Consider the procedure . procedure Loops2 (n, a positive integer) 1. for k = 1 t

Question

Consider the procedure .

procedure Loops2 (n, a positive integer) 1.

for k = 1 to nn -2 .

for i = k + 1 to n -1 .

for j = i + 1 to n .

print (i, j) i.

i.In terms of n, how many ordered pairs are printed?

ii. What is the runtime in notation of this procedure?

Consider the procedure procedure Loops2 (n, a positive integer) 1. for k 1 to n-2 2. for k+1 to n- 1 3. 4. for j i + l to n print (i.) i. In terms of n, how many ordered pairs are printed? ii. What is the runtime in notation of this procedure?

Explanation / Answer

Let me know if you have any doubts.

Solution i) (n-2)*(n-2)*(n-2)

Explanation:

procedure Loops2 (n, a positive integer)

for k = 1 to n -2 => this will run from 1 to (n-2) times, So total (n-2) times

for i = k + 1 to n -1   => this will run from 2 to (n-1) times, So total (n-2) times

for j = i + 1 to n => this will run from 3 to (n) times, So total (n-2) times

print (i, j) => Hence, Total number of times are: (n-2)*(n-2)*(n-2)

Solution ii) Runtime complexity: O(n3)

Explanation:

=> (n-2)*(n-2)*(n-2)

=> n3   ( 2 is constant)

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