Skip to main content

Design an algorithm to find the GCD and LCM of two positive numbers

Problem : Design an algorithm to find the GCD (Greatest Common Divisor) of two positive numbers and use the same to find LCM of two positive numbers.

Input: Two Numbers
Output : Least Common Multiple of two numbers.

Rem <-- NUM1 % NUM2  [% Modulus Operator]
Repeat While Rem <> 0
NUM1 <-- NUM2
NUM2 <-- Rem
Rem <-- NUM1 % NUM2
[End of While]
Return NUM2

Return (NUM1 * NUM2) / GCD( NUM1, NUM2)

NOTE : Here in this example algorithm, finding LCM in the main problem that can be divided into sub-problem like finding the GCD first and using the same to solve the main problem. So, GCD algorithm is written first and than the other algorithm that can be divided into small problems and algorithm can be written to solve such small problems.


Suppose that the LCM algorithm is called using 12 and 16. The algorithm calls GCD with 12 and 16.

Rem <-- 12 % 16 i.e. Rem= 12
Rem is not equal to zero. So, NUM1=16 and NUM2 =12
Rem<-- 16%12 i.e. Rem=4
Rem is not equal to Zero. So, NUM1 =12 and NUM2 =4
Rem <-- 12 % 4 i.e. Rem=0
Rem is equal to Zero.
So, GCD algorithm returns NUM2 that is 4.

When the algorithm LCM gets the result from GCD algorithm its find the expression value
(NUM1 * NUM2) / GCD (NUM1, NUM2).  i.e. (12 * 16)/4

It is equal to 48. 48 is the LCM of 12 and 16


Popular posts from this blog

difference between structure and union in C Language

In c language article we will see the difference between union and structure. Both are the user define datatype in c language. See the table which is mentioned below: ASP.NET Video Tutorial Series Structure Union1.The keywordstruct is used to define a structure 1. The keyword union is used to define a union. 2. When a variable is associated with a structure, the compiler allocates the memory for each member. The size of structure is greater than or equal to the sum ofsizes of its members. The smaller members may end with unused slack bytes. 2. When a variable is associated with a union, thecompiler allocates thememory by considering the size of the largest memory. So, size of union is equal to the size of largest member. 3. Each member within a structure is assigned unique storage area of location. 3. Memory allocated is shared by individual members of union. 4. The address of each member will be in ascending order This indicates that memory for each member will start at different offset v…

Difference between Linear search and Binary Search in c language

SQL Video Channel : Download all SQL Video

Binary Search Linear Search Works only on sorted items. such as  1,2,3,4,5,6  etc
Works on sorted as well as unsorted items. 12,4,5,3,2,1 etc Very efficient if the items are sorted Very efficient if the items are less and present in the beginning of the list. such as Suppose your list items are : 12,3,4,5,1 and you want to search 12 number then you get beginning in the list. Works well with arrays and not on linked lists. Works with arrays and linked lists.
Number of comparisons are less More number of comparisons are required if the items are present in the later part of the array or its elements are more.

Memory representation of Linked List Data Structures in C Language

Memory representation of Linked List

             In memory the linked list is stored in scattered cells (locations).The memory for each node is allocated dynamically means as and when required. So the Linked List can increase as per the user wish and the size is not fixed, it can vary.

               Suppose first node of linked list is allocated with an address 1008. Its graphical representation looks like the figure shown below:

      Suppose next node is allocated at an address 506, so the list becomes,

  Suppose next node is allocated with an address with an address 10,s the list become,

The other way to represent the linked list is as shown below:

 In the above representation the data stored in the linked list is “INDIA”, the information part of each node contains one character. The external pointer root points to first node’s address 1005. The link part of the node containing information I contains 1007, the address of next node. The last node …