LeetCode 62. Unique Paths

Question

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

Source: https://leetcode.com/problems/unique-paths/

Solution

// Time Complexity O(mn), Space Complexity O(mn)
public int uniquePaths(int m, int n) {
    if (m == 1 || n == 1) {
        return 1;
    }

    // dp[i][j] represents the number of unique paths from (0,0) to (i,j)
    int[][] dp = new int[m][n];
    for (int i = 0; i < m; i++) {
        dp[i][0] = 1;
    }
    for (int j = 0; j < n; j++) {
        dp[0][j] = 1;
    }
    for (int i = 1; i < m; i++) {
        for (int j = 1; j < n; j++) {
            // one grid can only be reached from top or left
            dp[i][j] = dp[i][j - 1] + dp[i - 1][j];
        }
    }
    return dp[m - 1][n - 1];
}

// Time Complexity: O(mn), Space Complexity: O(n)
// space complexity can be further optimized to O(min(m, n)) at the expense of readability
public int uniquePaths2(int m, int n) {
    if (m == 1 || n == 1) {
        return 1;
    }

    if (n > m) {
        return uniquePaths2(n, m);
    }

    // dp[j] represents the number of unique paths from (0,0) to (i,j)
    // i is the round number when dp[j] is written
    // the optimization direction is similar to fibonacci
    // we only use the last row of 2D dp array, so we can replace it with a 1D dp array
    int[] dp = new int[n];
    for (int j = 0; j < n; j++) {
        dp[j] = 1;
    }
    for (int i = 1; i < m; i++) {
        for (int j = 1; j < n; j++) {
            dp[j] = dp[j - 1] + dp[j];
        }
    }
    return dp[n - 1];
}