How to calculate Time Complexity for a given algorithm

nmallickMarch 30, 201054

One of my friends wanted to know "How to calculate the time complexity of a given algorithm". My obvious answer to him was... "Why do YOU want to calculate it?. There are tools available that do it for you!!" (E.g. Analyze menu in VS Team Suite, NDepend are a few). Well... I don't want to say what his answer was, but he wanted to know 🙂

In my personal view, it's a good to know "cool" thing, but not really required for ALL.

With that note, let me write a small program and calculate the time complexity for it.

Here is a sample code to remove an invalid character from an array.

   1: namespace TimeComplexity   2: {   3:     class Program   4:     {   5:         static void Main(string[] args)   6:         {   7:             char[] arr = { 'a', 'b', 'b', 'd', 'e' };   8:             char invalidChar = 'b';   9:             int ptr = 0, N = arr.Length;  10:             for (int i = 0; i < n; i++)  11:             {  12:                 if (arr[i] != invalidChar)  13:                 {  14:                     arr[ptr] = arr[i];  15:                     ptr++;  16:                 }  17:             }  18:    19:             for (int i = 0; i < ptr; i++)  20:             {  21:                 Console.Write(arr[i]);  22:                 Console.Write(' ');  23:             }  24:             Console.ReadLine();  25:         }  26:     }  27: }

Output for the above code will look like

a d e

Let's look at each loop one at a time. That's the key; you calculate the time complexity for each loop

   1: for (int i = 0; i < N; i++)   2: {   3:     if (arr[i] != invalidChar)   4:     {   5:         arr[ptr] = arr[i];   6:         ptr++;   7:     }   8: }

The above Code snippet contains a lot of basic operations which will be repeated. (That's why it's called a loop.. duh !!). The basic operations it contains are

int i=0;	This will be executed only once. The time is actually calculated to i=0 and not the declaration.
i<N;	This will be executed N+1 times
i++ ;	This will be executed N times
if(arr[i]!=invalidChar)	This will be executed N times
arr[ptr]=arr[i];	This will be executed N times (in worst case scenario)
ptr++;	This will be executed N times (in worst case scenario)

Note: I considered the worst case scenario and am calculating the Worst Case Time Complexity for the above code

So the number of operations required by this loop are

{1+(N+1)+N}+N+N+N = 5N+2

The part inside the curly braces is the time consumed by Loop alone (i.e.. for(int i =0;i<N;i++)), it is 2N+2. Keep this mind; it is usually the same (unless you have a non-default FOR loop)

Now for the second loop

   1: for (int i = 0; i < ptr; i++)   2: {   3:     Console.Write(arr[i]);   4:     Console.Write(' ');   5: }

Remember, a loop takes 2N+2 iterations. So, here it will take 2ptr+2 operations. Again, considering the worst case scenario ptr will be N so the above expression evaluates to (again) 2N+2. Then there are these additional 2 operations of Console.Write with will be executed N times each (Again, worst case scenario).

So the above code snippet will take

{1+(N+1)+N}+N+N = 4N+2

Oh.. I almost forgot the other statements

char[] arr = { 'a', 'b', 'b', 'd', 'e' };	This will be executed N times
char invalidChar = 'b';	This will be executed 1 time
int ptr = 0;	This will be executed 1 time
N = arr.Length;	This will be executed 1 time
Console.ReadLine();	This will be executed 1 time

Note: The character array initialization will actually execute N times. This is because you are assigning one character at a time.

So the rest of the code requires

N+4

Adding everything up I get

(N+4)+(5N+2)+(4N+2) = 10N+8

So the asymptotic time complexity for the above code is O(N), which means that the above algorithm is a liner time complexity algorithm.

There you have it, now you know how to calculate the time complexity of a simple program.

Comments (54)

sourabh sharma says:

January 25, 2011 at 5:28 pm

awesome !!!!! thank you for the elaborated explanation :):)
vineet kishor says:

May 18, 2011 at 10:54 pm

thanks body for give a full explanation to show how we can solve the complexity in very easy way…………..
Ni2l says:

June 11, 2011 at 8:25 pm

Thanx a lot, tht ws realy helpfull..
srija says:

July 8, 2011 at 4:24 am

can u give me explanation about program which is having labels at different levels
xyz says:

July 20, 2011 at 4:22 am

make web comfortable to all so that they can get ans
andy says:

July 31, 2011 at 12:39 am

Really helpful….but can u give an explanation for nested loops where d 2nd loop depends on the value of the outer loop variable?
007 says:

August 30, 2011 at 8:58 am

how the statement char[] arr = { 'a', 'b', 'b', 'd', 'e' }; executed N times? isn't it should be only one time ?
opposite says:

September 1, 2011 at 2:45 am

Come on, this is for a particular algorithm…but how can you write a program which calculates the time complexity of a given algoriythm??
sandy says:

September 14, 2011 at 10:20 am

can u give answr 4r dis..time complexity..

i=n;

while(i>=0)

{

x=x+2;

i=i/2;

}
raza says:

September 15, 2011 at 6:07 am

@Sandy since your coder halves the input every time the complexity is O(log n)
raza says:

September 15, 2011 at 6:07 am

@Sandy since your code halves the input every time so the complexity is O(log n)
Atul Rai says:

September 29, 2011 at 2:53 am

i=n;

while(i>=0)

{

x=x+2;

i=i/2;

}

logn is the complexity
priyanka says:

October 2, 2011 at 8:30 am

thanks for giving such array type example.
Tejashree says:

October 16, 2011 at 9:32 pm

amazing n easy 2 understand explanation..thnk:)
Rakesh Menon says:

October 21, 2011 at 7:06 am

Whats the complexity for nested 'for' loops??!??!?!?
aixa says:

October 22, 2011 at 1:19 am

O(N2) is the complexity for two nested for loops…2 is for square!
konda says:

October 22, 2011 at 8:18 pm

what is the time complexity of tower honoe prablem
ABCD says:

November 2, 2011 at 4:28 am

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November 2, 2011 at 4:31 am

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11
MANISH YADAV says:

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fools says:

January 16, 2012 at 9:39 pm

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GOOOOOOOOOOD says:

April 5, 2012 at 10:02 am

Good yara………………………………Thank
laa says:

April 11, 2012 at 9:50 pm

#include<stdio.h>

#include<conio.h>

#include<stdlib.h>

#include<string.h>

struct

{

char name[20],dtype[10],ename[10],native[10],treatment[10],docname[10],med[10],bgroup[10];

int c,age,date,mon,year,ph,ward,a,m,n,o,q,w,e,total,i,nn,L;

}s[10];

main()

{int c,i;

printf("************************WELCOME TO HOSPITAL **** ****************n");

A: ;

printf("choose:n");

printf("1.ENTER PATIENT DETAILS.n2.SHOW PATIENT DETAILS.n3.SEARCH PATIENT DETAILSn4.EXITn");

scanf("%d",&c);

switch(c)

{ case 1:

struct hos

{

char name[20],dtype[10],ename[10],native[10],treatment[10],docname[10],med[10],bgroup[10];

int c,age,date,mon,year,ph,ward,a,m,n,o,q,w,e,total,i,nn,l;

}s[10];

int q,w,e,m,n,o,total,l;

printf("ENTER NO PATIENT DETAILS TO BE ENTER:n");

scanf("%d",&l);

for(i=1;i<=l;i++)

{

printf("nENTER PATIENT NAME: ");

scanf("%s",&s[i].name);

printf("nENTER AGE: ");

scanf("%d",&s[i].age);

printf("nENTER JOIN DATE ddmmyyyy: ");

scanf("%d%d%d",&s[i].date,&s[i].mon,&s[i].year);

printf("ENTER CONTACT NUMBER: ");

scanf("%d",&s[i].ph);

printf("nENTER BLOOD GROUP: ");

scanf("%s",&s[i].bgroup);

printf("nENTER ESCORT NAME: ");

scanf("%s",&s[i].ename);

printf("nENTER NATIVE PLACE: ");

scanf("%s",&s[i].native);

printf("nENTER DISEASE TYPE: ");

scanf("%s",&s[i].dtype);

printf("nENTER TREATMENT TYPE: ");

scanf("%s",&s[i].treatment);

printf("nENTER CONSULTENT DOCTOR: ");

scanf("%s",&s[i].docname);

printf("nENTER WARD NUMBER: ");

scanf("%s",&s[i].ward);

printf("nENTER MEDICINES REQUIRED: ");

scanf("%s",&s[i].med);

printf("nMEDICINE COST: ");

scanf("%d",&s[i].a);

printf("nTOKEN FEE 100");

printf("nENTER DISCHARGE DATE: ");

scanf("%d%d%d",&s[i].m,&s[i].n,&s[i].o);

q=m-s[i].date;w=n-s[i].mon;e=o-s[i].year;

if(w==0)

total=q;

else

total=w*30+q;

}

goto A;

break;

case 2:

B:

for(i=1;i<=l;i++)

{

printf("***********PATIENT DETAILS******************");

printf("nPATIENT NAME: %sn",s[i].name);

printf("AGE: %dn",s[i].age);

printf("JOIN DATE : %d-%d-%dn",s[i].date,s[i].mon,s[i].year);

printf("CONTACT NUMBER: %dn",s[i].ph);

printf("BLOOD GROUP: %s+n",s[i].bgroup);

printf("NATIVE PLACE: %sn",s[i].native);

printf("DISEASE TYPE: %sn",s[i].dtype);

printf("TREATMENT TYPE: %sn",s[i].treatment);

printf("CONSULTENT DOCTOR: %sn",s[i].docname);

printf("WARD NUMBER: %dn",s[i].ward);

printf(" MEDICINES REQUIRED: %sn",s[i].med);

printf("MEDICINES CHARGES: 1900n");

printf("ACAMIDATION CHARGES: 5000n");

printf("TOTAL FEE:6900n");

printf("TOTAL DAYS :%d daysn",s[i].total);

}

goto A;

break;

case 3:

// for(i=0;i<=2;i++)

// {

int war;

int sa;

printf("nfor searching choose your option");

printf("n5.PATIENT NAME n6.WARD NUMBER n7.DATE OF ADMISSIONn8.CONTACT NUMBERn");

scanf("%d",&sa);

switch(sa)

{

case 5:

char p[10];

printf("enter patient name");

scanf("%s",&p);

if((strcmp(s[i].name,p)==0))

{

printf("***********PATIENT DETAILS******************");

printf("nPATIENT NAME: %sn",s[i].name);

printf("AGE: %dn",s[i].age);

printf("JOIN DATE : %d-%d-%dn",s[i].date,s[i].mon,s[i].year);

printf("CONTACT NUMBER: %dn",s[i].ph);

printf("BLOOD GROUP: %s+n",s[i].bgroup);

printf("NATIVE PLACE: %sn",s[i].native);

printf("DISEASE TYPE: %sn",s[i].dtype);

printf("TREATMENT TYPE: %sn",s[i].treatment);

printf("CONSULTENT DOCTOR: %sn",s[i].docname);

printf("WARD NUMBER: %dn",s[i].ward);

printf(" MEDICINES REQUIRED: %sn",s[i].med);

printf("MEDICINES CHARGES: 1900n");

printf("ACAMIDATION CHARGES: 5000n");

printf("TOTAL FEE:6900n");

printf("TOTAL DAYS :%d daysn",s[i].total);

}

else

printf("data not exist");

break;

case 6:

printf("enter ward number");

scanf("%d",&war);

if(s[i].ward==war)

{

printf("***********PATIENT DETAILS******************");

printf("nPATIENT NAME: %sn",s[i].name);

printf("AGE: %dn",s[i].age);

printf("JOIN DATE : %d-%d-%dn",s[i].date,s[i].mon,s[i].year);

printf("CONTACT NUMBER: %dn",s[i].ph);

printf("BLOOD GROUP: %s+n",s[i].bgroup);

printf("NATIVE PLACE: %sn",s[i].native);

printf("DISEASE TYPE: %sn",s[i].dtype);

printf("TREATMENT TYPE: %sn",s[i].treatment);

printf("CONSULTENT DOCTOR: %sn",s[i].docname);

printf("WARD NUMBER: %dn",s[i].ward);

printf(" MEDICINES REQUIRED: %sn",s[i].med);

printf("MEDICINES CHARGES: 1900n");

printf("ACAMIDATION CHARGES: 5000n");

printf("TOTAL FEE:6900n");

printf("TOTAL DAYS :%d daysn",s[i].total);

}

else

printf("data not exist");

break;

case 7:

int dd,mm,yy;

printf("ENTER DATE OF ADMISSION");

scanf("%d%d%d",&dd,&mm,&yy);

if(dd==s[i].date&&mm==s[i].mon&&yy==s[i].year)

{

printf("***********PATIENT DETAILS******************");

printf("nPATIENT NAME: %sn",s[i].name);

printf("AGE: %dn",s[i].age);

printf("JOIN DATE : %d-%d-%dn",s[i].date,s[i].mon,s[i].year);

printf("CONTACT NUMBER: %dn",s[i].ph);

printf("BLOOD GROUP: %s+n",s[i].bgroup);

printf("NATIVE PLACE: %sn",s[i].native);

printf("DISEASE TYPE: %sn",s[i].dtype);

printf("TREATMENT TYPE: %sn",s[i].treatment);

printf("CONSULTENT DOCTOR: %sn",s[i].docname);

printf("WARD NUMBER: %dn",s[i].ward);

printf(" MEDICINES REQUIRED: %sn",s[i].med);

printf("MEDICINES CHARGES: 1900n");

printf("ACAMIDATION CHARGES: 5000n");

printf("TOTAL FEE:6900n");

printf("TOTAL DAYS :%d daysn",s[i].total);

}

else

printf("data not exist");

break;

case 8:

int mo;

printf("ENTER MOBILE NUMBER");

scanf("%d",&mo);

if(s[i].ph==mo)

{

printf("***********PATIENT DETAILS******************");

printf("nPATIENT NAME: %sn",s[i].name);

printf("AGE: %dn",s[i].age);

printf("JOIN DATE : %d-%d-%dn",s[i].date,s[i].mon,s[i].year);

printf("CONTACT NUMBER: %dn",s[i].ph);

printf("BLOOD GROUP: %s+n",s[i].bgroup);

printf("NATIVE PLACE: %sn",s[i].native);

printf("DISEASE TYPE: %sn",s[i].dtype);

printf("TREATMENT TYPE: %sn",s[i].treatment);

printf("CONSULTENT DOCTOR: %sn",s[i].docname);

printf("WARD NUMBER: %dn",s[i].ward);

printf(" MEDICINES REQUIRED: %sn",s[i].med);

printf("MEDICINES CHARGES: 1900n");

printf("ACAMIDATION CHARGES: 5000n");

printf("TOTAL FEE:6900n");

printf("TOTAL DAYS :%d daysn",s[i].total);

}

else

printf("DATA NOT EXIST");

break;

getch();

}}}
laa says:

April 11, 2012 at 9:51 pm

caculate time complexcity for above program
gaandu says:

May 23, 2012 at 10:13 am

maa chudao bhosadi k…..chpta program to solve hota nahi..ye hoga
kau says:

June 2, 2012 at 12:45 am

how the statement char[] arr = { 'a', 'b', 'b', 'd', 'e' }; executed N times? isn't it should be only one time ?
sumit says:

June 4, 2012 at 4:36 am

awesome what a explanation of for loops……………….
Ali says:

July 29, 2012 at 9:46 pm

Wow..wonderful explanation………………
Vimal says:

September 11, 2012 at 9:11 am

Now that you have given the time complexity of "removing an invalid character", can you please show me how the hell the time complexity of a Heap sort is O(n log n)?
deeeppak says:

September 19, 2012 at 10:28 pm

dsdssssassssssssmmammnnmnanmnamnmmammakdkanannd
Jawad Baghii says:

September 25, 2012 at 4:09 am

Thanx a lot…this will definitely work for me…..JUG JUG GEO
sam says:

November 20, 2012 at 10:42 pm

thankx this is a really simple ans………:P
Student says:

February 18, 2013 at 8:08 pm

Thanks for posting this. It help a lot
nav says:

April 17, 2013 at 9:24 am

what is the complexity of my code.?????

try{

ReadExcel test = new ReadExcel();

test.setInputFile("C:/Users/PRIYA/Documents/NetBeansProjects/JavaApplication2/src/dbscan/Input.xls");

corrdinatMap= (HashMap)test.read();

System.out.println(">>>1");

if(corrdinatMap.containsKey("xCordinat"))

xAxis=(ArrayList)corrdinatMap.get("xCordinat");

if(corrdinatMap.containsKey("yCordinat"))

yAxis=(ArrayList)corrdinatMap.get("yCordinat");

System.out.println(">>>2");

xX = (Object[]) xAxis.toArray();

yY = (Object[]) yAxis.toArray();

System.out.println(">>>3");

if(xX.length > yY.length)

loopCount=xX.length;

else

loopCount=yY.length;

Class.forName("oracle.jdbc.driver.OracleDriver");

user="system";

password="priya";

URL="jdbc:oracle:thin:@localhost:1521:XE";

}

catch(Exception e){System.out.println("Exception>>1>>>"+e);}

for(int loop=0;loop< loopCount;loop++)

{

try{

tfx.setText(xX[loop].toString());

tfy.setText(yY[loop].toString());

x1 = Integer.parseInt(tfx.getText());

y1 = Integer.parseInt(tfy.getText());

System.out.println("aadPoints1>>>>"+x1+", "+y1);
moh kalid aayar noole says:

September 15, 2013 at 9:30 pm

thnz for ur giving full expntion of time complxty
Aysel says:

September 30, 2013 at 12:43 am

The most understandable answer i have ever got for complexity.. well done! 🙂
poppy says:

November 28, 2013 at 10:24 am

Is the complexity for the foll algorithm O(n)???

if(n<=2)

return 1;

else

return (A/n^0.5);
Pro says:

November 29, 2013 at 12:53 am

@Poppy..your program has a constant time complexity…represented as O(1).
msj says:

December 10, 2013 at 5:00 pm

how to decide that the following answer is asymtotic..tnx
Mohd haseen says:

May 7, 2014 at 12:15 am

Good but simple you should give another complex after this
Marry says:

December 21, 2014 at 7:03 pm

Anyone can help?

1. Dim C(n) As Integer

2. AA(C, 1, n)

3. For j As Integer = 1 To n

4. Dim i As integer = Sqrt(j)

5. print C(i)

6. Next

7. Sub AA (C As Integer(), m As Integer, n As Integer)

8. Dim p As Integer = (n-m+1)/3

9. If p > 5 Then

10. For i As Integer = 1 to 4

11. AA (C, m, m+p)

12. Next

13. For i As Integer = m To n

14. C(i)=C(i)+1

15. Next

16. End If

17. End Sub
Shashikant Mitkari says:

January 9, 2015 at 2:52 pm

@Laa:

Time complexity of your program is O(n)…

pretty simple.. 🙂
Blu says:

January 21, 2015 at 8:58 pm

Thanks.. Today is my exam and this really helped
Rana Nayab says:

March 16, 2015 at 5:29 am

i want to learn from starting to end any platform where i can learn algo and find conplexity
Civilized Gamer says:

May 17, 2015 at 12:37 am

How did you go from 10N + 8 to O(N)?
Phezi says:

August 30, 2015 at 3:14 am

a) What is an appropriate Big O expression for

for (int i = 0; i <= n -1; i++)

{

for (int j = i + 1; j <= n – 1; j++)

{

loop body;

}

}
Phezi says:

August 30, 2015 at 3:30 am

And:

for (int i = 0; i <= n-1; i += 2)

{

for (int j = n-1; j > 0; j–)

{

loop body;

}

}
Dee says:

September 13, 2015 at 10:56 am

I appreciate the explanation. It's very helpful
handa says:

November 5, 2015 at 11:52 pm

could you give the time complexity for this

int k=0;

for(int i=0;i<n;i++)

for(int j=i;j<n;j++)

k++
Raj says:

December 5, 2015 at 7:44 am

Good explanation!! Appreciate it!! But

What will b the running time fa

Sum= a[i];
Softek Information Services says:

January 3, 2016 at 10:11 am

How did you go from 10N + 8 to O(N)?
pruthviraj says:

January 26, 2016 at 7:54 pm

thanks for explanig in very easy langauge i really tell u i just understand little bit but i will make the practices for this session so if u r reading my comment then plz just text me on my no. i would like to help from u my mob no. is 8793666371 thank you so much…………….
Shahbaz Hassan says:

February 19, 2016 at 7:37 pm

Asalam o alikum. Your defining method was very good. Very good explanation

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