j]==number){
flag=false;
i--;
break;
}
}
}
if(flag==true){
tab[i]=number;
counter++;
}
i++;
}
for(i=0; i<counter-1; i++){
cout<<tab[i]<<", ";
}
}
I kinda meesed it up probably. I'm new to programing but this is my solution.
So. You said it has to take 20 digits, mine takes 10, after 10 inputs it will list the array. So if my input is "10, 10, 10, 12, 13, 15, 15, 18, 22, 22"
the output is: "10, 12, 13, 15, 18, 22"
I could probably erase half of the code but like i said I'm a beginner and i was writing it in a hurry.
EDIT: Ups, small mistake in the array declaration. I was testing it for 5 elements and forgot to make a bigger array.

Answer 4

The question asked to use one array, so I would loop over the numbers already in the array after reading in each number and check for duplicates. It would look something like this:
#include <iostream>
constexpr int ARRAY_SIZE = 20;
int main()
{
int userArray[ARRAY_SIZE], validNumbers = 0;
for(int i = 0; i < ARRAY_SIZE; i++)
{
int num;
bool isGood = true;
std::cin >> num;
//code to check if input is in range here
for(int j = 0; j < validNumbers; j++)
{
if(userArray[j] == num)
{
isGood = false;
break;
}
}
if(isGood)
{
userArray[validNumbers] = num;
validNumbers++;
}
}
//userArray now contains all unique numbers in the order that they were entered
for(int i = 0; i < validNumbers; i++)
{
std::cout << userArray[i];
if(i < validNumbers - 1)
{
std::cout << ' ';
}
}
std::cout << '\n';
}

Answer 5

inside of your for loop, instead of makign the cin go directly into the array, add another int variable and set it to equal the input
cin >> variable
then make a for loop that iterates through your current array to make sure its not a duplicate; use the line below as your for loop
for(int j = 0; j < i; j++)
if the number is already in the array, just continue. Else, add it in.
Also, instead of using a for loop in the beginning, I would use a while loop with
while(i < arraysize)
Then deincrement array size every time you see a duplicate and only increment i when you add an element