Matrix

Unique Paths

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). How many possible unique paths are there?

• Time: O(n)
• Space: O(n)

public int uniquePaths(int m, int n) {
    int[][] dp = new int[m][n];
    for (int i = 0; i < m; i++) {
        dp[i][0] = 1;
    }
    for (int i = 0; i < n; i++) {
        dp[0][i] = 1;
    }
    for (int i = 1; i < m; i++) {
        for (int j = 1; j < n; j++) {
            dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
        }
    }
    return dp[m - 1][n - 1];
}

// DFS
public int uniquePaths_2(int m, int n) {
    if(m == 1 || n == 1) return 1;
    return uniquePaths(m - 1, n) + uniquePaths(m, n - 1);
}

Unique Paths II

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

• Time: O(n)
• Space: O(n)

public int uniquePathsWithObstacles(int[][] grid) {
    int m = grid.length, n = grid[0].length;
    int[][] dp = new int[m][n];
    dp[0][0] = grid[0][0] == 1 ? 0 : 1;
    for (int i = 1; i < m; i++) {
        dp[i][0] = grid[i][0] == 1 ? 0 : dp[i - 1][0];
    }
    for (int i = 1; i < n; i++) {
        dp[0][i] = grid[0][i] == 1 ? 0 : dp[0][i - 1];
    }
    for (int i = 1; i < m; i++) {
        for (int j = 1; j < n; j++) {
            dp[i][j] = grid[i][j] == 1 ? 0 : dp[i - 1][j] + dp[i][j - 1];
        }
    }
    return dp[m - 1][n - 1];
}

Minimum Path Sum

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.

• Time: O(n)
• Space: O(n)

public int minPathSum(int[][] grid) {
    int m = grid.length, n = grid[0].length;
    for (int i = 1; i < m; i++) {
        grid[0][i] += grid[0][i - 1];
    }
    for (int i = 1; i < n; i++) {
        grid[i][0] += grid[i - 1][0];
    }
    for (int i = 1; i < m; i++) {
        for (int j = 1; j < n; j++) {
            grid[i][j] += Math.min(grid[i - 1][j], grid[i][j - 1]);
        }
    }
    return grid[m - 1][n - 1];
}