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Welcome back!

Are you good with directions? Let’s find out in this week’s coding challenge.

Today we have:

• Unique Paths problem

• What’s a Bloom Filter?

Read time: under 4 minutes

CODING CHALLENGE

Unique Paths

There is a robot on an m x n grid. The robot is initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either down or right at any point in time.

Given the two integers m and n, return the number of possible unique paths that the robot can take to reach the bottom-right corner.

The test cases are generated so that the answer will be less than or equal to 2 * 10^9.

Example 1:

Input: m = 3, n = 7
Output: 28

Example 2:

Input: m = 3, n = 2
Output: 3
Explanation: From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Down -> Down
2. Down -> Down -> Right
3. Down -> Right -> Down

Solve the problem here before reading the solution.

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SOLUTION

To solve this problem efficiently, we'll use dynamic programming (DP). The key insight is that the number of unique paths to reach any cell depends on the number of paths to reach the cells above and to the left of it.

We'll create a 2D DP table where each cell represents the number of unique paths to reach that cell from the top-left corner. We initialize the first row and first column with 1 since there's only one way to reach those cells (either moving right or down).

For each subsequent cell, the number of unique paths will be the sum of paths from the cell above and the cell to the left. This captures all possible ways to reach the current cell.

Time complexity is O(m*n) as we iterate through each cell in the grid once. Space complexity is also O(m*n) to store the DP table.

The solution works as follows:

1. Create a DP table of size m x n, initially filled with 1s

2. Iterate through cells starting from [1][1]

3. For each cell, calculate paths by summing paths from cell above and left

4. Return the value in bottom-right cell, which represents total unique paths

SYSTEM DESIGN

What’s a Bloom Filter?

Imagine you're building Netflix, and before showing users a movie, you need to check if it's available in your catalog. Querying the database for every movie check would be slow and expensive. You need a faster way to know if a movie exists before hitting the database. This is where Bloom filters come in.

A Bloom filter is a space-efficient data structure that quickly tells you if an item might exist in a set or definitely doesn't exist. It's like a quick preliminary check before doing an expensive database lookup.

Here's how it works: Bloom filter stores an array of bits. Initially, all the bits are set to 0. When you add an item to a Bloom filter, it uses hash functions to convert the item into several numbers. Each number corresponds to a position in the array of bits. The Bloom filter sets these positions to 1.

When checking if an item exists, the Bloom filter runs the same hash functions and checks if all corresponding positions are 1. If any position is 0, the item definitely doesn't exist. If all positions are 1, the item might exist.

Notice the word "might". Bloom filters can have false positives which means that it can say the item exists when it actually doesn’t. But it never has false negatives which means that if it says an item doesn’t exist, it’s always right. This makes them perfect for preliminary checks before expensive operations.

The size of the bit array and the number of hash functions affect the false positive rate. A larger array means fewer false positives but uses more memory. More hash functions reduce false positives up to a point, but increase computation time.

Here are some common uses of Bloom filters:

• Chrome uses them to check if a website might be malicious

• Medium uses them to avoid recommending articles you've already read

• Bitcoin uses them to help nodes sync efficiently

• Databases use them to avoid unnecessary disk lookups

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