62. Unique Paths (Medium)

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

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

Solutions

1.

class Solution {
    int[][] pathCount;

    public int uniquePaths(int m, int n) {
        pathCount = new int[m][n];
        for (int i = 0; i < m; i++) {
            Arrays.fill(pathCount[i], -1);
        }

        return walkThrough(m - 1, n - 1);
    }

    public int walkThrough(int m, int n) {
        if (m == 0 && n == 0) {
            return 1;
        }

        if (m < 0 || n < 0) {
            return 0;
        }

        if (pathCount[m][n] >= 0) {
            return pathCount[m][n];
        }

        int p1 = walkThrough(m - 1, n);
        int p2 = walkThrough(m, n - 1);

        pathCount[m][n] = p1 + p2;

        return p1 + p2;
    }
}

2.

class Solution {
    public int uniquePaths(int m, int n) {
        int[][] pathCount = new int[m][n];

        return getCount(pathCount, m, n);
    }

    private int getCount(int[][] pathCount, int m, int n) {
        int row = pathCount.length;
        int col = pathCount[0].length;

        // only one pathway to (0,0) when col.no=0
        for (int i = 0; i < row; i++) {
            pathCount[i][0] = 1;
        }

        // only one pathway to (0,0) when row.no=0
        for (int i = 0; i < col; i++) {
            pathCount[0][i] = 1;
        }

        for (int i = 1; i < row; i++) {
            for (int j = 1; j < col; j++) {
                pathCount[i][j] = pathCount[i - 1][j] + pathCount[i][j - 1];
            }
        }

        return pathCount[m - 1][n - 1];
    }
}

Incorrect Solutions

References