Solution to LeetCode 62 Unique Paths & LeetCode 63 Unique Paths II .
LeetCode 62
Unique Paths (Medium) [link]
1] Recursion with memoization.
Define f(i,j) as the number of paths from (0,0) to (i,j). There are two ways to reach (i,j): from (i-1,j) or from (i,j-1). The recurrence relation is f(i,j) = f(i-1,j) + f(i,j-1). The base case f(i,0) = 1 because there is only one way from i to 0, same as f(0,j) = 1. The final answer is f(m-1,n-1).
Time complexity O(mn). Space complexity O(mn).
class Solution(object):
def uniquePaths(self, m, n):
"""
:type m: int
:type n: int
:rtype: int
"""
# f = [[1 for i in range(n)] for j in range(m)]
dp = [[1] * n for _ in range(m)]
for i in range(1,m):
for j in range(1,n):
dp[i][j] = dp[i-1][j] + dp[i][j-1]
return dp[m-1][n-1]
2] Iteration with space optimization. Define f(j) of length n as the number of paths from (0,0) to (i,j). The recurrence relation is f(j) = f(j) + f(j-1). The base case is f(0) = 1. The final answer is f(n-1).
Time complexity O(mn). Space complexity O(n).
class Solution(object):
def uniquePaths(self, m, n):
"""
:type m: int
:type n: int
:rtype: int
"""
dp = [1] * n
for i in range(1, m):
for j in range(1, n):
dp[j] = dp[j] + dp[j - 1]
return dp[-1]
LeetCode 63
Unique Paths II (Medium) [link]
If there is obstacle at (i,j), f(i,j) = 0. We need to consider obstacles when constructing base cases as well as iterating dp array.
class Solution:
def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
m = len(obstacleGrid)
n = len(obstacleGrid[0])
dp = [[0] * n for _ in range(m)]
if obstacleGrid[0][0] == 1: return 0
dp[0][0] = 1
for i in range(1,n):
if obstacleGrid[0][i] == 0:
dp[0][i] = dp[0][i-1]
for j in range(1,m):
if obstacleGrid[j][0] == 0:
dp[j][0] = dp[j-1][0]
for i in range(1,m):
for j in range(1,n):
if obstacleGrid[i][j] == 0:
dp[i][j] = dp[i-1][j] + dp[i][j-1]
return dp[m-1][n-1]
