Question

num of times needed to run each line, please explain in detail.

for(j=1;j<n;j=j*3)

sum = sum + j;

for (i = 0; i <n; i++)

for (j=i; j<n;j++)

sum+=i*j;

Answer 1

FIRST ONE

for(j=1;j<n;j=j*3) // ceil(log₃ n) times

sum = sum + j; // ceil(log₃ n) times

Possible numbers of j that can be inside the for loop are
1 3 9 27 81 243 . . .

They are are in powers of three
So, answer is ceil of log₃ n

O(ceil(log₃ n))

SECOND ONE

for (i = 0; i <n; i++) // n times

for (j=i; j<n;j++) // n(n+1)/2 times

sum+=i*j; // n(n+1)/2 times

First for loop runs for n times [0, 1, 2, 3, ... n-1]
Second for loop runs for

n times, n-1 times, n-2 times, ... 2 times, 1 time
n + (n-1) + (n-2) + ... + 2 + 1 = n(n+1)/2

Time complexity = O(n²)

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