Unique Paths II – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
You are given an m x n
integer array grid
. There is a robot initially located at the top-left corner (i.e., grid[0][0]
). The robot tries to move to the bottom-right corner (i.e., grid[m-1][n-1]
). The robot can only move either down or right at any point in time.
An obstacle and space are marked as 1
or 0
respectively in grid
. A path that the robot takes cannot include any square that is an obstacle.
Return the number of possible unique paths that the robot can take to reach the bottom-right corner.
The testcases are generated so that the answer will be less than or equal to 2 * 109
.
Example 1:
Input: obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]] Output: 2 Explanation: There is one obstacle in the middle of the 3x3 grid above. There are two ways to reach the bottom-right corner: 1. Right -> Right -> Down -> Down 2. Down -> Down -> Right -> Right
Example 2:
Input: obstacleGrid = [[0,1],[0,0]] Output: 1
Constraints:
m == obstacleGrid.length
n == obstacleGrid[i].length
1 <= m, n <= 100
obstacleGrid[i][j]
is0
or1
.
C++ Unique Paths II LeetCode Solution
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
int m = obstacleGrid.size() , n = obstacleGrid[0].size();
vector<vector<int>> dp(m+1,vector<int>(n+1,0));
dp[0][1] = 1;
for(int i = 1 ; i <= m ; ++i)
for(int j = 1 ; j <= n ; ++j)
if(!obstacleGrid[i-1][j-1])
dp[i][j] = dp[i-1][j]+dp[i][j-1];
return dp[m][n];
}
};
Java Unique Paths II LeetCode Solution
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int width = obstacleGrid[0].length;
int[] dp = new int[width];
dp[0] = 1;
for (int[] row : obstacleGrid) {
for (int j = 0; j < width; j++) {
if (row[j] == 1)
dp[j] = 0;
else if (j > 0)
dp[j] += dp[j - 1];
}
}
return dp[width - 1];
}
Python 3 Unique Paths II LeetCode Solution
def uniquePathsWithObstacles1(self, obstacleGrid):
if not obstacleGrid:
return
r, c = len(obstacleGrid), len(obstacleGrid[0])
dp = [[0 for _ in xrange(c)] for _ in xrange(r)]
dp[0][0] = 1 - obstacleGrid[0][0]
for i in xrange(1, r):
dp[i][0] = dp[i-1][0] * (1 - obstacleGrid[i][0])
for i in xrange(1, c):
dp[0][i] = dp[0][i-1] * (1 - obstacleGrid[0][i])
for i in xrange(1, r):
for j in xrange(1, c):
dp[i][j] = (dp[i][j-1] + dp[i-1][j]) * (1 - obstacleGrid[i][j])
return dp[-1][-1]
Array-1180
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