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.

 

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

 

Constraints:

• 1 <= m, n <= 100

Solution

/**
 * @param {number} m
 * @param {number} n
 * @return {number}
 */
var uniquePaths = function(m, n) {
    const dp = Array(m)
        .fill(null)
        .map(() => Array(n).fill(1));

    for (let i = 1; i < m; i++) {
        for (let j = 1; j < n; j++) {
            dp[i][j] = dp[i][j - 1] + dp[i - 1][j];
        }
    }
    return dp[m - 1][n - 1];
};

We will implement a DP solution. Let dp[i][j] be the number of unique paths given an (i + 1) x (j + 1) grid.

• For the base case:
  • If i = 0, then dp[0][j] = 1 since there's 1 possible path (move only down).
  • If j = 0, then dp[i][0] = 1 since there's again 1 possible path (move only right).
• For the recursive case, the number of paths to reach (i, j) is the number of paths to reach (i - 1, j), since we can move down one space to reach (i, j), plus the number of paths to reach (i, j - 1), since we can move right one space to reach (i, j).