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?
This is a dynamic programming problem. We create a DP table, where each entry represents the number of ways to get to that grid position. The top row and leftmost column are all obviously "1". For any other entry, there are 2 possible ways of getting there, either right from its top, or right from its left.
Putting the idea together, we may calculate any entry dynamically as such:
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];Finally, we return dp[m-1][n-1].
An alternative method is to use "combinations" (i.e. n-choose-k), specifically, "(m-1 + n-1) choose min(m-1, n-1)".
class Solution {
public:
int uniquePaths(int m, int n) {
vector<vector<int>> dp(m, vector<int>(n, 1));
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (i != 0 && j != 0) {
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
}
return dp[m - 1][n - 1];
}
};```