output different elements from two arrays - c

Question

I am trying to output different elements from two arrays. So if i have an array A: {9, 0, 1} and B is {0, 8, 1}, I need to output an element which included in the first set, but are not included in the second :9 Can not think how I should compare all elements from the first array with the second one.
#include <stdio.h>
#include <stdlib.h>
int main()
{
int a[10],b[10],c,n,i,j;
printf("enter a number: ");
scanf("%d",&n);
for(i=0;i<n;i++){
printf("Enter a[%d]: ",i+1);
scanf("%d",&a[i]);
}
printf("\n");
for(j=0;j<n;j++){
printf("Enter b[%d]: ",j+1);
scanf("%d",&b[j]);
}
for (i = 0; i < n; i++) {
printf("%d ", a[i]); }
printf("\n");
for (i = 0; i < n; i++) {
printf("%d ", b[i]); }
printf("\n");
return 0;
}
I'd like to show my thoughts but i think it's stupid:
for(i=0;i<n;i++){
for(j=0;j<n;j++){
if(a[i]!= b[j]){
c=a[i];
}
}
printf("%d ",c);
}

Answer 1

This can be easily solved using Binary search. Follow the simple steps.
Step 1: Sort the second array.
Step 2: For each element of the first array, binary search it in the second array, if its not present , print it, otherwise dont.
The time complexity is O(m log n), where m is length of first array and n is length of second array.

Answer 2

If you want a more efficient solution, as suggested by #Sumeet Singh, you can sort the second array with qsort, then find similar elements from the first array with bsearch(binary search).
Your current solution is O(N^2) time, which will be very slow with large n, but you can achieve more efficiency with this approach.
Here is some code I wrote up with demonstrates this:
#include <stdio.h>
#include <stdlib.h>
#define NNUMBERS 10
void get_array_input(int array1[], int array2[], size_t *n);
void search_elements(int array1[], int array2[], size_t n);
void print_arrays(int array[], size_t n);
int cmp_func(const void *a, const void *b);
int main(void) {
int array1[NNUMBERS], array2[NNUMBERS];
size_t n;
/* input from user */
get_array_input(array1, array2, &n);
printf("\nFirst array: ");
print_arrays(array1, n);
printf("\nSecond array: ");
print_arrays(array2, n);
/* sorting the second array */
qsort(array2, n, sizeof(*array2), cmp_func);
printf("\nSorted Second array: ");
print_arrays(array2, n);
/* the search begins */
search_elements(array1, array2, n);
return 0;
}
void get_array_input(int array1[], int array2[], size_t *n) {
size_t i;
printf("Enter n: ");
if (scanf("%zu", n) != 1) {
printf("Invalid n value.\n");
exit(EXIT_FAILURE);
}
for (i = 0; i < *n; i++) {
printf("Enter array1[%zu]: ", i);
if (scanf("%d", &array1[i]) != 1) {
printf("Invalud array value.\n");
exit(EXIT_FAILURE);
}
}
for (i = 0; i < *n; i++) {
printf("Enter array2[%zu]: ", i);
if (scanf("%d", &array2[i]) != 1) {
printf("Invalud array value.\n");
exit(EXIT_FAILURE);
}
}
}
void search_elements(int array1[], int array2[], size_t n) {
size_t i;
void *key;
printf("\nElements in first array which are not in second array: ");
for (i = 0; i < n; i++) {
key = bsearch(&array1[i], array2, n, sizeof(*array2), cmp_func);
if (!key) {
printf("%d ", array1[i]); /* not found, so print it */
}
}
printf("\n");
}
void print_arrays(int array[], size_t n) {
size_t i;
for (i = 0; i < n; i++) {
printf("%d ", array[i]);
}
printf("\n");
}
/* cmp function needed for qsort and bsearch */
/* many ways to write these */
int cmp_func(const void *a, const void *b) {
const int *num1 = (const int *)a;
const int *num2 = (const int *)b;
if (*num1 > *num2) {
return +1;
} else if (*num1 < *num2) {
return -1;
}
return 0;
}
Input:
Enter n: 3
Enter array1[0]: 9
Enter array1[1]: 0
Enter array1[2]: 1
Enter array2[0]: 0
Enter array2[1]: 8
Enter array2[2]: 1
Output:
First array: 9 0 1
Second array: 0 8 1
Sorted Second array: 0 1 8
Elements in first array which are not in second array: 9

Answer 3

You are on the right path. You are taking each value from the first array and comparing to each value in the second.
What you need to do now is to only print a[i] if there isn't any b[j] such that they are the same. The easiest way is to set a flag (say, unique=1). You can give this flag any name you find suitable, but in this case I'm thinking it says that the number a[i] is "unique" to the array a. So in this case you start with the premise that, yes, you won't find a[i] in the arrayb, and then you try to disprove your assumption. If at any time of you search you find an instance of a[i] == b[j], then your premise was wrong, so you set unique=0.
After you have compared this a[i] against all elements in b, you review your flag. And you print the appropriate message depending on whether you found this element in b or not.
Note that this assumes that the same value doesn't appear twice in a.

Answer 4

I have edited your code a little bit and this code gives you desired output:
#include <stdio.h>
int main(void){
int a[10],b[10],c,n,i,j;
int counter=0;
printf("enter a number: ");
scanf("%d",&n);
for(i=0;i<n;i++){
printf("Enter a[%d]: \n",i+1);
scanf("%d",&a[i]);
}
printf("\n");
for(j=0;j<n;j++){
printf("Enter b[%d]: \n",j+1);
scanf("%d",&b[j]);
}
for(i=0;i<n;i++){
counter=0;
for(j=0;j<n;j++){
if(a[i]!=b[j]){
counter++;
}
}
if(counter == n){
printf("%d ",a[i]);
}
}
return 0;
}
Let's explain this code a little bit:
In the last nested for loop, outer loop takes one element from array a. Inner loop gets every element of array b in order to compare it to taken element from array a. If none of the elements of array b is equal to a's taken element, counter will be equal to n(array size). Then we can print this element taken from a(it means there is no match between this taken element and array b's all of elements.