class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int R = obstacleGrid.length;
int C = obstacleGrid[0].length;
// If the starting cell has an obstacle, then simply return as there would be
// no paths to the destination.
if (obstacleGrid[0][0] == 1) {
return 0;
}
// Number of ways of reaching the starting cell = 1.
obstacleGrid[0][0] = 1;
// Filling the values for the first column
for (int i = 1; i < R; i++) {
obstacleGrid[i][0] = (obstacleGrid[i][0] == 0 && obstacleGrid[i - 1][0] == 1) ? 1 : 0;
}
// Filling the values for the first row
for (int i = 1; i < C; i++) {
obstacleGrid[0][i] = (obstacleGrid[0][i] == 0 && obstacleGrid[0][i - 1] == 1) ? 1 : 0;
}
// Starting from cell(1,1) fill up the values
// No. of ways of reaching cell[i][j] = cell[i - 1][j] + cell[i][j - 1]
// i.e. From above and left.
for (int i = 1; i < R; i++) {
for (int j = 1; j < C; j++) {
if (obstacleGrid[i][j] == 0) {
obstacleGrid[i][j] = obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1];
} else {
obstacleGrid[i][j] = 0;
}
}
}
// Return value stored in rightmost bottommost cell. That is the destination.
return obstacleGrid[R - 1][C - 1];
}
}
