Introduction to complexity of algorithm

Table of Contents

What is an algorithm?
Why do you need to evaluate an algorithm?
Counting number of instructions:
Asymptotic behaviour :
How will you compare algorithms?
Big O Notation:
Rules of thumb for calculating complexity of algorithm:
Logarithmic complexity
Exercise:

“How will you calculate complexity of algorithm” is very common question in interview.How will you compare two algorithm? How running time get affected when input size is quite large? So these are some question which is frequently asked in interview.In this post,We will have basic introduction on complexity of algorithm and also to big o notation

What is an algorithm?

An algorithm is step by step instructions to solve given problem.

Lets take a simple example.You want to write an algorithm for listening particular song.

1) Search for song on computer.

2) Is song available?

i.If Yes,Then listen that song.

ii.If no,download that song and then listen that song.

So we are solving a problem by step by step procedure.This step by step instructions is called Algorithm.

Why do you need to evaluate an algorithm?

You need to evaluate an algorithm so that you can find most optimize algorithm for solving given problem and also considering various factors and constraints.

For example:

You want to go from city A to City B.Then there are various choices available i.e. by flight,bus or train.So you need to choose among different options depending on your budget and urgency.

Counting number of instructions:

Number of instructions can be different for different programming languages.

Lets count number of instruction for searching a element in a array.


int n=array.length
for (int i = 0; i < n; i++) {

 if(array[i]==elementToBeSearched)
 return true;
 }
 return false;

int n=array.length

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

if(array[i]==elementToBeSearched)

return true;

}

return false;

Let’s assume our processor takes one instruction for each of the operation:

For assigning a value to a variable
For comparting two values
Multiply or addition
Return statement

In Worst case:

If element which we want to search is last element in sorted array then it will be worst case here.
So :


if(array[i]==elementToBeSearched) ,i++ and i<n <b>will be executed n times</b>
int n=array.length,i=0,return true or false <strong>will be executed one time.
</strong>

if(array[i]==elementToBeSearched) ,i++ and i<n <b>will be executed n times</b>

int n=array.length,i=0,return true or false <strong>will be executed one time.

</strong>

Hence f(n)=3n+3

Asymptotic behaviour :

Here We will see how f(n) performs with larger value of n.Now in above function, we have two parts i.e. 3n and 3. Here you can note two points:

As n grows larger, we can ignore constant 3 as it will be always 3 irrespective of value of n. It makes sense as you can consider 3 as initialization constant and different language may take different time for initialization.So other function remains f(n)=3n.
We can ignore constant multiplier as different programming language may compile the code differently. For example array look up may take different number of instructions in different languages. So what we are left with is f(n)=n

How will you compare algorithms?

You can compare algorithms by its rate of growth with input size n

Lets take a example.For solving same problem, you have two functions:
f(n) =4n^2+2n+4 and f(n) =4n+4

For f(n) =4n^2²+2n+4
so here
f(1)=4+2+4
f(2)=16+4+4
f(3)=36+6+4
f(4)=64+8+4
….
As you can see here contribution of n^2²increasing with increasing value of n.So for very large value of n,contribution of n^2 will be 99% of value on f(n).So here we can ignore low order terms as they are relatively insignificant as described above.In this f(n),we can ignore 2n and 4.so
n^2+2n+4 ——–>n^2

For f(n) =4n+4
so here
f(1)=4+4
f(2)=8+4
f(3)=12+4
f(4)=16+4
….
As you can see here contribution of n increasing with increasing value of n.So for very large value of n,contribution of n will be 99% of value on f(n).So here we can ignore low order terms as they are relatively insignificant.In this f(n),we can ignore 4 and also 4 as constant multiplier as seen above so
4n+4 ——–>nSo here n is highest rate of growth.
Point to be noted :
We are dropping all the terms which are growing slowly and keep one which grows fastest.

Big O Notation:

This notation is used for theoretical measure of execution of an algorithm. It gives tight upper bound of a given function. Generally it is represented as f(n)=O(g(n)) and it reads as “f of n is big o of g of n”.

Formal definition:

f(n) = O(g(n)) means there are positive constants c and n0, such that 0 ≤ f(n) ≤ cg(n) for all n ≥ n0. The values of c and n0 must not be depend on n.

When you say O(g(n)) , It means it will never be worst than g(n). Having said that it means O(g(n)) includes smaller or same order of growth as g(n).

So O(n) includes O(n),O(logn) and O(1).

So O(g(n)) is a good way to show complexity of algorithm.

Lets take some example and calculate value for c and n0.
1. f(n)=4n+3
Writing in a form of f(n)<=c*g(n) with f(n)=4n+3 and g(n)=5n

4n+3<=5n for n0=3 and c=5.

or 4n+3<=6n for n0=2 and c=6
Writing in a form of f(n)<=c*g(n) with f(n)=4n+3 and g(n)=6n
so there can be multiple values for n0 and c for which f(n)<=c g(n) will get satisfied.

2. f(n)=4n^2+2n+4
Writing in a form of f(n)<=c*g(n) with f(n)=4n^2+2n+4 and g(n)=5n^2
4n^2+2n+4<=5n^2 for n0=4 and c=5

Rules of thumb for calculating complexity of algorithm:

Simple programs can be analyzed using counting number of loops or iterations.

Consecutive statements:

We need to add time complexity of consecutive statements.


                   int m=0; // executed in constant time c1
                   m=m+1;  // executed in constant time c2

int m=0; // executed in constant time c1

m=m+1; // executed in constant time c2

f(n)=c1+c2;
So O(f(n))=1

Calculating complexity of a simple loop:

Time complexity of a loop can be determined by running time of statements inside loop multiplied by total number of iterations.


         int m=0; // executed in constant time c1
         // executed n times
         for (int i = 0; i < n; i++) {
             m=m+1;  // executed in constant time c2
  }

int m=0; // executed in constant time c1

// executed n times

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

m=m+1; // executed in constant time c2

}

f(n)=c2*n+c1;
So O(n)=n

Calculating complexity of a nested loop:

It is product of iterations of each loop.


          int m=0; executed in constant time c1
          // Outer loop will be executed n times 
   for (int i = 0; i < n; i++) {
          // Inner loop will be executed n times
  for(int j = 0; j < n; j++)
  {
   m=m+1; executed in constant time c2
  }
 }

int m=0; executed in constant time c1

// Outer loop will be executed n times

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

// Inner loop will be executed n times

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

{

m=m+1; executed in constant time c2

}

f(n)=c2*n*n + c1
So O(f(n))=n^2

If and else:

When you have if and else statement, then time complexity is calculated with whichever of them is larger.


           int countOfEven=0;//executed in constant time c1
        int countOfOdd=0; //executed in constant time c2
           int k=0; //executed in constant time c3
//loop will be executed n times  
  for (int i = 0; i < n; i++) {
          if(i%2==0) //executed in constant time c4
            { countOfEven++; //executed in constant time c5
              k=k+1; //executed in constant time c6
            }
          else
            countOfOdd++; //executed in constant time c7
 }

int countOfEven=0;//executed in constant time c1

int countOfOdd=0; //executed in constant time c2

int k=0; //executed in constant time c3

//loop will be executed n times

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

if(i%2==0) //executed in constant time c4

{ countOfEven++; //executed in constant time c5

k=k+1; //executed in constant time c6

}

else

countOfOdd++; //executed in constant time c7

}

f(n)=c1+c2+c3+(c4+c5+c6)*n
So o(f(n))=n

Logarithmic complexity

Lets understand logarithmic complexity with the help of example.You might know about binary search.When you want to find a value in sorted array, we use binary search.


public  int binarySearch(int[] sorted, int first, int last, int elementToBeSearched) {
     int iteration=0;
     while (first < last) {
      iteration++;
      System.out.println("i"+iteration);
         int mid = (first + last) / 2;  // Compute mid point.
         System.out.println(mid);
         if (elementToBeSearched < sorted[mid]) { last = mid; // repeat search in first half. } else if (elementToBeSearched > sorted[mid]) {
             first = mid + 1;  // Repeat search in last half.
         } else {
             return mid;     // Found it. return position
         }
     }
     return -1;    // Failed to find element
 }

public int binarySearch(int[] sorted, int first, int last, int elementToBeSearched) {

int iteration=0;

while (first < last) {

iteration++;

System.out.println("i"+iteration);

int mid = (first + last) / 2; // Compute mid point.

System.out.println(mid);

if (elementToBeSearched < sorted[mid]) { last = mid; // repeat search in first half. } else if (elementToBeSearched > sorted[mid]) {

first = mid + 1; // Repeat search in last half.

} else {

return mid; // Found it. return position

}

return -1; // Failed to find element

}

Now let’s assume our soreted array is:


int[] sortedArray={12,56,74,96,112,114,123,567};

int[] sortedArray={12,56,74,96,112,114,123,567};

and we want to search for 74 in above array. Below diagram will explain how binary search will work here.

When you observe closely, in each of the iteration you are cutting scope of array to the half. In every iteration, we are overriding value of first or last depending on soretedArray[mid].
So for
0th iteration : n
1th iteration: n/2
2nd iteration n/4
3rd iteration n/8.
Generalizing above equation:
For ith iteration : n/2ⁱ

So iteration will end , when we have 1 element left i.e. for any i, which will be our last iteration:
1=n/2ⁱ;
2ⁱ=n;
after taking log
i= log(n);
so it concludes that number of iteration requires to do binary search is log(n) so complexity of binary search is log(n)
It makes sense as in our example, we have n as 8 . It took 3 iterations(8->4->2->1) and 3 is log(8).
So If we are dividing input size by k in each iteration,then its complexity will be O(logk(n)) that is log(n) base k.

Lets take an example:


  int m=0; 
         // executed log(n) times
         for (int i = 0; i < n; i=i*2) {
             m=m+1; 
  }

int m=0;

// executed log(n) times

for (int i = 0; i < n; i=i*2) {

m=m+1;

}

Complexity of above code will be O(log(n)).

Exercise:

Lets do some exercise and find complexity of given code:


         int m=0; 
         for (int i = 0; i < n; i++) {
             m=m+1; 
  }

int m=0;

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

m=m+1;

}

Ans:


         int m=0; 
         // Executed n times
         for (int i = 0; i < n; i++) {
             m=m+1; 
  }

int m=0;

// Executed n times

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

m=m+1;

}

Complexity will be O(n)


        int m=0;
         for (int i = 0; i < n; i++) {
             m=m+1; 
  }
   for (int i = 0; i < n; i++) {
       for(int j = 0; j < n; j++)
             m=m+1; 
  }
}

int m=0;

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

m=m+1;

}

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

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

m=m+1;

}

Ans:


        int m=0;
       // Executed n times
         for (int i = 0; i < n; i++) {
             m=m+1; 
  }
    // outer loop executed n times
   for (int i = 0; i < n; i++) {
 // inner loop executed n times
       for(int j = 0; j < n; j++)
             m=m+1; 
  }

int m=0;

// Executed n times

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

m=m+1;

}

// outer loop executed n times

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

// inner loop executed n times

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

m=m+1;

}

}

}

Complexity will be :n+n*n —>O(n^2)


        int m=0;
       
    // outer loop executed n times
   for (int i = 0; i < n; i++) {
 // middle loop executed n/2 times
       for(int j = n/2; j < n; j++)
          for(int k=0;k*k < n; k++ )
             m=m+1; 
  }
        }
}

int m=0;

// outer loop executed n times

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

// middle loop executed n/2 times

for(int j = n/2; j < n; j++)

for(int k=0;k*k < n; k++ )

m=m+1;

}

Ans:


        int m=0;
       
    // outer loop executed n times
   for (int i = 0; i < n; i++) {
 // middle loop executed n/2 times
       for(int j = n/2; j < n; j++)
      // inner loop executed log(n) times
          for(int k=0;k*k < n; k++ )
             m=m+1; 
  }
       }
}

int m=0;

// outer loop executed n times

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

// middle loop executed n/2 times

for(int j = n/2; j < n; j++)

// inner loop executed log(n) times

for(int k=0;k*k < n; k++ )

m=m+1;

}

Complexity will be n*n/2*log(n)–> n^2log(n)


 int m=0;
       
    
   for (int i = n/2; i < n; i++) {
       for(int j = n/2; j < n; j++)
          for(int k=0;k < n; k++ )
             m=m+1; 
  }

int m=0;

for (int i = n/2; i < n; i++) {

for(int j = n/2; j < n; j++)

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

m=m+1;

}

Ans:


 int m=0;
       
    // outer loop executed n/2 times
   for (int i = n/2; i < n; i++) {
 // middle loop executed n/2 times
       for(int j = n/2; j < n; j++)
      // inner loop executed n times
          for(int k=0;k < n; k++ )
             m=m+1; 
  }

int m=0;

// outer loop executed n/2 times

for (int i = n/2; i < n; i++) {

// middle loop executed n/2 times

for(int j = n/2; j < n; j++)

// inner loop executed n times

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

m=m+1;

}

Complexity will be n/2*n/2*n –> n^3

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Introduction to complexity of algorithm

What is an algorithm?

Why do you need to evaluate an algorithm?

Counting number of instructions:

Asymptotic behaviour :

How will you compare algorithms?

Big O Notation:

Rules of thumb for calculating complexity of algorithm:

Logarithmic complexity

Exercise:

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What is an algorithm?

Why do you need to evaluate an algorithm?

Counting number of instructions:

Asymptotic behaviour :

How will you compare algorithms?

Big O Notation:

Rules of thumb for calculating complexity of algorithm:

Logarithmic complexity

Exercise:

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