**source:**https://leetcode.com/problems/unique-paths/**C/C++**

**Solution to LeetCode**problem

**62**.

**Unique Paths**.

## Problem

There is a robot on an `m x n`

grid. The robot is initially located at the **top-left corner** (i.e., `grid[0][0]`

). The robot tries to move to the bottom-right corner (i.e., `grid[m - 1][n - 1]`

). The robot can only move either down or right at any point in time.

Given the two integers `m`

and `n`

, return *the number of possible unique paths that the robot can take to reach the bottom-right corner*.

The test cases are generated so that the answer will be less than or equal to `2 * 10`

.^{9}

## Examples

**Example 1:**

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:

- Right -> Down -> Down
- Down -> Down -> Right
- Down -> Right -> Down

## Constraints

`1 <= m, n <= 100`

## Solution

This can be solved using **dynamic programing** (either recursive or iterative).

- We can use a map to memoize the results, but that makes it slower, so, we will allocate a vector of vectors (representing the board).
- The iterative version is a shorter solution.

### Solution 1:

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class Solution {
vector<vector<int>> dp;
private:
int paths(int m, int n, int r, int c) {
if (r+1 == m && c+1 == n)
return 1;
if (r == m || c == n)
return 0;
if (dp[r][c] != 0)
return dp[r][c];
dp[r][c] = 0;
dp[r][c] += paths(m, n, r + 1, c);
dp[r][c] += paths(m, n, r, c + 1);
return dp[r][c];
}
public:
int uniquePaths(int m, int n) {
dp = vector(m, vector<int>(n, 0));
int r = paths(m, n, 0, 0);
return r;
}
};

### Solution 2:

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class Solution {
public:
int uniquePaths(int m, int n) {
dp = vector(m, vector<int>(n, 1));
for (int r=1; r<m; r++) {
for (int c=1 ; c<n; c++)
dp[r][c] = dp[r][c-1] + dp[r-1][c];
}
return dp[m-1][n-1];
}
};