LeetCode: 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?

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

Note: m and n will be at most 100.

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We can use DP to solve this problem. Assume grid[i][j] is the total paths to (i, j), it will be grid[i-1][j] + grid[i][j-1]. We can get to this spot from its left side or its upside if i and j are greater than 0. Otherwise we can only get there either only from its left side or only from its upside.

C++ Code:

/*
 * func: unique_path
 * goal: find the numbers of all unique paths from start to the end
 * @param m: the number of rows in the grid
 * @param n: the number of columns in the grid
 * return: the total numbers
 */
/*
 * Use dynamic programming, the path to the current spot is the path to its left
 * plus path to its upside.
 * complexity: time O(m*n), space O(m*n)
 */
int unique_path(int m, int n){
    if(m < 0 || n < 0){
        return 1;
    }
    int **grid =  new int* [m];
    for(int i=0; i<m; i++){
        grid[i] = new int[n];
    }
    grid[0][0]=1;
    for(int i=0; i<m; i++){
        for(int j=0; j<n; j++){
            if(i == 0 && j > 0){
                grid[i][j] = grid[i][j-1];
            }else if(i > 0 && j == 0){
                grid[i][j] = grid[i-1][j];
            }else if(i > 0 && j > 0){
                grid[i][j] = grid[i-1][j] + grid[i][j-1];
            }
        }
    }
    int result = grid[m-1][n-1];
    for(int i=0; i<m; i++){
        delete[] grid[i];
    }
    delete [] grid;
    return result;
}

Python Code:

# func: find the number of total unique paths to the target spot
# @param m: the number of rows
# @param n: the number of columns
# @return: total numbers
# complexity: time O(m*n), space O(m*n)
def unique_paths(m, n):
    if m < 0 or n < 0:
        return 1
    grid = [[1] * n for _ in xrange(m)]
    for i in xrange(m):
        for j in xrange(n):
            if i > 0 and j == 0:
                grid[i][j] = grid[i-1][j]
            elif i == 0 and j > 0:
                grid[i][j] = grid[i][j-1]
            elif i > 0 and j > 0:
                grid[i][j] = grid[i-1][j] + grid[i][j-1]

    return grid[m-1][n-1]