62 Unique Paths II – Medium¶
Problem:¶
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 3×3 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.
Thoughts:¶
This problem is almost identical to Unique Paths problem. The only difference is that there are conditions when assigning value to d[i][j].
d[i][j] = 0 if board[i][j] == 1
d[i][j] = d[i-1][j] +d[i][j-1] othwerwise
Plus, now d[i][0] and d[0][j] is not sure to be 1.
Solutions:¶
public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int m = obstacleGrid.length;
int n = obstacleGrid[0].length;
int[][] paths = new int[m][n];
if (obstacleGrid[0][0] == 0){
paths[0][0] = 1;
}
else{
paths[0][0] = 0;
}
for (int i = 1; i < m; i ++){
if (obstacleGrid[i][0] == 0 && paths[i-1][0] == 1){
paths[i][0] = 1;
}
else{
paths[i][0] = 0;
}
}
for (int j = 1; j < n; j ++){
if (obstacleGrid[0][j] == 0 && paths[0][j-1] == 1){
paths[0][j] = 1;
}
else{
paths[0][j] = 0;
}
}
for (int i = 1; i < m ; i ++){
for (int j = 1; j < n; j ++){
if (obstacleGrid[i][j] == 1){
paths[i][j] = 0;
}
else{
paths[i][j] = paths[i-1][j] + paths[i][j-1];
}
}
}
return paths[m-1][n-1];
}
}