# 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?

![](https://assets.leetcode.com/uploads/2018/10/22/robot_maze.png)\
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
```

## Analysis

标准的二维动态规划问题

**状态**：`dp[i][j]` - 从起点到达(i, j)的路径数目\
**状态转移方程**: `dp[i][j] = dp[i - 1][j] + dp[i][j - 1];` - 左侧`(i, j - 1)`和上方 `(i - 1, j)`位置的路径之和，因为从左侧和上方到达`(i, j)`均只有一种路径\
**初始条件**: `dp[i][0] = 1 (i = 0, ... m - 1)`, `dp[0][j] = 1 (j = 0, ..., n - 1)`

**答案**：`dp[m - 1][n - 1]` 即终点位置

同时另一种思路是转化为排列组合问题，即：需要走 m + n - 2 步，其中 m - 1向下，n - 1向右。也就是求解组合数：

`C(m + n - 2, n - 1)`

## Solution

2D DP - O(mn) space, O(mn) time - (0ms, 100% AC)

```java
class Solution {
    public int uniquePaths(int m, int n) {
        int[][] dp = new int[m][n];
        // initialization
        for (int i = 0; i < m; i++) {
            dp[i][0] = 1;
        }
        for (int j = 0; j < n; j++) {
            dp[0][j] = 1;
        }
        // state transfer function
        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];
    }
}
```

C++ Ref: <https://leetcode.com/problems/unique-paths/discuss/22981/My-AC-solution-using-formula>

```cpp
class Solution {
    public:
        int uniquePaths(int m, int n) {
            int N = n + m - 2;// how much steps we need to do
            int k = m - 1; // number of steps that need to go down
            double res = 1;
            // here we calculate the total possible path number 
            // Combination(N, k) = n! / (k!(n - k)!)
            // reduce the numerator and denominator and get
            // C = ( (n - k + 1) * (n - k + 2) * ... * n ) / k!
            for (int i = 1; i <= k; i++)
                res = res * (N - k + i) / i;
            return (int)res;
        }
    };
```

## Reference

<https://leetcode.com/problems/unique-paths/discuss/22954/0ms-5-lines-DP-Solution-in-C++-with-Explanations>

<https://leetcode.com/problems/unique-paths/discuss/22981/My-AC-solution-using-formula>