path: root/04-13.md

[\<- 04/10](04-10.md)

---

## So far ...

We have discussed contents in Chapter 1
1. Concept of data Structure
2. Algorithm Efficiency - big O notation

# Search

## First Data Structure

- You've seen it before: an array!
- How is an array laid out?

|[] |[] |[] |[] |[] |
|-|-|-|-|-|
|0|1|...|...|m-1|

- a series of data laid in **consecutive** memory locations
- picture of slots... starting at "zero"
- m slots, so the end is at "m-1"

## Arrays - Precondition

- Pre-condition - n existing elements
	- If we want to store n (0 <= n <= m) elements in the array, we place them in slots "0" through "n-1"

|x|x|x|x|x|[]|[]|[]|
|-|-|-|-|-|--|--|--|
|0|1|...|...|n-1|...|...|m-1|

## Arrays - Insert

- Your first operation: insert
	- It's easiest to add a new element at the end of the array

|x|x|x|x|x|[]|[]|[]|[]|
|-|-|-|-|-|--|--|--|--|
|0|1|...|...|n-1|n|...|...|m-1|

- To add the element x, we could do

```
a[n] = x;
n++;
```

Or, alternatively
```
a[n++] = x;
```

- What is the big-O runtime of this operation?
	- O(1) (no repeating operations)

## Arrays - Access/Retrieve

- Second operation: Access/Retrieve
	- **Indexing** - Retrieve an element through indexing: O(1)
	- What about **searching a given value**?

|x|x|x|x|x|[]|[]|[]|
|-|-|-|-|-|--|--|--|
|0|1|...|...|n-1|...|...|m-1|

- How to retrieve an element through an index
```
x = a[i];
```

- How to search for a given value?
	- We'll cover this soon

## Let's Play a Game

Let's look for Matt in this array

|?|?|?|?|?|
|-|-|-|-|-|
|0|1|2|3|4|

Logically, we would check each index for Matt

- Check the first index (i=0)

|Chris|?|?|?|?|
|-|-|-|-|-|
|0|1|2|3|4|

- Comapare "Chris" to "Matt"
	- They do not match, so now we move on to the next value

|Chris|Rachael|?|?|?|
|-|-|-|-|-|
|0|1|2|3|4|

- Keep Checking

|Chris|Rachael|Ryan|?|?|
|-|-|-|-|-|
|0|1|2|3|4|

- Keep Checking

|Chris|Rachael|Ryan|Mike|?|
|-|-|-|-|-|
|0|1|2|3|4|

- Keep Checking

|Chris|Rachael|Ryan|Mike|Brian|
|-|-|-|-|-|
|0|1|2|3|4|

- Unsuccessful search - but What's the runtime?
	- it takes n comparisons -> O(n)
	- if we were looking for Chris, it would only take one comparison
		- this is called the best case O(1)
	- If we were looking for Brian, it would take five comparisons (or n comparisons)
		- this is called the worst case O(n)
	- How many comparisons will a successful search require on average? (Keep this question in mind)

## Sequential Search

- Searching an array: start with the first element and proceed to the last element
- Code:

```
int found = 0;
//assume the array 'a' is 'n' elements long
//assume 'x' is the search term

for(i = 0; i < n; i++){
	if(a[i] == x){
		found = 1;
		break;
	}
}
```

- The break statement is important so that the big-O is O(1) instead of O(n) for best case
	- More efficient
- Reminder: to use booleans (found = false) include stdbool.h

## Sequential Search: how to stop

- We can stop after finding first match
- Two ways to do that
	- Add a "break" inside the loop when found (see the previous example)
		- break means exit the innermost loop
	- make the for loop look like this: `for(i=0; i<n && !found; i++)`
- You **should NOT** combine them. You **lose points** in clarity if you do that in the assignment. Pick which approach you like, or use a while, but do not use both techniques

---

The Big O notation?
- It depends on the sequence to search (i.e. input)
- The best case? O(1)
	- search is located at the beginning of the array
- The worst case? O(n)
	- search is at the end or not in the array

On average, how many tries to find a given item?
- On average it takes you n/2 tries

Why?
- In general, if you want an average, you add up everything and divide by the number of something... so search in 1 step, 2 steps,... n steps gives us (1+2+3...+n)/n -> n/2 is a nice shortcut

- So what's the Big-O notation
	- O(n/2) -> O(n)

## Arrays - Sequential Search - Summary

- This searching algorithm is called "linear search" or "sequential search"
- It's big-O runtime is O(n)
	- best case O(1)
	- average case O(n)
	- worst case O(n)
- The military wouldn't be happy if you told them that *on average* the missile defense system stops the missiles
	- The worst case big-O is the one we care most!
- Can we improve this?

## Arrays Probability Search

- A game again!!

|Chris|Rachael|Ryan|Mike|Brian|
|-|-|-|-|-|
|0|1|2|3|4|

- How to improve the search algorithm?
	- Let's say we want to find Mike
	- We can add a swap to move Mike closer to the beginning in case he is searched for again

|Chris|Rachael|Mike|Ryan|Brian|
|-|-|-|-|-|
|0|1|2|3|4|

- If we look for Mike again, it will take one less comparison to find him

|Chris|Mike|Rachael|Ryan|Brian|
|-|-|-|-|-|
|0|1|2|3|4|

- If we look for Mike a third time, it will take one more less comparison to find him

|Mike|Chris|Rachael|Ryan|Brian|
|-|-|-|-|-|
|0|1|2|3|4|

- Now that Mike is at the front, he is the best case

---

- Note
	- Be careful not to swap the first person with the non-existent preceding person
- This is called a "move to front heuristic"
	- Heuristics are things that probably help, but not certain they would
	- This is in the book called "probability search"
- Example: having an application in your cache and therefore it runs faster
- Will this improve unsuccessful searches?
	- Worst case is still n. The average time probably improves, but is still O(n)

- You will do sequential search in Project 3

## Arrays - Another Search Type

Let's play a game:
- Pick a number from 1-100
- Tell a guesser whether he us too high or too low
- I claim I can guess in 7 or less

1. Guess 50
	- "50 is too high" -> now we know the number is 1-49
	- In one guess, the range is cut in half

2. Guess 25
	- "25 is too low" -> now we know the number is 26-49
	- In two guesses, the range is cut down to 1/4 of the original range

## Binary Search

- Idea: we halve the space over which we search
- At each step, we eliminate 1/2 of the remaining possibilities
- Maximum number of steps required is O(log(n)) if the original size of the space was n
	- Theoretically faster than sequential search
- This algorithm is called **binary search**
- Pre-condition:
	- It has to be **sorted**, otherwise we couldn't get rid of an entire sub-range

## Binary Search - Code

- side note: be clear on counter/variable names

```
//return true if 'x' is in 'a[]', false if not; 'n' is the length of 'a[]'
bool bsearch(int a[], int n, int x){

	int lo, hi, mid;
	lo = 0;
	hi = n-1;

	while(lo <= hi){
		mid = (lo + hi)/2; //updates every iteration
		if(a[mid] == x) return true;
		if(a[mid] > x) hi = mid-1;
		else lo = mid+1;
	}

	return false;

}
```

## Binary Search - Big-O

- If you are halving each time, how long does the loop run?
	- O(log(n))
	- O(log(n)) beats the pants off linear O(n)
	- log2(1000000) = 20
	
- What is the big-O runtime for binary search?
	- Best case? O(1)
	- Worst case? O(log(n))
	- Of course, we care the most about the worst case
- If you were working with strings... you'd just use strcmp() instead of comparisons

## Binary Search? Sequential Search?

Is binary search always better than sequential search?
Answer: It depends
- How about if the sequence is not sorted?

---

[04/15 ->](04-15.md)