I want to check if two integer type arrays have the same set of digits. For example, if array 1 is 5 1 2 3 3 4 6 1, and array 2 is 1 2 3 4 5 6, the program returns 1. If any number from either array isn't in the second one, the program returns a 0.
I tried doing something like this, but I can't get it to work:
#include <stdio.h>
int main()
{
int i, j, a[8]={5, 1, 2, 3, 3, 4, 6, 1}, b[6]={1, 2, 3, 4, 5, 6}, x=0;
for(i=0; i<6; i++)
{
for(j=0; j<8; j++)
{
if(a[j]==b[i])
{
x=1;
continue;
}
else
{
x=0;
break;
}
}
}
return x;
}
EDIT: Thank you Some programmer dude
#include <stdio.h>
void sort(int arr[], int n)
{
int i, j, a;
for (i=0; i<n; i++)
{
for (j=i+1; j<n; j++)
{
if (arr[i]>arr[j])
{
a=arr[i];
arr[i]=arr[j];
arr[j]=a;
}
}
}
}
int main()
{
int i, j, k;
int a[8]={5, 1, 2, 3, 3, 4, 6, 1};
int b[6]={1, 2, 3, 4, 5, 6};
int na=8, nb=6;
for(i=0; i<na; i++) // removing duplicates from a
{
for(j=i+1; j<na; j++)
{
if(a[i]==a[j])
{
for(k=j; k<na; k++)
{
a[k]=a[k+1];
}
na--;
j--;
}
}
}
for(i=0; i<nb; i++) // removing duplicates from b
{
for(j=i+1; j<nb; j++)
{
if(b[i]==b[j])
{
for(k=j; k<nb; k++)
{
b[k]=b[k+1];
}
nb--;
j--;
}
}
}
sort(a, na);
sort(b, nb);
if(na!=nb)
return 0;
for(i=0; i<na; i++)
{
if(a[i]!=b[i])
return 0;
}
return 1;
}
You have several ways you can approach this, you can use two sets of nested loops swapping the order you loop over the two arrays validating each element is found in the other. Two full sets of nested loops are needed as you have a 50/50 chance any single outlier will be contained in either of the arrays. This is the brute-force method and has the potential worst-case number of iterations.
Since an outlier is what drove the need for looping with one arrays as outer and the other inner and then swapping a repeating, e.g. to catch 5, 1, 2, 3, 3, 4, 6, 1 and 1, 2, 3, 4, 5, 6, 7, if you can catch the outlier with another method that requires fewer iterations you can make your algorithm more efficient.
An outlier would be detected in a comparison of the min and max from each array, and to find min and max only requires a single linear traversal of each array. Much better than the worst-case nested loop over all elements.
The min and max check provide a way to shorten your work, but do not eliminate the need to press forward with a second set of nested loops if the result is inconclusive at that point. Why? Consider the following sets, where the min and max are equal, but one element within the range is not included in both arrays, e.g.:
int a[] = { 5, 1, 2, 3, 3, 4, 6, 112 },
b[] = { 1, 2, 3, 4, 5, 6, 7, 112 };
The only way the 7 will be detected is by nested loop with the array containing 7 being the outer loop.
So you could write a short function to test for the common set as:
#include <stdio.h>
#include <limits.h>
int commonset (int *a, int *b, int sza, int szb)
{
int maxa = INT_MIN, maxb = INT_MIN,
mina = INT_MAX, minb = INT_MAX;
for (int i = 0; i < sza; i++) { /* find max/min of elements of a */
if (a[i] > maxa)
maxa = a[i];
if (a[i] < mina)
mina = a[i];
}
for (int i = 0; i < szb; i++) { /* find max/min of elements of b */
if (b[i] > maxb)
maxb = b[i];
if (b[i] < minb)
minb = b[i];
}
if (maxa != maxb || mina != minb) /* validate max & mins equal or return 0 */
return 0;
for (int i = 0; i < sza; i++) { /* compare of each element between arrays */
int found = 0;
for (int j = 0; j < szb; j++)
if (a[i] == b[j]) {
found = 1;
break;
}
if (!found)
return 0;
}
for (int i = 0; i < szb; i++) { /* compare of each element between arrays */
int found = 0;
for (int j = 0; j < sza; j++)
if (a[j] == b[i]) {
found = 1;
break;
}
if (!found)
return 0;
}
return 1;
}
Adding a short example program:
int main (void) {
int a[] = { 5, 1, 2, 3, 3, 4, 6, 1 },
sza = sizeof a / sizeof *a,
b[] = { 1, 2, 3, 4, 5, 6 },
szb = sizeof b / sizeof *b,
result;
result = commonset (a, b, sza, szb);
if (result)
puts ("arrays have common set of numbers");
else
puts ("arrays have no common set of numbers");
return result;
}
Example Use/Output
$ ./bin/arr_commonset
arrays have common set of numbers
$ echo $?
1
With b[] = { 1, 2, 3, 4, 5, 6, 7 }:
$ ./bin/arr_commonset
arrays have no common set of numbers
$ echo $?
0
With a[] = { 5, 1, 2, 3, 3, 4, 6, 112 } and b[] = { 1, 2, 3, 4, 5, 6, 7, 112 }:
$ ./bin/arr_commonset
arrays have no common set of numbers
$ echo $?
0
There are probably even ways to combine the two and shave off a few iterations, and, if you have a guaranteed range for your input sets, you can use a simple frequency array for each and then two simple linear iterations would be needed to increment the element that corresponds to the index for each value in the array, and then a third linear iteration over both frequency arrays comparing that like indexes either both are non-zero or both are zero to confirm the common set -- that is left to you.
Look things over and let me know if you have any further questions.