Pseudocode Quiz 1

++i and i++ both will increment i for next iteration in for loop.

When the function(6, 8) call since, a is not greater than b and b!=0 so else if condition will becomes true and it return 6 + function(6,7)...
Again when the function(6,7) will call it return (6+function(6,6)).. in this way a remains same and function will call by decrements the value of b when b =0 then it return 0 and recursively a will be added b times, hence we get output (6+6+6+6+6+6+6+6 = 48).

Input m = 12, n = 4
m = m - 1 ; // m = 11
n = n + 1 ; // n = 5
m = m + n ; // m = 16
n = n - 1 ; // 4
Print m,n // 16 4

Initially, i=8 and j=9
When x=1 , j>i // i = 9, j=9
When x=2, j is not greater than i //i=9, j=10
When x=3 , j>i // i = 10, j=10
When x=4, j is not greater than i //i=10, j=11
When x=3 , j>i // i = 11, j=11

In the given code for loop will run for the value of x from 1 to 5, hence the complexity of the code will be O(x).

function(9) -> function(8)+function(7)
function(8) -> function(7)+function(6)
function(7) -> function(6)+function(5)
function(6) -> function(5)+function(4)
function(5) -> function(4)+function(3)
function(4) -> function(3)+function(2)
function(3) -> function(2)+function(1)
function(2) -> function(1)+function(0)
function(1) -> 1
function(0) -> 0
So, recursively, function(2) -> 1+0 => 1
function(3) => 1+1 =>2
function(4) => 2+1 => 3
function(5) => 2+3 => 5
function(6) => 5+3 => 8
function(7) => 8+5 => 13
function(8) => 13+8 => 21
function(9) => 21+13 => 34

The given pseudo code pseudo code represents the code to print the fibonacci series, so the complexity of the code will be O(n) as the for loop will run from 2 to n.

The given piece of the code represents the Post-order tree Traversal
In this traversal method, the root node is visited last, hence the name. First we traverse the left subtree, then the right subtree and finally the root node.

Pre-order Traversal

In this traversal method, the root node is visited first, hence the name. First we traverse the the root node, then left subtree and finally right sub tree.

Since the ASCII value of f is 102 so we get the output as 102.