How to Calculate Time Complexity of an Algorithm + Solved Questions (With Notes)
CodeWithHarry
Before Solving Some Questions of Time Complexity I will tell you some tricks to get rid of time
complexity. After that we will do some set of questions. which will make you a very good grasp in
such questions. due to the time complexity of any algorithm when you have to nd it so what is the
rst step that you do and at the same time how to approach this problem. In this way, whatever
instructions are going on here , it is taking almost ( k ) time. We believe that these operations are all (k
) time consuming This for loop , that is , how much time is being taken for this fragment It seems
( kn ) , okay So before this ( int i ) would have been written here, ( int k=0 ) would be written here. The
third technique that I want to tell you is this : That break the code into fragments. The rst fragment
turned out to be this one , with a little bit of initialization. It took constant time because it is not such
that if the value of ( n ) increases, then its time will increases.
I will go for ( n = 100 ) to determine whether i will be going for n = 1000. I will accept it in ( k4) and ( n *
k4 ) I will do it in k4 and (n* k4), and it will happen O (n²) If you do it ( k=0 ) , ( k < n ) and if you look at
it , it will come out O ( n²) Okay, it will not (N²) ok remember you this thing. There will be some code
on it which will take ( k1 ) Now I have become so smart, by doing questions , and you will be done
too That (k1) it is will going to be non-dominant , if constant is being added then we will remove it. So
once the value of ( i ) will be zero (0) and then the value of ( j ) will run for. Then ( j=1) will become Then
( 0,2) Then (i=0 , and j=0 ) will then run for Okay. The value of ( i ) will be zero ( 0 ) for ( n ) times
running then the value of (i ) will become ( 1 ) , it will run again (n ) times then it will go on till n. When
( n) is running out, watch carefully , watch very closely. Then later I will ask the question, then I am
telling if it is not done. value of i will be ( 1) , ( n ), it will be n-1 because I am taking the index ( i=0 )
then (i=n-1) will be and here is ( n- 1 )
I told you guys If it 's not clear to you why it will work ( n² ) times So I 'd say let 's go look at it for 3
and 3 and print here (i , j ) and make a count variable and count it , how many times it is running You
write ( c ) program , write in Python, write in Java, write in Python and write in the Java. But when
there are 2 loops inside one , then that will run for n² times. And if another loop is given inside it , then
it will run ( n³ ) times. If there is a double for loop, then it becomes straight (n² ) I have handpicked
some questions which I am going to give to you guys here. And I have also given their programs to
you. So you see here I have opened this folder in visual studio code. So it 's saying that Find the time
complexity ( Func1 ) function in the program shown in program1. c as follows. Even if you come from
another programming language nothing is going to be change. The time of (F1+F2+F3) will be that I
will take as the overall time of the whole function. The time is not depending on Array 's length so i 'll
accept it ( k1 ) and I can not accept ( F2 ) as ( k ) , i will accept it as k2 * n. So now if I nd a total time
complexity that is , if I T ( n ) come out , then what will it be ? T (n ) will be done as T (N) =F1 +F2 +F3.
The answer is O (length) The algorithm it is talking in terms of length over here. The time complexity
of the Func function in the program from program2. c as follows. Find the , nd the , I have written
ne here , let me correct it I wrote ne above also , nd this man , this is nd Never mind , these small
mistakes are made. If you look carefully above it is written ( n ) times, if you write 1 for ( n times as
( 1+1+1 +1 + n) So what will happen if I write (n²) so it is done O ( n²) So anytime you see the double
for loop , and it 's going to [ 0, n ] , [ 0 , n ] and [ 0 n ] i. e. for 0 , ( n) times it will run OK , so this will
run 0 for (n) times, then 1 for n times then it will again 2 for n. times. And there is some constant work
going on inside it so it means The average processing time is T ( n ) so T ( 6 ) is a small number. If I
directly solve for T (6 ) I will try to solve the problem directly for T. only. And if this is a simple problem,
CodeWithHarry
Before Solving Some Questions of Time Complexity I will tell you some tricks to get rid of time
complexity. After that we will do some set of questions. which will make you a very good grasp in
such questions. due to the time complexity of any algorithm when you have to nd it so what is the
rst step that you do and at the same time how to approach this problem. In this way, whatever
instructions are going on here , it is taking almost ( k ) time. We believe that these operations are all (k
) time consuming This for loop , that is , how much time is being taken for this fragment It seems
( kn ) , okay So before this ( int i ) would have been written here, ( int k=0 ) would be written here. The
third technique that I want to tell you is this : That break the code into fragments. The rst fragment
turned out to be this one , with a little bit of initialization. It took constant time because it is not such
that if the value of ( n ) increases, then its time will increases.
I will go for ( n = 100 ) to determine whether i will be going for n = 1000. I will accept it in ( k4) and ( n *
k4 ) I will do it in k4 and (n* k4), and it will happen O (n²) If you do it ( k=0 ) , ( k < n ) and if you look at
it , it will come out O ( n²) Okay, it will not (N²) ok remember you this thing. There will be some code
on it which will take ( k1 ) Now I have become so smart, by doing questions , and you will be done
too That (k1) it is will going to be non-dominant , if constant is being added then we will remove it. So
once the value of ( i ) will be zero (0) and then the value of ( j ) will run for. Then ( j=1) will become Then
( 0,2) Then (i=0 , and j=0 ) will then run for Okay. The value of ( i ) will be zero ( 0 ) for ( n ) times
running then the value of (i ) will become ( 1 ) , it will run again (n ) times then it will go on till n. When
( n) is running out, watch carefully , watch very closely. Then later I will ask the question, then I am
telling if it is not done. value of i will be ( 1) , ( n ), it will be n-1 because I am taking the index ( i=0 )
then (i=n-1) will be and here is ( n- 1 )
I told you guys If it 's not clear to you why it will work ( n² ) times So I 'd say let 's go look at it for 3
and 3 and print here (i , j ) and make a count variable and count it , how many times it is running You
write ( c ) program , write in Python, write in Java, write in Python and write in the Java. But when
there are 2 loops inside one , then that will run for n² times. And if another loop is given inside it , then
it will run ( n³ ) times. If there is a double for loop, then it becomes straight (n² ) I have handpicked
some questions which I am going to give to you guys here. And I have also given their programs to
you. So you see here I have opened this folder in visual studio code. So it 's saying that Find the time
complexity ( Func1 ) function in the program shown in program1. c as follows. Even if you come from
another programming language nothing is going to be change. The time of (F1+F2+F3) will be that I
will take as the overall time of the whole function. The time is not depending on Array 's length so i 'll
accept it ( k1 ) and I can not accept ( F2 ) as ( k ) , i will accept it as k2 * n. So now if I nd a total time
complexity that is , if I T ( n ) come out , then what will it be ? T (n ) will be done as T (N) =F1 +F2 +F3.
The answer is O (length) The algorithm it is talking in terms of length over here. The time complexity
of the Func function in the program from program2. c as follows. Find the , nd the , I have written
ne here , let me correct it I wrote ne above also , nd this man , this is nd Never mind , these small
mistakes are made. If you look carefully above it is written ( n ) times, if you write 1 for ( n times as
( 1+1+1 +1 + n) So what will happen if I write (n²) so it is done O ( n²) So anytime you see the double
for loop , and it 's going to [ 0, n ] , [ 0 , n ] and [ 0 n ] i. e. for 0 , ( n) times it will run OK , so this will
run 0 for (n) times, then 1 for n times then it will again 2 for n. times. And there is some constant work
going on inside it so it means The average processing time is T ( n ) so T ( 6 ) is a small number. If I
directly solve for T (6 ) I will try to solve the problem directly for T. only. And if this is a simple problem,
