Problem

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?

Above is a 7 x 3 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

Example 1:

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 -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right

Example 2:

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

Solution

/**
 * @param {number} m
 * @param {number} n
 * @return {number}
 */
var uniquePaths = function(m, n) {
  var dp = Array(m);
  if (!m || !n) return 0;
  for (var i = 0; i < m; i++) {
    dp[i] = Array(n);
    for (var j = 0; j < n; j++) {
      if (j > 0 && i > 0) dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
      else if (j > 0 && i === 0) dp[i][j] = dp[i][j - 1];
      else if (j === 0 && i > 0) dp[i][j] = dp[i - 1][j];
      else dp[i][j] = 1;
    }
  }
  return dp[m - 1][n - 1];
};

Explain:

dp[i][j] 代表到达该点的路径数量。该点可以从左边点或上边点到达 也就是 dp[i][j] = dp[i - 1][j] + dp[i][j - 1] 。

Complexity:

• Time complexity : O(m*n).
• Space complexity : O(m*n).