# Unique Paths LeetCode Solution

Here, We see Unique Paths LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

# List of all LeetCode Solution

## Problem Statement

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 * 109`.

```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:
1. Right -> Down -> Down
2. Down -> Down -> Right
3. Down -> Right -> Down
```

## Unique Paths Leetcode Solution C++

``````class Solution {
public:
int uniquePaths(int m, int n) {
if(m <=0 || n <= 0) return 0;
long long res = 1;
for(int i = n; i < m+n-1 ; i++){
res = res * i / (i- n + 1);
}
return (int)res;
}
};```Code language: PHP (php)```

## Unique Paths Leetcode Solution Java

``````class Solution {
public int uniquePaths(int m, int n) {
if(m == 1 || n == 1)
return 1;
m--;
n--;
if(m < n) {
m = m + n;
n = m - n;
m = m - n;
}
long res = 1;
int j = 1;
for(int i = m+1; i <= m+n; i++, j++){
res *= i;
res /= j;
}
return (int)res;
}
}```Code language: PHP (php)```

## Unique Paths Leetcode Solution JavaScript

``````var uniquePaths = function(m, n) {
let perRow = Array(n).fill(1);
let pathCount = Array.from( Array(m).fill( perRow ) );
for(let y = 1 ; y < m ; y++){
for(let x = 1 ; x < n ; x++){
pathCount[y][x] = pathCount[y][x-1] + pathCount[y-1][x]
}
}
return pathCount[m-1][n-1]
};```Code language: JavaScript (javascript)```

## Unique Paths Leetcode Solution Python

``````class Solution(object):
def uniquePaths(self, m, n):
return math.factorial(m+n-2)/math.factorial(m-1)/math.factorial(n-1)``````
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