Blind 75.25 - Unique Paths

Unique paths.

Problem description

Taken directly from LeetCode.

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 * 109.

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

Constraints:

• 1 <= m, n <= 100

Solution type

Caching via DP.

Solution

This is a DP classic and I think might’ve been the first problem that was showcased to us in DP. Actually no, I think staircase came first. Either way, the recurrence is just \(\textrm{ways}(i, j) = \textrm{ways}(i - 1, j) + \textrm{ways}(i, j -1)\). If \(i = 0\) or \(j = 0\), that’s a base case where \(ways(i, j) = 1\), since there will only be one way to reach that cell (go straight right or down from the origin).

Not much more going on with the code here.

def uniquePaths(self, m: int, n: int) -> int:
    memo = [[1 for _ in range(n)] for _ in range(m)]
    for i in range(1, m):
        for j in range (1, n):
            memo[i][j] = memo[i - 1][j] + memo[i][j - 1]
    return memo[-1][-1]