Check if array is subset of another array

Question

I have two arrays.

1. 1,3,5,7,9
2. {3,5} or {1.9}. (left to right order)

So the second array is a subset of the first array But not a subset if the second array is {5.3} or, {9.1} (right to left order.)

My code is

#include <stdio.h>
void subset(int set11[], int set12[])
{
    int length1;
    int set1[] = {1, 5, 3, 9, 7, 0, 5};
    length1 = sizeof(set1) / sizeof(int);

    int length2;
    int set2[] = {5, 9};
    length2 = sizeof(set2) / sizeof(int);

    int i = 0;
    int j = 0;
    int count = 0;
    for (i = 0; i < length1; i++)
    {
        if (set1[i] == set2[j])
        {

            count = 1;
        }
    }
    printf(" is \n");
    if (count == 1)
    {
        printf("is subset");
    }
    else
    {
        printf("not subset");
    }
}

int main()
{
    int set11[] = {1, 5, 3, 9, 7};
    int set12[] = {5, 9};
    subset(set11, set12);
    printf("");

    return 0;
}

I get output in all cases only not subset.

Answer 1

Applied some changes in logic. refer comments.

#include <stdio.h>
#include <stdbool.h>
void subset(int set11[], int set12[])
{
    int length1;
    int set1[] = {1,3,5,7,9};
    length1 = sizeof(set1) / sizeof(int);

    int length2;
    int set2[] = {3,5};
    length2 = sizeof(set2) / sizeof(int);

    int i = 0;
    bool isSubset = true;   //will indicate weather the second array is subset or not
//    int j = 0;    //
    int startPosition = 0;  // will decide the starting position for searching in  the main array.  {set 2}
//    int count = 0; //not needed; we will represent it with bool variable 'isSubset'.
    for (i = 0; i < length2; i++)   //Iterating through the subset array
    {
        bool isFound = false;
        for (int j=startPosition;j<length1;j++){        //Iterating through the original array {set 1}
            if (set2[i]==set1[j]){  //if element from second array is found in first array then...
                isFound = true;     //found the element
                startPosition = j+1;        //increasing the starting position for next search in the first array.
                printf("\t%d found at %d\n",set2[i],j);
                break;
            }
        }
        if(isFound==false){     //if not found then this is not subarray.
            printf("\t%d not found\n",set2[i]);
            isSubset = false;
            break;
        }
    }
//    printf(" is \n");
    if (isSubset)
    {
        printf("subset");
    }
    else
    {
        printf("not subset");
    }
}

int main()
{
    int set11[] = {1, 5, 3, 9, 7};
    int set12[] = {5, 9};
    subset(set11, set12);
    printf("");

    return 0;
}

Answer 2

You can run a nested loop to get hold of all the matching elements in the subset array

for (i = 0; i < length1; i++) 
{
    for(k = 0; k < length2; k++) 
    {
        if (set1[i] == set2[k])
        {
        count == 1;
        }
    }
}

Outer loop is for the for the First array Inner loop to check for the element at any position in the subset/second array

set1[] = {1, 5, 3, 9, 7, 0, 5}; set2[] = {5, 9};

If we run a nested loop we will get all the subsets regardless of their positions in the second array(set2)

it wont matter if the set2 is {5,9} or {9,5} in any case the counter variable will increase.

Answer 3

#include <stdio.h>
 
int main () {

int []={1,2,3,7,8};    // add element after 3 --> 4,5,6 (condition that i don't know position of 3 in array)


for(int i=0,i<10;i++)
{
    printf("%d\n",n[i]);
}
   return 0;
}

i want output 1,2,3,4,5,6,7,8 but remember in case i don't know the position of 3 or 7 in array