# 62. Unique Paths
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 7 x 3 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
Example 1:
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 -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right
Example 2:
Input: m = 7, n = 3
Output: 28
# Solution
Approach 1: Recursion.
Approach 2: Combination.
Approach 3: DP.
# Code (Python)
Approach 1:
Approach 2:
Approach 3:
# Code (C++)
Approach 1:
// Time Limit Exceeded.
class Solution {
public:
int uniquePaths(int m, int n) {
if (m == 1 || n == 1)
return 1;
return uniquePaths(m - 1, n) + uniquePaths(m, n - 1);
}
};
Approach 2:
// C((m-1)+(n-1), (m-1)) = (m-1+n-1)!/(m-1)!(n-1)!
class Solution {
public:
int uniquePaths(int m, int n) {
long res = 1;
for (int i = m; i <= m + n - 2; ++i)
{
res = res * i / (i - m + 1);
}
return res;
}
};
Approach 3:
// DP.
class Solution {
public:
int uniquePaths(int m, int n) {
long paths[m][n]; // 2D array can't be applied with "= {0}".
paths[m-1][n-1] = 1;
for (int i = m - 1; i >= 0; --i)
{
for (int j = n - 1; j >= 0; --j)
{
if (i < m - 1 || j < n - 1)
paths[i][j] = 0;
if (i + 1 < m)
paths[i][j] += paths[i+1][j];
if (j + 1 < n)
paths[i][j] += paths[i][j+1];
}
}
return paths[0][0];
}
};