Unique Paths

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- Tag: #dynamicprogramming, #treedp

A robot is located at the top-left corner of a m x n grid (marked ‘Start’ in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish’ in the diagram below).

How many possible unique paths are there?

Above is a 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

## https://code.dennyzhang.com/unique-paths class Solution(object): ## Basic Ideas: Dynamic programming ## For the previous step of finish point, it should come from either up or left ## f(i, j) = f(i-1, j) + f(j, j-1) ## ## Complexity: Time O(m*n), Space O(m) def uniquePaths(self, m, n): """ :type m: int :type n: int :rtype: int """ if m == 0 or n == 0: return 0 dp = [1 for i in range(n)] for i in range(1, m): for j in range(1, n): dp[j] = dp[j-1] + dp[j] return dp[-1] ## Basic Ideas: Dynamic programming ## For the previous step of finish point, it should come from either up or left ## f(i, j) = f(i-1, j) + f(j, j-1) ## ## Complexity: Time O(m*n), Space O(m*n) def uniquePaths_v1(self, m, n): """ :type m: int :type n: int :rtype: int """ if m == 0 or n == 0: return 0 matrix = [None]*m for i in range(m): matrix[i] = [1]*n for i in range(1, m): for j in range(1, n): matrix[i][j] = matrix[i-1][j] + matrix[i][j-1] return matrix[-1][-1]