每日一题 day48 （DP topic）

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本文探讨了如何解决机器人在网格中找到唯一路径的问题。首先介绍了使用排列（permutation）的方法，然后通过动态规划（DP）的两种实现进行优化。第一种DP方法通过建立二维数组记录每一步的选择，而第二种DP优化则通过维护一维数组减少空间复杂性。示例展示了不同方法的计算过程和结果。

problem

62. Unique Paths
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:
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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

approach `permutation`

class Solution {
public:
    int uniquePaths(int m, int n) {
        if(m == 1 || n == 1)
            return 1;
        m--, n--;
        if(m < n) swap(m, n);
        long res = 1;
        for(int i = m+1, j=1; i <= m+n; i++, j++){       // Instead of taking factorial, keep on multiply & divide
            res *= i;
            res /= j;
        }
        return (int)res;
    }
};

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approach 1 `DP`

class Solution {
public:
    int uniquePaths(int m, int n) {
        vector<vector<int>> dp = vector<vector<int>>(m, vector<int>(n, 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];
    }
};

approach 2 `DP improve`

class Solution {
public:
    int uniquePaths(int m, int n) {
        vector<int> cur(n, 1);
        for (int i = 1; i < m; i++) 
            for (int j = 1; j < n; j++) 
                cur[j] += cur[j - 1];
        return cur[n - 1];
    }
};