Unique Path I (Medium)

Dynamic Programming

public int uniquePaths(int m, int n) {
    if(m==0 || n==0) return 0;
    if(m==1 || n==1) return 1;

    int[][] dp = new int[m][n];

    //left column
    for(int i=0; i<m; i++){
        dp[i][0] = 1;
    }

    //top row
    for(int j=0; j<n; j++){
        dp[0][j] = 1;
    }

    //fill up the dp table
    for(int i=1; i<m; i++){
        for(int j=1; j<n; j++){
            dp[i][j] = dp[i-1][j] + dp[i][j-1];
        }
    }

    return dp[m-1][n-1];
}

Unique Path II (Medium)

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 as1and0respectively in the grid.

For example,

There is one obstacle in the middle of a 3x3 grid as illustrated below.

public int uniquePathsWithObstacles(int[][] obstacleGrid) {
    if(obstacleGrid==null||obstacleGrid.length==0)
        return 0;

    int m = obstacleGrid.length;
    int n = obstacleGrid[0].length;

    if(obstacleGrid[0][0]==1||obstacleGrid[m-1][n-1]==1) 
        return 0;


    int[][] dp = new int[m][n];
    dp[0][0]=1;

    //left column
    for(int i=1; i<m; i++){
        if(obstacleGrid[i][0]==1){
            dp[i][0] = 0;
        }else{
            dp[i][0] = dp[i-1][0];
        }
    }

    //top row
    for(int i=1; i<n; i++){
        if(obstacleGrid[0][i]==1){
            dp[0][i] = 0;
        }else{
            dp[0][i] = dp[0][i-1];
        }
    }

    //fill up cells inside
    for(int i=1; i<m; i++){
        for(int j=1; j<n; j++){
            if(obstacleGrid[i][j]==1){
                dp[i][j]=0;
            }else{
                dp[i][j]=dp[i-1][j]+dp[i][j-1];
            }

        }
    }

    return dp[m-1][n-1];
}