Follow up for "Unique Paths":
Now consider if some obstacles are added to the grids. How many unique paths would there be?
An obstacle and empty space is marked as
1
and 0
respectively in the grid.
For example,
There is one obstacle in the middle of a 3x3 grid as illustrated below.
[ [0,0,0], [0,1,0], [0,0,0] ]
The total number of unique paths is
2
.
Note: m and n will be at most 100.
Answer:class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
int m= obstacleGrid.size();
int n= obstacleGrid[0].size();
int dp[m][n];
//Initialize the edge cases for the dp table.
if(obstacleGrid[0][0]==0){
dp[0][0]=1;
}else{
dp[0][0]=0;
}
for(int i=1;i<m;++i){
if(obstacleGrid[i][0]==1){
dp[i][0]=0;
}
else{
dp[i][0]=dp[i-1][0];
}
}
for(int j=1;j<n;++j){
if(obstacleGrid[0][j]==1){
dp[0][j]=0;
}
else{
dp[0][j]=dp[0][j-1];
}
}
//Recursion cases for dp table
for(int i=1; i<m;++i){
for(int j=1;j<n;++j){
if(obstacleGrid[i][j]==0){
dp[i][j]= dp[i-1][j]+dp[i][j-1];
}else{
dp[i][j]= 0;
}
}
}
return dp[m-1][n-1];
}
};
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