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Dynamic Programming - Easy Level - Question 2

Dynamic Programming - Easy Level - Question 2


Leetcode 62 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?

Constraints:

1 <= m, n <= 100

It's guaranteed that the answer will be less than or equal to 2 * 10^9.


Analysis:


We can apply the same idea in Question1 to the current one, which is from 1D to 2D.

One of the key information is: the robot can only go down or right. So if we define f[i][j] means the total number of ways to reach this position, then

f[i][j] = f[i-1][j] + f[i][j-1]

The first term on the right side is for "go right", and the second for "go down".

After having the state definition and transition formula, we just need to figure out the initial states' values.

Obviously the first row and the first column should always be 1, since there are only one way to reach them.

To simplify the code, usually we can define an extra row & column, and then just need to define f[1][1] = 1, which is enough for the initialization.


See the code below:


class Solution {
public:
    int uniquePaths(int m, int n) {
        vector<vector<int>> dp(m+1, vector<int>(n+1, 0));
        dp[1][1] = 1;
        for(int i=1; i<=m; ++i) {
            for(int j=1; j<=n; ++j) {
                if(i==1 && j==1) continue;
                dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }
        }
        return dp[m][n];
    }
};

Both time and space complexity is O(m*n). The space complexity can be further reduced down to O(n) with the so-called "rolling array" idea, which is explained in the following,

The scanning order for the code above is top-down, and then for each row, it is from left to right. For each dp[i][j], we need its left neighbor and top neighbor. So if we reduce the 2D dp to 1D, we still need to scan from left to right (not the reverse). Why?

In the 2D dp case, when we calculate dp[i][j], dp[i][j-1] has been calculated at the current row; if we use 1D dp, scanning from left to right will use the updated dp[j-1], but scanning from right to left will use the un-updated dp[j-1], or the one from the previous loop. In this question, we need the updated one, so scan from left to right is needed.


See the code below:


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

Note:

Like the Fibonacci sequence, this question also has directly mathematical formula: it is just like the permutation of  m color1 boxes with n color2 boxes, which is (m+n)!/(m!*n!). Such as small world, isn't it?


Upper Layer

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