finding common numbers

BB89 · 11-09-2009, 12:08 AM

I am trying to find the numbers in the two arrays that are in common.

Say they are:

int A[3] = {1, 2, 5};
int B[4] = {1, 5, 9, 10};

Answer should say A \ B = {1, 5}

Code:

#include <stdio.h>

int main (void)
{
    int A[3] = {1, 2, 5};
    int B[4] = {1, 5, 9, 10};

    int x, z;

   for( int x = 0; x < 3; x++ )
    for( int z = 0; z < 4; z++ )

}

I have declared the arrays and I am think the rest is correct.

I'm not sure how to find the common numbers or how to associate the loops with the arrays.

New to C and Arrays.

RockyMarrone · 11-09-2009, 12:28 AM

What you have to do

Two arrays lets say A and B

Code:

    for i = 0  to n - 1 of A
        for j = 0 to n -1 of B
           if value of A[i] == B[j] then add to answer array

before adding to the answer array just check its already there or not

BB89 · 11-09-2009, 12:55 PM

Code:

#include <stdio.h>

int main (void)
{
    int A[3] = {1, 2, 5};
    int B[4] = {1, 5, 9, 10};

    int x, z;

   for( int x = 0; x < 3; x++ )
    for( int z = 0; z < 4; z++ )
        if( A[x] == B[z] )
            printf("A \ B = {%d, %d\n}", A[x], B[z] );

    return 0;
}

My output: A \ B = {1,1}
A \ B = {5, 5}

The numbers are correct.

What's wrong.

zalezog · 11-09-2009, 01:01 PM

When
A[x] == B[z]
then

Code:

printf("A \ B = {%d, %d\n}", A[x], B[z] );

A[x] and B[z] are same you need to print any one.

EDIT : You are declaring x and z twice.

BB89 · 11-09-2009, 01:34 PM

Ahh, makes since.

So

Code:

 printf("A \ B = {%d}", A[x] );

Why is it printing it twice?

It doing this, I dont understand it.

A \ B = {1,1}
A \ B = {5, 5}

Im wanting jusr A \ B = {1, 5}

iMalc · 11-10-2009, 02:41 AM

Quote:

Originally Posted by RockyMarrone

What you have to do

Two arrays lets say A and B

Code:

    for i = 0  to n - 1 of A
        for j = 0 to n -1 of B
           if value of A[i] == B[j] then add to answer array

before adding to the answer array just check its already there or not

If A and B each held 10000 numbers or more, would you still use the same algorithm?

The arrays are sorted. If it's a safe assumption that they are always sorted then there's a much quicker approach. Try studying the C++ standard library code for set_intersection.

Can we also assume there are no duplicates?

RockyMarrone · 11-10-2009, 03:02 AM

Quote:

Originally Posted by iMalc

If A and B each held 10000 numbers or more, would you still use the same algorithm?

The arrays are sorted. If it's a safe assumption that they are always sorted then there's a much quicker approach. Try studying the C++ standard library code for set_intersection.

Can we also assume there are no duplicates?

Ok iMac Please tell me if the numbers are unsorted then the algo

**laserlight** · 11-10-2009, 03:05 AM

Quote:

Originally Posted by RockyMarrone

Ok iMac Please tell me if the numbers are unsorted then the algo

... would be to sort them and then use an algorithm that works when they are sorted.

RockyMarrone · 11-10-2009, 03:09 AM

Quote:

Originally Posted by laserlight

... would be to sort them and then use an algorithm that works when they are sorted.

So how many compressions will be there less the one above of yours or more ????
And which will be faster

**laserlight** · 11-10-2009, 04:10 AM

Quote:

Originally Posted by RockyMarrone

So how many compressions will be there less the one above of yours or more ????
And which will be faster

I think you mean "comparisons" instead of "compressions". Unfortunately, I do not really understand your sentence. What algorithms are you trying to compare?

RockyMarrone · 11-10-2009, 04:20 AM

Quote:

Originally Posted by laserlight

I think you mean "comparisons" instead of "compressions". Unfortunately, I do not really understand your sentence. What algorithms are you trying to compare?

Then without knowing the algo how can u write

Quote:

Quote:
Originally Posted by RockyMarrone
Ok iMac Please tell me if the numbers are unsorted then the algo
... would be to sort them and then use an algorithm that works when they are sorted.

Brain_Child · 11-10-2009, 05:23 AM

Quote:

Originally Posted by BB89

Ahh, makes since.

So

Code:

 printf("A \ B = {%d}", A[x] );

Why is it printing it twice?

It doing this, I dont understand it.

A \ B = {1,1}
A \ B = {5, 5}

Im wanting jusr A \ B = {1, 5}

you have your printf("a\b...}"); in the middle of a for loop. so everytime you find a matching pair you are printing the entire line. you need to take it out of the for loop and print it only once. you can either stick the statement before the for loops, or after the for loops.

also i cant see an easy way to print matching values inside nested for loops (due to the required format of the output, particularly the comma). you'll need to store the matching values somewhere and print them later, its about 5 extra lines.

(you should also fix up your indenting. either 3 spaces or 4 spaces. pick one and use it all the time)

**laserlight** · 11-10-2009, 08:07 AM

Quote:

Originally Posted by RockyMarrone

Then without knowing the algo how can u write

Ah, but I do know the algorithm that iMalc was talking about

BB89 · 11-10-2009, 08:21 AM

I figured it out.

Thanks for the help.

**laserlight** · 11-10-2009, 08:22 AM

Quote:

Originally Posted by BB89

I figured it out.

Great. What did you use in the end?

11-09-2009, 12:08 AM	#1
BB89 Registered User Join Date: Sep 2009 Location: Dallas, Texas Posts: 70	finding common numbers I am trying to find the numbers in the two arrays that are in common. Say they are: int A[3] = {1, 2, 5}; int B[4] = {1, 5, 9, 10}; Answer should say A \ B = {1, 5} Code: #include <stdio.h> int main (void) { int A[3] = {1, 2, 5}; int B[4] = {1, 5, 9, 10}; int x, z; for( int x = 0; x < 3; x++ ) for( int z = 0; z < 4; z++ ) } I have declared the arrays and I am think the rest is correct. I'm not sure how to find the common numbers or how to associate the loops with the arrays. New to C and Arrays.
BB89 is offline

11-09-2009, 12:28 AM	#2
RockyMarrone Registered User Join Date: Oct 2009 Location: While(1) Posts: 368	What you have to do Two arrays lets say A and B Code: for i = 0 to n - 1 of A for j = 0 to n -1 of B if value of A[i] == B[j] then add to answer array before adding to the answer array just check its already there or not
RockyMarrone is offline

11-09-2009, 01:01 PM	#4
zalezog Registered User Join Date: Mar 2009 Posts: 38	When A[x] == B[z] then Code: printf("A \ B = {%d, %d\n}", A[x], B[z] ); A[x] and B[z] are same you need to print any one. EDIT : You are declaring x and z twice. Last edited by zalezog; 11-09-2009 at 01:07 PM.
zalezog is offline

11-09-2009, 01:34 PM	#5
BB89 Registered User Join Date: Sep 2009 Location: Dallas, Texas Posts: 70	Ahh, makes since. So Code: printf("A \ B = {%d}", A[x] ); Why is it printing it twice? It doing this, I dont understand it. A \ B = {1,1} A \ B = {5, 5} Im wanting jusr A \ B = {1, 5}
BB89 is offline

11-10-2009, 08:21 AM	#14
BB89 Registered User Join Date: Sep 2009 Location: Dallas, Texas Posts: 70	I figured it out. Thanks for the help.
BB89 is offline

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