Well, this problem is similar to Unique Paths. The introduction of obstacles only changes the boundary conditions and make some points unreachable (simply set to 0).
Denote the number of paths to arrive at point (i, j) to be P[i][j], the state equation is P[i][j] = P[i - 1][j] + P[i][j - 1] if obstacleGrid[i][j] != 1 and 0 otherwise.
Now let's finish the boundary conditions. In the Unique Paths problem, we initialize P[0][j] = 1, P[i][0] = 1 for all valid i, j. Now, due to obstacles, some boundary points are no longer reachable and need to be initialized to 0. For example, if obstacleGrid is like [0, 0, 1, 0, 0], then the last three points are not reachable and need to be initialized to be 0. The result is [1, 1, 0, 0, 0].
Now we can write down the following (unoptimized) code. Note that we pad the obstacleGrid by1 and initialize dp[0][1] = 1 to unify the boundary cases.
1 class Solution { 2 public: 3 int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) { 4 int m = obstacleGrid.size(), n = obstacleGrid[0].size(); 5 vector<vector<int> > dp(m + 1, vector<int> (n + 1, 0)); 6 dp[0][1] = 1; 7 for (int i = 1; i <= m; i++) 8 for (int j = 1; j <= n; j++) 9 if (!obstacleGrid[i - 1][j - 1]) 10 dp[i][j] = dp[i - 1][j] + dp[i][j - 1]; 11 return dp[m][n]; 12 } 13 };
Well, the code is accepted but it has some obvious redundancy. There are two major concerns:
- Each time when we update
path[i][j], we only needpath[i - 1][j](at the same column) andpath[i][j - 1](at the left column), so it is unnecessary to maintain the fullm*nmatrix. Maintaining two columns is enough. - There are some cases that the loop can be terminated earlier. Suppose
obstacleGrid = [[0, 1, 0, 0], [0, 1, 0, 0], [0, 1, 0, 0]], then we can see that it is impossible to reach the bottom-right corner after updating the second column since the number of paths to reach each element in the second column is0.
Taken these into considerations, we write down the following optimized code.
1 class Solution { 2 public: 3 int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) { 4 int m = obstacleGrid.size(); 5 int n = obstacleGrid[0].size(); 6 vector<int> pre(m, 0); 7 vector<int> cur(m, 0); 8 for (int i = 0; i < m; i++) { 9 if (!obstacleGrid[i][0]) 10 pre[i] = 1; 11 else break; 12 } 13 for (int j = 1; j < n; j++) { 14 bool flag = false; 15 if (!obstacleGrid[0][j]) { 16 cur[0] = pre[0]; 17 if (cur[0]) flag = true; 18 } 19 else cur[0] = 0; 20 for (int i = 1; i < m; i++) { 21 if (!obstacleGrid[i][j]) { 22 cur[i] = cur[i - 1] + pre[i]; 23 if (cur[i]) flag = true; 24 } 25 else cur[i] = 0; 26 } 27 if (!flag) return 0; 28 swap(pre, cur); 29 } 30 return pre[m - 1]; 31 } 32 };
Further inspecting the above code, keeping two vectors only serve for the purpose of recoveringpre[i], which is simply cur[i] before its update. So we can use only one vector and the space is further optimized.
1 class Solution { 2 public: 3 int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) { 4 int m = obstacleGrid.size(); 5 int n = obstacleGrid[0].size(); 6 vector<int> cur(m, 0); 7 for (int i = 0; i < m; i++) { 8 if (!obstacleGrid[i][0]) 9 cur[i] = 1; 10 else break; 11 } 12 for (int j = 1; j < n; j++) { 13 bool flag = false; 14 if (obstacleGrid[0][j]) 15 cur[0] = 0; 16 else flag = true; 17 for (int i = 1; i < m; i++) { 18 if (!obstacleGrid[i][j]) { 19 cur[i] += cur[i - 1]; 20 if (cur[i]) flag = true; 21 } 22 else cur[i] = 0; 23 } 24 if (!flag) return 0; 25 } 26 return cur[m - 1]; 27 } 28 };