该代码的运行时复杂度是多少？

Question

What is the run-time complexity of this Code?

#include <cstdio>
#include <cstring>
int main()
{
    int n, i, l, j, c, ans = 25;
    scanf("%d", &n);
    char x[21], y[21];
    scanf("%s", &x);
    l = strlen(x);
    for(i=1; i<n; i++)
    {
        scanf("%s", &y);
        c = 0;
        for(j=0 ;j<l; j++)
            if(x[j] == y[j])
                c++;
            else
                break;
        strcpy(x, y);
        if(ans > c)
            ans = c;
    }
    printf("%d", ans);
    return 0;
}

My lecturer tells me that the complexity of this code is O(n*n) but I'm not convince with this answer cause the inner loop runs string length times.

Answer 1

The runtime of main() is composed of the runtime of some constant-time statements and the runtime of the i -loop:

T_main(n, l) ∈ O(1) + T_fori(n, l)

The i -loop runs exactly (n - 1) times and is composed of some constant-time statements and the runtime of the j -loop:

T_fori(n, l) ∈ (n - 1) * (O(1) + T_forj(n, l))

The runtime of the j-loop depends on the data. In the best case, the first character of x and y are different, thus:

T_forj_best(n, l) ∈ 1 * O(1)

In the worst case, x and y have l equal first characters, thus:

T_forj_worst(n, l) ∈ l * O(1) = O(l)

This yields:

T_fori_best(n, l)  ∈ (n - 1) * (O(1) + O(1)) = O(n)
T_fori_worst(n, l) ∈ (n - 1) * (O(1) + O(l)) = O(n * l)

and

T_main_best(n, l)  ∈ O(1) + O(n)     = O(n)
T_main_worst(n, l) ∈ O(1) + O(n * l) = O(n * l)