assigning running time
It's hard to say without seeing the preceding code. If seq is initially empty the the run time of the outer loop is $O(n2)$ and the first nested loop is $O(k)$ and the second nested loop is $O(n1-k)$. The total running time would be $O(n2)*(O(k)+O(n1-k))$ which is $O(n^2)$.

However if seq has some initial elements, the worst case running time for the algorithm is $O(\infty)$. The problem is with the nested while loop seq[k]&gt;seq2[j]. If any of the first n elements in seq are &gt; seq2[j] the loop will not increment and run endlessly. Also you may run into array index out of range errors.

For example if seq = 2 , 1 , 0 and seq2 = 1 , 2 , 3. Then seq[0] &gt; seq2[0] would be equivalent to 2 &gt; 1 which i s false and k would never increment.
The solution would be to use

n1 = sizeof(seq)/sizeof(seq[0])  # number of elements in seq
int k = 0;
for(; k &lt; n1 ; k++){             # iterate over seq
    if( seq[k] &gt;= seq2[j] )      # until an we find an element in seq &gt;= seq2[j]
        break;