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# How To Solve Leetcode 62. Unique Paths

Nhut Nguyen
Â·Jun 21, 2022Â·

## Problem statement

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?

### 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
``````

### Example 3

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

### Example 4

``````Input: m = 3, n = 3
Output: 6
``````

### Constraints

• `1 <= m, n <= 100`.

• It's guaranteed that the answer will be less than or equal to `2*10^9`.

## Solution 1: Recursive

At each point, the robot has two ways of moving: right or down. Let `P(m,n)` is the wanted result. Then you have a recursive relationship:

``````P(m,n) = P(m-1, n) + P(m, n-1)
``````

If the grid has only one row or only one column, then there is only one possible path.

``````P(1, n) = P(m, 1) = 1.
``````

We have a recursive implementation.

### Code

``````#include <iostream>
#include <vector>
using namespace std;
int uniquePaths(int m, int n) {
if (m == 1 || n == 1) {
return 1;
}
return uniquePaths(m - 1, n) + uniquePaths(m, n - 1);
}
int main() {
std::cout << uniquePaths(3,7) << std::endl;
std::cout << uniquePaths(7,3) << std::endl;
std::cout << uniquePaths(3,2) << std::endl;
std::cout << uniquePaths(3,3) << std::endl;
}
``````
``````Output:
28
28
3
6
``````

### Complexity

• Runtime: `O(2^m + 2^n)`, where `m*n` is the size of the grid.

• Extra space: `O(2^m + 2^n)`.

## Solution 2: Dynamic programming

The recursive implementation repeats a lot of computations.

For example, `uniquePaths(2,2)` was recomputed in both `uniquePaths(2,3)` and `uniquePaths(3,2)` when you compute `uniquePaths(3,3)`.

One way of storing what has been computed is by using dynamic programming.

### Code

``````#include <iostream>
#include <vector>
using namespace std;
int uniquePaths(int m, int n) {
vector<vector<int> > dp(m, vector<int>(n,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];
}
int main() {
std::cout << uniquePaths(3,7) << std::endl;
std::cout << uniquePaths(7,3) << std::endl;
std::cout << uniquePaths(3,2) << std::endl;
std::cout << uniquePaths(3,3) << std::endl;
}
``````
``````Output:
28
28
3
6
``````

### Complexity

• Runtime: `O(m*n)`, where `m*n` is the size of the grid.

• Extra space: `O(m*n)`.

## Solution 3: Reduced dynamic programming

You can rephrase the relationship inside the loop like this:

"new value" = "old value" + "previous value";

Then you do not have to store all values of all rows.

### Code

``````#include <iostream>
#include <vector>
using namespace std;
int uniquePaths(int m, int n) {
vector<int> dp(n, 1);
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
dp[j] += dp[j - 1];
}
}
return dp[n - 1];
}
int main() {
std::cout << uniquePaths(3,7) << std::endl;
std::cout << uniquePaths(7,3) << std::endl;
std::cout << uniquePaths(3,2) << std::endl;
std::cout << uniquePaths(3,3) << std::endl;
}
``````
``````Output:
28
28
3
6
``````

### Complexity

• Runtime `O(m*n)`.

• Memory `O(n)`.

## Final thought

I am wondering if there is some mathematics behind this problem. Please share your finding if you find a formula for the solution to this problem.

## References

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