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).
Input: m = 3, n = 7
Output: 28
Example 2:
Input: m = 3, n = 2
Output: 3
Explanation:
From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Down -> Down
2. Down -> Down -> Right
3. Down -> Right -> Down
Example 3:
Input: m = 7, n = 3
Output: 28
Example 4:
Input: m = 3, n = 3
Output: 6
Solution : (Using DP)
Approach:
For 1st row and column there is only 1 possible path
For Other, it is sum of left and top
class Solution
{
public:
int uniquePaths(int m, int n)
{
int a[m][n];
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
{
if (i == 0 || j == 0)
{
a[i][j] = 1;
}
else
{
a[i][j] = a[i][j - 1] + a[i - 1][j];
}
}
}
return a[m - 1][n - 1];
}
};
Time Complexity: O(MN) , Space Complexity: O(MN)
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).
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
Input: obstacleGrid = [[0,1],[0,0]]
Output: 1
Now consider if some obstacles are added to the grids. How many unique paths would there be?
Example 1:
Input:
[
[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
Solution:
class Solution
{
public:
int uniquePathsWithObstacles(vector<vector<int>> &obstacleGrid)
{
int n = obstacleGrid.size();
int m = obstacleGrid[0].size();
int arr[n][m];
if (obstacleGrid[0][0] == 1)
{
arr[0][0] = 0;
}
else
{
arr[0][0] = 1;
}
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
if (i == 0 && j == 0)
{
continue;
}
if ((i == 0 || j == 0) && obstacleGrid[i][j] != 1)
{
if (i == 0)
{
if (arr[i][j - 1] == 0)
{
arr[i][j] = 0;
}
else
{
arr[i][j] = 1;
}
}
else if (j == 0)
{
if (arr[i - 1][j] == 0)
{
arr[i][j] = 0;
}
else
{
arr[i][j] = 1;
}
}
}
else if (obstacleGrid[i][j] == 1)
{
arr[i][j] = 0;
}
else
{
arr[i][j] = arr[i][j - 1] + arr[i - 1][j];
}
}
}
return arr[n - 1][m - 1];
}
};
