[Leetcode 26] 62 Unique Paths

Problem:

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 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

 

Analysis:

It is a simple DP problem. Assume Smn is the number of unique paths when the grid is m*n.

Then Smn = S(m-1)n + Sm(n-1). If simply coding it, it will has Time Limit Exceeded error due to the many recomputations.

 S(23, 12) = S(22, 12) + S(23, 11)

　　　　　  = S(21, 12) + S(22, 11) + S(22, 11) + S(23, 10)    // rep 1

　　　　    = S(20, 12) + S(21, 11) + S(21, 11) + S(22, 12) +S(21, 11) + S(22, 10) + S(22, 10) + S(23, 9)  //rep 3

　　　　    = ....

The time complxity here may be expotential.

So we need to use a bottom-up method which compute base cases first and store the result into the table for future reference.

The time complexity is only O(m*n)

 

Code:

class Solution {
public:
    int uniquePaths(int m, int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int tab[m][n];
        
        for (int i=0; i<m; i++) 
            for (int j=0; j<n; j++) {
                if (i == 0 || j == 0)
                    tab[i][j] = 1;
                else
                    tab[i][j] = -1;
            }
        
        for (int i=1; i<m; i++) {
            for (int j=1; j<n; j++) {
                if (tab[i][j] == -1) 
                    tab[i][j] = tab[i-1][j] + tab[i][j-1];
            }
        }
        
        return tab[m-1][n-1];
    }
};

 

Attention: