Use dynamic programming where `dp[i][j]` represents the number of unique paths to reach cell `(i, j)` from the start, initializing `dp[0][0] = 1` if the start cell is not an obstacle.

For each cell `(i, j)`, if it's an obstacle (`obstacleGrid[i][j] == 1`), set `dp[i][j] = 0`. Otherwise, `dp[i][j] = dp[i-1][j] + dp[i][j-1]` (sum of paths from top and left cells).

Optimize space by using a 1D array (rolling array) to store only the current and previous row, reducing space complexity from O(mn) to O(n).