How to write a c++ code that takes 10 integers and type them without repetition ?

Question

Thanks

Answer 1

#include <iostream.h>

void main(){
int nos[10], j=0, i=0;
int repeat=-1;
for(i=0; i<=10; i++){
// to input nos. in array
cout<<"Enter the no.";
cin>>nos[i] //nos are stored
}
cout<<"The nos you typed are";
for(i=0!; i<=10; i++){
//comparing no with old no.
for(j=0; j<i; j++){
if( nos[I]==nos[j])
repeat=i;
break;
}
//this will check its repeat or not
if(i!=repeat) cout<<nos[i];
}
}

// it can be by flags too

thanks to Abhishek for suggestions
c++ has inbuilt feature to do this so px16 has the normal code

Answer 2

#include <iostream>
using namespace std;

int main() {

int arr[10];
bool flag = false;

for(int i=0;i<10;i++)
{
cin >> arr[i] ;
}

for(int i=0;i<10;i++)
{
flag = false;
for(int j=i-1;j>=0;j--)
{
if(arr[i] == arr[j])
{
flag = true;
break;
}
}

if(!flag)
cout << arr[i] << " ";
}

return 0;
}

Plz VOTE UP if this helped :)

Answer 3

#include <iostream>
using namespace std;
#include <set>
using std::set;
int main() {
set<int> nums;
for (int i = 0; i < 10; i++){
int v;
cin >> v;
nums.insert(v);
}
for (auto v : nums)
cout << v << endl;;
return 0;
}
There is used std::set to make numbers unique. If you need to keep order, you may use set to verify if number was already provided and push back number to result std::vector or list.
set<int> nums;
vector<int> result;
for (int i = 0; i < 10; i++){
int v;
cin >> v;
if (nums.find(v) == nums.end())
result.push_back(v);
nums.insert(v);
}
for (auto v : result)
cout << v << endl;;

Answer 4

Abhishek thanks for your suggestion the loop statement has been corrected it was to compare all no it's j is i-1
[6, 2, 6]
the idea is , in loop it will check nos[2] is equal to nos[0] or not if yes it will mark this element as repeat now after check loop ends it will not print nos[2 ] like nos[0] and nos[1] as nos[2] is marked as repeat.
the small loop should check all previous nos and report repetition.

Answer 5

yes please suggest how to optimize

Answer 6

i am not good in c++ but i recommend u to use int array, the length is 10 , then every time u need e for loop, in oder to check the value, because u do not wanna same value, so u need if, i hope , i can help u

Answer 7

@sandeep chatterjee in your code at some point 'j' value will become same as 'i' so your below condition will always be true.

if( nos[I]==nos[j])
repeat=i;

which will set your repeat to i thus you will not get any output

Answer 8

@sandeep chatterjee I hope you know if i==j then nos[i] will be equal to nos[j] since both i and j have same index value

Answer 9

@sandeep chatterjee I see now you have changed condition in second loop condition now it is correct but you can still optimize it

Answer 10

@sandeep chatterjee

for(j=0; j<i; j++){
if( nos[I]==nos[j]){
repeat=i;
break;
}
}

break will stop the loop once it finds the number is a repeat

Answer 11

good suggestion ok Abhishek that's good but I think it will work without it too rather it can be extended to count the missing no. if we add a counter that counts missing no.
I mean there is no problem in small program to mark repeat again repeat should be reported at least one time.
yes break or continue will give fast result but I have left this for counter