62. Unique Paths

Problem

Tags: Math, Dynamic Programming, Combinatorics

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 * 10^9.

Example 1:

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

Constraints:

• 1 <= m, n <= 100

Code

C

// 62. Unique Paths (11/21/54352)
// Runtime: 0 ms (93.32%) Memory: 6.04 MB (0.00%) 

int uniquePaths (int m, int n) {
    int32_t* counts = calloc(m * n, sizeof(int32_t));
    
    counts[0] = 1;
    for (size_t y = 0; y < m; y++) {
        for (size_t x = 0; x < n; x++) {
            if (!(x == 0 && y == 0)) {
                counts[y * n + x] = (x > 0 ? counts[y * n + x - 1] : 0) + (y > 0 ? counts[(y - 1) * n + x] : 0);
            }
        }
    }
    
    int32_t result = counts[m * n - 1];
    free(counts);
    return result;
}