Accenture Pseudo Code Quiz-1a

The code will acquire a space of 2n+8.
There are n elements in the array and the int data type acquires 2 bytes, so there will be 2n bytes acquired.
The int array occupies 4 bytes.
The sum variable will occupy 2 bytes.
The i variable will occupy 2 bytes.
so total=2n+4+2+2=2n+8 bytes

function(8,9) will return function(9,8)
function(9,8) will return 9+function(9,7)
function(9,7) will return 9+9+function(9,6)
function(9,6) will return 9+9+9+function(9,5)
function(9,5) will return 9+9+9+9+function(9,4)
function(9,4) will return 9+9+9+9+9+function(9,3)
function(9,3) will return 9+9+9+9+9+9+function(9,2)
function(9,2) will return 9+9+9+9+9+9+9+function(9,1)
function(9,1) will return 9+9+9+9+9+9+9+9+function(9,0)

function(9,0) will return 9+9+9+9+9+9+9+9+0

So the result is 72.

m = m + 1, so m=9+1=10

n = n - 1, so n=6-1=5

m = m + n, so m=10+5=15

Hence the output is 15

In this pseudo code, the for loop operates from n = 6 until n < 9.

In the first iteration we have the value of n = 6, sum = 0 + 6 = 6.

In the next iteration we have the value of n = 7, sum = 6 + 7 = 13.

In the third iteration we have the value of n = 8, sum = 13 + 8 = 21.

The next iteration wouldn't be executed since the condition of the loop n < 9 becomes false. And thus the loop terminates.

So the final answer we get is 21.

There are three variables i, j, and k declared. Value initialized for k is 8, In this code, we are moving with nested for loop.

Here I value is 1, for loop will check the condition i<=1 condition gets true. Now, moving with inner for loop j value will be 1 condition gets true j<=1.so, it prints K+1. Then j value will be incremented by 1(2<+1) inner for loop condition gets false.

On the next iteration, i value will be incremented by 1, here the updated i value is 2 (2<=1) condition get false. So the answer is 9.

if(1<3 && 2<4): Since this is true. So, return funn(2,3) .

funn (2,3) will return funn(3,4)

funn (3,4) will return a+b=3+4=7

Hence, 7 will be the output

a=2,b=6

Since both a,b is greater than 0. Hence funn will return 2+6+funn(3,0)+funn(4,0)+funn(5,0)

funn(3,0) will return a+b=3+0

funn(4,0) will return a+b=4+0

funn(5,0) will return a+b=5+0

So, 2+6+3+4+5=20

hence output will be 20

For loop will iterate from count=0 to count=8

Count=0: print "#", print count, so 0 will be printed

Count=1: print "#", print count, so 1 will be printed

Count=2: print "#", print count, so 2 will be printed

Count=3: print "#", print count, so 3 will be printed

Count=4: print "#", print count, so 4 will be printed

Count=5: print "#", print count, so 5 will be printed

Count=6 :print "#", print count, so 6 will be printed

Count =7: print"#", then if condition will become true, hence now next iteration will work.

Count=8: print"#", then if condition will become true, hence now next iteration will work.

Count=9: Condition will become become false. Now exit from for loop.

In a postorder traversal, first the left child is visited, then the right child and finally the parent.

Learn more about Inorder Postorder preorder here

Inorder Preorder postorder example here

Inorder

• Traverse the left sub-tree, (recursively call inorder(root -> left).


• Visit and print the root node.


• Traverse the right sub-tree, (recursively call inorder(root -> right).

• Direction : Clockwise direction


• Rule : LCR i.e. Left ,Center(root), Right

Preorder

• Visit and print the root node.


• Traverse the left sub-tree, (recursively call Preorder(root -> left).


• Traverse the right sub-tree, (recursively call Preorder(root -> right)

• Direction : Anti-Clockwise direction


• Rule : CLR i.e. Center(root) ,left, Right

Postorder

• Traverse the left sub-tree, (recursively call Postorder(root -> left).


• Traverse the right sub-tree, (recursively call Postorder(root -> right).


• Visit and print the root node

• Direction : Anti-Clockwise direction


• Rule : LRC i.e. Left, Right, Center(root)

Since all the elements lesser than a given node will be towards the left, iterating to the leftmost leaf of the root will give the minimum element in a binary search tree.

Read more about Binary Search Tree here

Binary search tree in Data Structures (Introduction)

• BST: Introduction to Binary Search Tree


• BST: Binary Search Tree Program


• BST: Search a node in Binary Search Tree