Chapter 2 Big O notation questions and explanations (package meeting), bignotation
I have listed some difficult questions or error-prone questions. I have not explained them too much, but if you have any questions, please trust me or comment on them, I will try to answer questions or questions in real time.
Int x = 0;
For (int I = 4 * n; I> = 1; I --)
X = x + 2 * I;
O (n)
The loop runs O (n) times and does O (1) work per iteration.
Int z = 0;
Int x = 0;
For (int I = 1; I <= n; I = I * 3 ){
Z = z + 5;
Z ++;
X = 2 * x;
}
O (log n)
Think about the values of I as the loop progresses. it will take on the series of values 1, 3, 9, 27, 81,243 ,..., 3 k. since I is tripling on each iteration, it takes on successive powers of three.
The loop clearly only does O (1) work per iteration, so the main
Question here is how many total iterations there will be. The loop
Will stop when I> n. If we let k be some arbitrary iteration of
Loop, the value of I on iteration k will be 3 k. The loop stops when
3 k> n, which happens when k> log3 n. Therefore, the number
Iterations is only O (log n)
Int y = 0;
For (int j = 1; j * j <= n; j ++)
J <= (n) ^ 1/2 y ++;
O (√ n)
Notice that j is still growing linearly, but the loop runs as long
J2 ≤ n. This means that as soon as j exceeds √ n, the loop will
Stop. Therefore, there will only be O (√ n) iterations of the loop,
And since each one does O (1) work, the total work done is O (√ n)
Int B = 0; // constant
For (int I = n; I> 0; I --)
For (int j = 0; j <I; j ++)
B = B + 5;
O (n ^ 2)
The most accurate answer wocould be O (n2)
Int y = 1;
Int j = 0;
For (j = 1; j <= 2n; j = j + 2)
Y = y + I;
Int s = 0;
For (I = 1; I <= j; I ++)
S ++;
O (n)
Int B = 0;
For (int I = 0; I <n; I ++)
For (int j = 0; j <I * n; j ++)
B = B + 5;
The inner loop will run 0 + n + 2n + 3n + 4n +... + n (n-1) = n (0 + 1
+ 2 +... + n-1) times, so the total work done is O (n3). You
Shouldn't multiply by the number of times the outer loop runs because
You're already summing up your SS all iterations. The most accurate
Runtime wocould be O (n3)
Int t = 0;
For (int I = 1; I <= n; I ++)
For (int j = 0; j * j <4 * n; j ++)
For (int k = 1; k * k <= 9 * n; k ++)
T ++;
Look at the second loop. This actually runs O (√ n) times using
Same logic as one of the earlier parts. That third inner loop also
Runs O (√ n) times, and so the total work done will be O (n2)
Int a = 0;
Int k = n * n;
While (k> 1)
{
For (int j = 0; j <n * n; j ++)
{A ++ ;}
K = k/2;
}
The outer loop starts with k initialized to n2, but notice that k is
Halved on each iteration. This means that the number of iterations
The outer loop will be log (n2) = 2 log n = O (log n), so the outer
Loop runs only O (log n) times. That inner loop does do O (n2) work, so
The total runtime is O (n ^ 2 log n)
Int I = 0, j = 0, y = 0, s = 0;
For (j = 0; j <n + 1; j ++)
Y = y + j;
For (I = 1; I <= y; I ++)
S ++;
O (n ^ 3)
