[SOLVED] Print if number occurs more than once in array

srinietrx · 09-22-2015, 09:58 AM

I written program to count the number of times element in array appears.
And if it appears more than once I need to print.
Code is

Code:

#include <iostream>
#include <stdio.h>

using namespace std;
int main()
{
	int arr[6] = {5, 3, 7, 5, 1, 3};
	int count = 0,i,j;
	for(i = 0;i < 6;i++)
	{
		for(j = 0;j<6;j++)
		{
			if(arr[i] == arr[j])
				count++;
		}
		if(count > 1)
		cout << "Element " << arr[i] << " occured " << count << " times " << endl;
		count = 0;
	}
	return 0;
}

output is

Code:

Element 5 occured 2 times 
Element 3 occured 2 times 
Element 5 occured 2 times 
Element 3 occured 2 times

But in this case 5 and 3 are printing as many times as it is present in array.
What condition can be written so duplicates are not printed?
i.e
My expected output is

Code:

Element 5 occured 2 times 
Element 3 occured 2 times

danielbmartin · 09-22-2015, 10:51 AM

A gentle suggestion: it is helpful for you to identify your language of choice.
You may do this with tags and/or a specific reference in your post.

Daniel B. Martin

millgates · 09-22-2015, 12:13 PM

I think the easiest way to modify your code to do that is to check whether j > i in your inner loop, i.e. if you find a number equal to arr[i], but before it, break the inner loop and continue to the next iteration of the outer one.
Other than that, there's a couple of other options, such as

- sorting the array before counting
- remembering which numbers you already encountered
- using an associative data structure such as a map

rtmistler · 09-22-2015, 12:21 PM

Quote:

Originally Posted by srinietrx

... in this case 5 and 3 are printing as many times as it is present in array.
What condition can be written so duplicates are not printed?

Homework?

Your outer loop encounters 5 and 3 twice, hence why it prints those statements twice.

One thing you can consider is to code it such that you remember and store elements which have been detected to be repeated. For each new element processed in your outer loop, check whether or not they appear in your remembered list of repeats. If so, then skip checking any further.

srinietrx · 09-22-2015, 01:56 PM

Thanks
@millgates j > i in innerloop solves the problem
But I will try to implement using map.It is opportunity to learn STL

@rtmistler :- I was thinking about that logic you are saying.Thanks for hint.