What is the Big-O running time for this code? Explainyour answer. for (int i=0;

Question

What is the Big-O running time for this code? Explainyour answer.
   for (int i=0; i<num; i++)    System.out.println(i+1);


What is the Big-O running time for this code? Explain your answer.        for (int i=0; i<num; i++)    for (int j=0; j<num2; j++)    System.out.println((i+1) * (j+1) ); What is the Big-O running time for this code? Explainyour answer.
   for (int i=0; i<num; i++)    System.out.println(i+1);


What is the Big-O running time for this code? Explain your answer.        for (int i=0; i<num; i++)    for (int j=0; j<num2; j++)    System.out.println((i+1) * (j+1) );

Explanation / Answer

for (int i=0; i<num; i++)    System.out.println(i+1);

The for loop runs "num" times
suppose num= n, i.e. the for loop runs n times
Therefore the complexity of the given code snippet isO(n)


b)       for (int i=0; i<num; i++)    for (int j=0; j<num2; j++)    System.out.println((i+1) * (j+1) );

The outer for loop runs "num" times

For every iteration of the outer for loop...the inner for loop runs"num2" times.
Therefore the total number of times the third line of the codesnippet runs is "num * num2" times

let n = maximum( num, num2)

Therefore the total number of times the third statement is executedis < n* n
                                                                                                    < n2

Therefore the complexity of the given code snippet isO(n2)

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