# Unique Paths Problem Solution

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 * 109`.

```Input: m = 3, n = 7
Output: 28```

## Problem Solution In Python

``````class Solution(object):
def uniquePaths(self, m, n):

dp = [[0] * m for _ in range(n)]

for i in range(n):
for j in range(m):
if (i == 0) or (j == 0): dp[i][j] = 1
else: dp[i][j] = dp[i][j-1] + dp[i-1][j]

return dp[-1][-1]
``````

## Problem Solution In Java

``````class Solution {
public int uniquePaths(int m, int n) {
int[][] dp = new int[m][n];
for(int i = 0;i < m;i++)
for(int j = 0;j < n;j++)
dp[i][j] = i == 0 || j == 0 ? 1 : dp[i - 1][j] + dp[i][j - 1];
return dp[m - 1][n - 1];
}
}
``````

## Problem Solution In C++

``````class Solution {
public:
int uniquePaths(int m, int n) {
int dp[m][n];
for(int i=0;i<n;i++){
dp[0][i]=1;
}
for(int i=1;i<m;i++){
dp[i][0]=1;
}
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];
}
};
``````

## Problem Solution In C

``````int solve(int **count, int m, int n, int row, int col) {
if (row == m || col == n) {
return 0;
}

if (count[row][col] != -1) {
return count[row][col];
}

if (row == m - 1 && col == n - 1) {
count[row][col] = 1;
return 1;
}

int down = solve(count, m, n, row + 1, col);
int right = solve(count, m, n, row, col + 1);
int uniquePaths = down + right;
count[row][col] = uniquePaths;
return uniquePaths;
}

int uniquePaths(int m, int n){
int **count = malloc(m * sizeof(int *));
for (int i = 0; i < m; ++i) {
count[i] = malloc(n * sizeof(int));
memset(count[i], -1, n * sizeof(int));
}

int uniquePaths = solve(count, m, n, 0, 0);

for (int i = 0; i < m; ++i) {
free(count[i]);
}
free(count);
return uniquePaths;
}``````

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