Five-digit number is given. Determine whether the number contains at least two identical digits

Question

This is my code, written in C++ :

#include <iostream>
using namespace std;

int main()
{
    int n,a,b,c,d,f;
    cin >> n;
    a=n/10000;
    b=n/1000%10;
    c=n/100%10;
    d=n/10%10;
    f =n%10;
    if(a==b && c==d && b!=c ||
       a==c && b==d && c!=b ||
       a==d && b==c && d!=b ||
       a==f && f==b && f!=d )
        cout << "YES";
    else
        cout << "NO";
    return 0;
}

This doesn’t work… Can someone please help?

I’m voting to close this question because no question asked. — Peter, Oct 31 '21 at 06:20
@Emily, please describe what problem you're facing and what's your question. — Md. Faisal Habib, Oct 31 '21 at 06:22
Have you considered another approach like reading the number as a `std::string` and then filling a `std::map` with digits as the key and the number of occurrences as the values? (or old-school, just use an array of 10 `char` as a *frequency array* and map digits to each index) — David C. Rankin, Oct 31 '21 at 06:23
You can find further information on using a *frequency array* here [How to remove duplicate char in string in C](https://stackoverflow.com/a/63255183/3422102) — David C. Rankin, Oct 31 '21 at 06:31
Why are you doing things like `b!=c`? If you give a number like `55555`, it will produce "NO" while there clearly are two identical numbers. :-) — Dominique, Oct 31 '21 at 08:19

score 3 · Answer 1 · edited Oct 31 '21 at 08:14

3

This should work

#include <iostream>
using namespace std;

int main(){
    int n, a, b, c, d, f;
    cin >> n;

    a = n / 10000;
    b = n / 1000 % 10;
    c = n / 100 % 10;
    d = n / 10 % 10;
    f = n % 10;

    if( a==b || b==c || c==d ||
        a==c || b==d || c==f ||
        a==d || b==f || d==f ||
        a==f )
        cout << "YES";
    else
        cout << "NO";
    
    return 0;
}

edited Oct 31 '21 at 08:14

Dharman

30,962
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answered Oct 31 '21 at 08:09

PreethiSamanthaBennet

233
1
7

Can you explain what you changed? – Jeremy Caney Oct 31 '21 at 08:33
I just changed the “if condition”. Please compare this with the code in the question. Will help in better understanding! – PreethiSamanthaBennet Oct 31 '21 at 09:13

score 1 · Answer 2 · answered Oct 31 '21 at 06:24

Read the input as a string and then scan for duplicate integers with a table.

std::string n;
int digits[10] = {0}; // the number of occurrences of all digits from 0..9
bool dupes = false;

cin >> n;

for (char c : n) {
   if ((c >= '0') && (c <= '9')) {
       int index = c - '0';
       digits[index]++;
       if (digits[index] > 1) {
           dupes = true;
       }
   }
}

if (dupes) {
   cout << "YES";
}
else {
   cout << "NO";
}

score 1 · Answer 3 · answered Oct 31 '21 at 06:52

This code will work for Integer input

#include<bits/stdc++.h>
using namespace std;

int main(){
  int n;
  vector<int>vec;
  set<int>s;
  cin>>n;
  while(n!=0)
  {
    int re=n%10;
    n=n/10;
    vec.push_back(re);
    s.insert(re);
  }
  if ((vec.size()-s.size())>=1)
    cout<<"YES"<<endl;
  else
    cout<<"NO"<<endl;
  return 0;


}

Five-digit number is given. Determine whether the number contains at least two identical digits

3 Answers3