### ALGOS

Python library to visualize data structures and **ALGO**rithm**S**. It can be used at terminals and also in Jupyter Notebooks.

Having the ability to visualize what our code is doing can help enhance our understanding of data structures and search algorithms, and, as a bonus, may even aid in identifying bugs.

### EXAMPLE:

The following is a LeetCode problem (No. 5). It is called the Last Stone Weight and it is considered easy. The statement goes as follows:

You are given an array of integers stones where stones[i] is the weight of the ith stone.

We are playing a game with the stones. On each turn, we choose the heaviest two stones and smash them together. Suppose the heaviest two stones have weights x and y with x <= y. The result of this smash is:

- If x == y, both stones are destroyed, and
- If x != y, the stone of weight x is destroyed, and the stone of weight y has new weight y - x.

At the end of the game, there is at most one stone left. Return the weight of the last remaining stone. If there are no stones left, return 0.

### How to implement ALGOS?

This is the original code solution:

```
def lastStoneWeight(stones):
while len(stones) > 1:
stones.sort()
y, x = stones[-1], stones[-2]
stones.pop()
stones.pop()
if y-x:
stones.append(y-x)
if stones:
return stones[0]
return 0
```

Implementing ALGOS looks like this:

```
import search1D
def lastStoneWeight(stones):
while len(stones) > 1:
stones.sort()
# --------- ALGOS starts here ---------
state = search1D.OneDimSearch(stones, None, 1)
state.print_visited_indices(range(len(stones)), None, len(stones)-2, len(stones)-1)
# --------- ALGOS ends here ---------
y, x = stones[-1], stones[-2]
stones.pop()
stones.pop()
if y-x:
stones.append(y-x)
# --------- ALGOS starts here ---------
state = search1D.OneDimSearch(stones, None, 1)
state.print_visited_indices(range(len(stones)), None, len(stones)-2, len(stones)-1)
# --------- ALGOS ends here ---------
if stones:
return stones[0]
return 0
```

ALGOS is called at each moment to capture a 'snapshot' of the current state of our array. It is called twice - first to identify the position of the last two numbers, and then to remove them and append their difference.

LeetCode offers the following explanation for its example:

Input: stones = [2,7,4,1,8,1] Output: 1 Explanation:

- We combine 7 and 8 to get 1 so the array converts to [2,4,1,1,1] then,
- We combine 2 and 4 to get 2 so the array converts to [2,1,1,1] then,
- We combine 2 and 1 to get 1 so the array converts to [1,1,1] then,
- We combine 1 and 1 to get 0 so the array converts to [1] then that's the value of the last stone.

By implementing ALGOS, the user can see the following process:

which shows the same reduction process.

You can find the files here: GitHub.