62. Unique Paths && 63. Unique Paths II

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

How many possible unique paths are there?

Above is a 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

 

Subscribe to see which companies asked this question

Hide Tags

 Array Dynamic Programming

Hide Similar Problems

 (M) Unique Paths II (M) Minimum Path Sum (H) Dungeon Game

 

 

public class Solution {
    public int uniquePaths(int m, int n) {
        int[][] count = new int[m][];
        for(int i = 0; i<m; ++i)
        {
            count[i] = new int[n];
            count[i][0] = 1;
        }  
        for(int i = 0; i<n; ++i)
        {
            count[0][i] = 1;
        }
        for(int r = 1; r<m; ++r)
            for(int c = 1; c<n; ++c)
                count[r][c] = count[r][c-1] + count[r-1][c];
        
        return count[m-1][n-1];
    }
}

 

 

63. Unique Paths II

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 as 1 and 0 respectively in the grid.

For example,

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

[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]

The total number of unique paths is 2.

Note: m and n will be at most 100.

 

Subscribe to see which companies asked this question

Hide Tags

 Array Dynamic Programming

Hide Similar Problems

 (M) Unique Paths

 

 

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        if(obstacleGrid[0][0] == 1) //This is very important and fixes bugs.
            return 0;
            
        int row = obstacleGrid.length;
        int col = obstacleGrid[0].length;
        
        boolean firstColumnOb = false;
        for(int r = 0; r<row; ++r)
        {
            for(int c = 0;c<col; ++c)
            {
                if(r == 0)
                {
                    if(obstacleGrid[r][c] == 0)
                        obstacleGrid[r][c] = 1;
                    else
                    {
                        for(;c<col; ++c)
                        {
                            obstacleGrid[r][c] = 0;
                        }
                    }
                }
                else if(c == 0)
                {
                    if(!firstColumnOb && obstacleGrid[r][c] == 0)
                        obstacleGrid[r][c] = 1;
                    else
                    {
                        firstColumnOb = true;
                        obstacleGrid[r][c] = 0;
                    }
                }
                else
                {
                    if(obstacleGrid[r][c] == 0)
                        obstacleGrid[r][c] = obstacleGrid[r-1][c] + obstacleGrid[r][c-1];
                    else
                        obstacleGrid[r][c] = 0;
                }
            }
        }
        
        return obstacleGrid[row-1][col-1];
    }
}