How to solve Leetcode 63. Unique Paths II

Problem statement

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 test cases are generated so that the answer will be less than or equal to 2 * 10^9.

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] is 0 or 1.

Solution: Dynamic programming in place

Let us find the relationship between the positions.

If there is no obstacle at position (row = i, col = j), the number of paths np[i][j] that the robot can take to reach this position is:

np[i][j] = np[i - 1][j] + np[i][j - 1]

As long as there is no obstacle in the first row, np[0][j] = 1. Otherwise, np[0][k] = 0 for all k >= j0, where (0, j0) is the position of the first obstacle in the first row.
Similarly, as long as there is no obstacle in the first column, np[i][0] = 1. Otherwise, np[k][0] = 0 for all k >= i0, where (i0, 0) is the position of the first obstacle in the first column.

Code

#include <vector>
#include <iostream>
using namespace std;
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
    const int row = obstacleGrid.size();
    const int col = obstacleGrid[0].size();
    vector<vector<int>> np(row, vector<int>(col, 0));    
    for (int i = 0; i < row && obstacleGrid[i][0] == 0; i++) {
        np[i][0] = 1;
    }    
    for (int j = 0; j < col && obstacleGrid[0][j] == 0; j++) {
        np[0][j] = 1;
    }
    for (int i = 1; i < row; i++) {
        for (int j = 1; j < col; j++) {
            if (obstacleGrid[i][j] == 0) {
                np[i][j] = np[i - 1][j] + np[i][j - 1];
            }
        }
    }
    return np[row - 1][col - 1]; 
}
int main() {
    vector<vector<int>> obstacleGrid = {{0,0,0},{0,1,0},{0,0,0}};
    cout << uniquePathsWithObstacles(obstacleGrid) << endl;
    obstacleGrid = {{0,1},{0,0}};
    cout << uniquePathsWithObstacles(obstacleGrid) << endl;
}

Output:
2
1

Complexity

Runtime: O(m*n), where m x n is the size of the grid.
Extra space: O(m*n).

References

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Nhut Nguyen