Question

What is execution time growth rate Big O of this code?

int maxValue = m[0][0];         
for (int i = 0; i < N; i++)         
{                       
    for (int j = 0; j < N; j++)         
    {                       
        if ( m[i][j] > maxValue )           
        {                       
            maxValue = m[i][j];         
        }                       
    }                       
}                   
cout << maxValue << endl;           
int sum = 0;                    
for (int i = 0; i < N; i++)         
{                       
    for (int j = 0; j < N; j++)         
    {                       
        sum = sum + m[i][j];            
    }                       
}                           
cout << sum << endl;                            

Answer 1

O(N^2) because you have two nested “for” loops running from 0 to N.