# Olog n types of Algorithms

suggest changeLet’s say we have a problem of size n. Now for each step of our algorithm(which we need write), our original problem becomes half of its previous size(n/2).

So at each step, our problem becomes half.

Step | Problem | —— | —— | 1 | n/2 | 2 | n/4 | 3 | n/8 | 4 | n/16 |

When the problem space is reduced(i.e solved completely), it cannot be reduced any further(n becomes equal to 1) after exiting check condition.

- Let’s say at kth step or number of operations:

**problem-size** = 1

- But we know at kth step, our problem-size should be:

**problem-size** = n/2k

- From 1 and 2:

n/2<sup>k</sup> = 1 or n = 2<sup>k</sup>

- Take log on both sides

log<sub>e</sub> n = k log<sub>e</sub>2 or k = log<sub>e</sub> n / log<sub>e</sub> 2

- Using formula
**logx m / logx n = logn m**

k = log2 n

or simply k = log n

Now we know that our algorithm can run maximum up to log n, hence time complexity comes as

O( log n)

A very simple example in code to support above text is :

for(int i=1; i<=n; i=i*2) { // perform some operation }

So now if some one asks you if n is 256 how many steps that loop( or any other algorithm that cuts down it’s problem size into half) will run you can very easily calculate.

k = log2 256

k = log2 2 8 ( => logaa = 1)

k = 8

Another very good example for similar case is **Binary Search Algorithm**.

int bSearch(int arr[],int size,int item){ int low=0; int high=size-1; while(low<=high){ mid=low+(high-low)/2; if(arr[mid]==item) return mid; else if(arr[mid]<item) low=mid+1; else high=mid-1; } return –1;// Unsuccessful result }