How to calculate Time Complexity for a given algorithm

Artikkeli
03/30/2010

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

Anonymous
January 25, 2011
awesome !!!!! thank you for the elaborated explanation :):)
Anonymous
May 18, 2011
thanks body for give a full explanation to show how we can solve the complexity in very easy way..............
Anonymous
June 11, 2011
Thanx a lot, tht ws realy helpfull..
Anonymous
July 07, 2011
can u give me explanation about program which is having labels at different levels
Anonymous
July 19, 2011
make web comfortable to all so that they can get ans
Anonymous
July 30, 2011
Really helpful....but can u give an explanation for nested loops where d 2nd loop depends on the value of the outer loop variable?
Anonymous
August 30, 2011
how the statement char[] arr = { 'a', 'b', 'b', 'd', 'e' }; executed N times? isn't it should be only one time ?
Anonymous
August 31, 2011
Come on, this is for a particular algorithm...but how can you write a program which calculates the time complexity of a given algoriythm??
Anonymous
September 14, 2011
can u give answr 4r dis..time complexity.. i=n; while(i>=0) { x=x+2; i=i/2; }
Anonymous
September 14, 2011
@Sandy since your coder halves the input every time the complexity is O(log n)
Anonymous
September 14, 2011
@Sandy since your code halves the input every time so the complexity is O(log n)
Anonymous
September 28, 2011
i=n; while(i>=0) { x=x+2; i=i/2; } logn is the complexity
Anonymous
October 02, 2011
thanks for giving such array type example.
Anonymous
October 16, 2011
amazing n easy 2 understand explanation..thnk:)
Anonymous
October 20, 2011
Whats the complexity for nested 'for' loops??!??!?!?
Anonymous
October 21, 2011
O(N2) is the complexity for two nested for loops...2 is for square!
Anonymous
October 22, 2011
what is the time complexity of tower honoe prablem
Anonymous
November 01, 2011
SDEGFFJDHSGFDHGSFDHSFGDHGSFDHS DJGHKDJFGHDF DFJKHGKJDFGH DKJFGHKFDJG
Anonymous
November 01, 2011
fdgdfklhgdf fglkhflkg fghfgh fgh gfh fg hf gh gf hfg h gfh gf hf gh fgh fg hfg hgf h fgh fg hfg h gfh fg hfg h gfh fg h fgh gf hgf h fg hf gh fg hfg h fg h fgh fg h gfh fg h fg hfg hfg h fgh gfh gf h fgh fgh fgh fgh fg h fgh gf hf gh fgh hjfgh hjkjljk jkl jkl jl jk ljk l jkl jk l jl jkl jk l jkl jk lkj ljk
Anonymous
November 01, 2011
1 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 111111111 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1
Anonymous
December 06, 2011
ase sa as as sas sa sas as sa ss sa as asas dd df ad 1 1 1 1 1 1 1 3e 2 2 2 xdasd sd ds dds ads a 3 w d sd s s wwad a as ds s s s s sadsa sa s ddfr f a a g aa f f gg ag a a a a a F F G SGFD FD FD DF DFSDFAGF F F FD DF DF F FD FD DF FD DF F DF DF DF DF FD FDDF DFDFASS SD SD D D D D F FD DF DF G G H D S S SD S S S DASDS DS DS Z C C C C CX CX C CX C CX CX XC C CX CX X ZZZ ZX X X X XZ VC C CVX Z XCZ XZ XZ XZ XZ X Z XZ X XZ XZ X X XZ C VC CV CV X C Z XC V CX Z Z Z CVCVVZ ZC XZ XC X C XC XZ CZ ZC ZC CZ C C Z
Anonymous
January 16, 2012
bgdgdg dfg sdfg sdf gsdfgsdfgsdfgsdg dfgsdfg dgsdf gsdf gsdf gsdfg sd gsdfg sdfg sd gsd gsdf gsd g sdfg dg sdfg sd gdg df gsd gsdf gsdf g db b dg sdfg dgwe rtw erter t wert wert wer twer twer t wert wet we te t wert wet wert
Anonymous
April 05, 2012
Good yara....................................Thank
Anonymous
April 11, 2012
#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=w30+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); &nb
Anonymous
April 11, 2012
caculate time complexcity for above program
Anonymous
May 23, 2012
maa chudao bhosadi k.....chpta program to solve hota nahi..ye hoga
Anonymous
June 01, 2012
how the statement char[] arr = { 'a', 'b', 'b', 'd', 'e' }; executed N times? isn't it should be only one time ?
Anonymous
June 03, 2012
awesome what a explanation of for loops...................
Anonymous
July 29, 2012
Wow..wonderful explanation..................
Anonymous
September 11, 2012
The comment has been removed
Anonymous
September 19, 2012
dsdssssassssssssmmammnnmnanmnamnmmammakdkanannd
Anonymous
September 24, 2012
Thanx a lot...this will definitely work for me.....JUG JUG GEO
Anonymous
November 20, 2012
thankx this is a really simple ans.........:P
Anonymous
February 18, 2013
Thanks for posting this. It help a lot
Anonymous
April 17, 2013
The comment has been removed
Anonymous
September 15, 2013
thnz for ur giving full expntion of time complxty
Anonymous
September 29, 2013
The most understandable answer i have ever got for complexity.. well done! :)
Anonymous
November 28, 2013
Is the complexity for the foll algorithm O(n)??? if(n<=2) return 1; else return (A/n^0.5);
Anonymous
November 28, 2013
@Poppy..your program has a constant time complexity...represented as O(1).
Anonymous
December 10, 2013
how to decide that the following answer is asymtotic..tnx
Anonymous
May 06, 2014
Good but simple you should give another complex after this
Anonymous
December 21, 2014
Anyone can help?

Dim C(n) As Integer
AA(C, 1, n)
For j As Integer = 1 To n
Dim i As integer = Sqrt(j)
print C(i)
Next
Sub AA (C As Integer(), m As Integer, n As Integer)
Dim p As Integer = (n-m+1)/3
If p > 5 Then
For i As Integer = 1 to 4
AA (C, m, m+p)
Next
For i As Integer = m To n
C(i)=C(i)+1
Next
End If
End Sub

Anonymous
January 09, 2015
@Laa: Time complexity of your program is O(n)... pretty simple.. :)
Anonymous
January 21, 2015
Thanks.. Today is my exam and this really helped
Anonymous
March 15, 2015
i want to learn from starting to end any platform where i can learn algo and find conplexity
Anonymous
May 16, 2015
How did you go from 10N + 8 to O(N)?
Anonymous
August 29, 2015
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; } }
Anonymous
August 29, 2015
And: for (int i = 0; i <= n-1; i += 2) { for (int j = n-1; j > 0; j--) { loop body; } }
Anonymous
September 13, 2015
I appreciate the explanation. It's very helpful
Anonymous
November 05, 2015
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++
Anonymous
December 04, 2015
Good explanation!! Appreciate it!! But What will b the running time fa Sum= a[i];
Anonymous
January 03, 2016
How did you go from 10N + 8 to O(N)?
Anonymous
January 26, 2016
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................
Anonymous
February 19, 2016
Asalam o alikum. Your defining method was very good. Very good explanation

Jaa

How to calculate Time Complexity for a given algorithm

Comments

Lisäresursseja