# Clear and simple solution, time O(mn), space O(n), compared with UniquePath

• Compared to UniquePath, we can get some clues. I don't want to judge n or m who is smaller to save space. I just want to make the solutions as simple and clear as possible.

``````1  1  1  1  1  1  1
|  |  |  |  |  |
1--2--3--4--5--6--7
|  |  |  |  |  |
1--3--6--10-15-21-28
``````

UniquePath

cur[0] = 1 is the start palce.

``````class Solution {
public:
int uniquePaths(int m, int n) {
int i = 0, j = 0;
vector<int> cur(n, 0);
cur[0] = 1;
for(i = 0; i < m; i++){
for(j = 1; j < n; j++)
cur[j] += cur[j - 1];
}
return cur[n - 1];
}
};
``````

UniquePath II

We need to judge whether the start point and end point is valid. We also need to judge obstacleGrid[i][0], so j should starts at 0.

``````class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int i = 0, j = 0, m = obstacleGrid.size();
if(m == 0) return 0;
int n = obstacleGrid[0].size();
vector<int> cur(n, 0);
if(obstacleGrid[0][0] == 1 || obstacleGrid[m - 1][n - 1] == 1)
return 0;
cur[0] = 1;
for(i = 0; i < m; i++){
for(j = 0; j < n; j++){
if(obstacleGrid[i][j] == 1)
cur[j] = 0;
else if(j > 0)
cur[j] += cur[j - 1];
}
}
return cur[n - 1];
}
};``````

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