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 * 109.
Idea:
Dynamic Programing
dp[m][n] = unique paths of m x n grid
// base case:
dp[1][1] = 1;
// dp[i][j] = top(unique paths) + left(unique paths)
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];Solution:
/**
* @param {number} m
* @param {number} n
* @return {number}
*/
var uniquePaths = function(m, n) {
// one row or one column dp[i][1] = dp[1][j] = 1; Let's fill them all 1
let dp = Array.from(new Array(m), () => new Array(n).fill(1));
for (let i = 0; i < m; i++) {
for (let j = 0; j < n; j++) {
// dp[i][j] = top(unique paths) + left(unique paths)
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
return dp[m - 1][n - 1];
}