C program to remove duplicate values from an array

#include <stdio.h>

main()

{

               int a[10]={1,2,1,1,2,3,2,3,3,4},b[10],i,j=0,c,flag;

               b[0]=a[0];

               for(i=1;i<10;i++)

               {

                              flag=0;

                              for(c=0;c<=j;c++)

                              {

                                             if(a[i]==b[c])//  check whether a[i] is present in entire b array

                                             {

                                                            flag=1;

                                                            break;

                                             }

                              }

                              if(flag == 0)

                              {

                                             b[++j]=a[i];

                              }

               }

               printf("the array elements after removing duplicate values are\n");

               for(i=0;i<=j;i++)

                              printf("%d\t",b[i]);

}

Output:

Time complexity :

In worst case : Worst case occurs when there are no duplicate values in 'a' array

b[0]= a[0]

i.e, a[1]  is present in b array or not  --1 comparison as b array contains only one element at that moment

Since there are no duplicate elements in 'a' array then b[1] will contain a[1] i.e, b contains 2 elements now.

then 

a[2] is checked if it is present in b[0] or b[1]---doing 2 comparisons 

a[3] is checked if it is present in b[0] or b[1] or b[2]---doing 3 comparisons

:

:

a[n-1] is checked if it is present in b[0] or b[1] or......b[n-2] --- doing n-1 comparisons

adding all the comparisons 

1+2+3+....n-1 .

this is nothing but sum of first n-1 natural number i.e, = ((n-1)*(n-2))/2 = n^2 

Hence Time complexity of above program is O(n^2)

In Best Case:  occurs when all the array elements in 'a' array are same

b[0]=a[0]

a[1] is compared with b[0] for equality ...since a[1] and b[0] are same, b array will contain only one element.....1 comparison is done

a[2] is compared with b[0] for equality  ...1 comparison is done

a[3] is compared with b[0] for equality  ...1 comparison is done

:

:

a[n-1] is compared with b[0] for equality  ...1 comparison is done

Adding all comparisons

1+1+1+......n-1 times.

that is n-1 comparisons

Hence Best Time complexity is 'n'