**Learn Binary Search in a new Way!**

Binary search is the most popular method to find or search an element (numbers) from an array (data structure). In **Linear search** we did only one thing: we start from the beginning and search it through the end. It’s the best method if that required element is in the beginning but for the last element searching, it’s the worst method.

**LINEAR SEARCH**

**e.g. **array of size 10 with a[10]={1,3,4,5,8,9,7,2,1,11}

If we have to search 3 then by linear search algorithm we can search it in only 2 iterations but if we want to search 11 then a number of iteration will be 10 which is very large.

Linear Search Algorithm

```
Linear_Search(Array a, size n, search_value s){
for(Int i=0; i<n; i++){
if(a[i]==s)
return i;
}
}
```

Due to this reason, we choose a new method that is binary search method.

In Binary search, we search a particular item by comparing the middle most item of the array. If a get a match, then the index of the item is returned. But if the middle item is greater than the item, then the item is searched in sub-array to the right of the middle item otherwise the item is searched in the sub-array to the left of the middle item. This process continues on the sub-array as well until the size of sub-array reduces to zero.

Binary Search Algorithm

```
// must be sorted array
Binary_Search(Array a, size n, search_value s){
low=0;
high=n-1;
while(low<=high){
mid=(low+high)/2;
if(a[mid]<s)
low=mid+1;
else if(a[mid]>s)
high=mid-1;
else if(a[mid]==s)
return mid;
}
return -1;
}
// mid is the answer
```

You can learn basic binary search from these links:

## Finding first or Last occurrence of a number using Binary Search

in the above array if we apply simple binary search then what will happen let’s discuss.

- if the searching value s=6, then how to decide which 6, because it has 4 occurrences at index 2,3,5,9
- if we use binary search algorithm

if search value s=6;

here index low=0, index (size of array-1) high=10

so mid=5;

the array at mid index a[5]=6;

and we were searching 6 so the index is 5. Our answer will be 5 but It’s not the first occurrence or last of 6!

##### a[6]={2,4,10,10,10,18,20}

index 0 1 2 3 4 5 6

if s=10

low=0; high=6; mid=3;

A[3]=10; which is s so index is 3

Due to repetition we did few modifications in algorithm, for the first occurrence we will approach this algorithm

```
// must be sorted array
Binary_Search_New(Array a, size n, search_value s){
low=0;
high=n-1;
result=-1
while(low<=high){
mid=(low+high)/2;
if(a[mid]<s)
low=mid+1;
else if(a[mid]>s)
high=mid-1;
else if(a[mid]==s) {
result=mid; high=mid-1;}
}
return result;
}
```

in this algorithm what we did ?

we set the value of low=0 and high=size-1; and also we took a variable result and assign -1 as default if number not found in array.

in while loop we found the mid value form low+high/2

because we have to find 1st occurrence so we checked if the number is in the middle we assign result=mid. but it might happen same number is in that array and it is in the beginning of an array (from 0 index to mid index.)

For the Last Occurrence we will do the following:

```
// must be sorted array
Binary_Search_New(Array a, size n, search_value s){
low=0;
high=n-1;
result=-1
while(low<=high){
mid=(low+high)/2;
if(a[mid]<s)
low=mid+1;
else if(a[mid]>s)
high=mid-1;
else if(a[mid]==s)
{ result=mid; low=mid+1;}
}
return ressult;
}
```

we can check that how these algorithms are working using above examples. Try these! We will continue this in next article till then enjoy coding and say I have a coder in me.

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