8.2 Robot in a Grid

Simple dynamic problem, we can reduce the memory space we used by only using O(n) space instead of O(mn) space.

O(mn) space code:

    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {        // write your code here        int m = obstacleGrid.size();        int 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];    }

O(n) space;

    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {        int m = obstacleGrid.size();        int n = obstacleGrid[0].size();        vector<int> dp(n,0);        for(int j = 0; j<n; ++j){            if(obstacleGrid[0][j] != 1) dp[j] = 1;            else break;// if there is a 1: which means the grid after it cannot be reach at initialization time.        }        for(int i = 1; i<m; ++i){            dp[0] = (obstacleGrid[i][0] == 1) ? 0 : dp[0];            for(int j = 1; j<n; ++j){                dp[j] = (obstacleGrid[i][j] == 1) ? 0 : dp[j] + dp[j-1];            }        }        return dp[n-1];    }