62. Unique Paths - cocoder39/coco39_LC GitHub Wiki

62. Unique Paths

time is O(m * n), memory is O(m * n). memory can be optimized to O(n)

int uniquePaths(int m, int n) {
        if (m == 0 || n == 0) {
            return 0;
        }
        vector<vector<int>> dp(m, vector<int>(n));
        
        for (int i = 0; i < m; i++) {
            dp[i][0] = 1;
        }
        for (int j = 0; j < n; j++) {
            dp[0][j] = 1;
        }
        
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp[m - 1][n - 1];
    }

int uniquePaths(int m, int n) {
        if (m == 0 || n == 0) {
            return 0;
        }
        
        vector<vector<int>> dp(2, vector<int>(n));
        for (int i = 0; i < n; i++) {
            dp[0][i] = 1;
        }

        int pre = 0, cur = 1;
        for (int i = 1; i < m; i++) {
            dp[pre][0] = 1;
            dp[cur][0] = 1;
            for (int j = 1; j < n; j++) {
                dp[cur][j] = dp[pre][j] + dp[cur][j - 1];
            }
            swap(pre, cur);
        }
        return dp[pre][n - 1];
    }

also can be solved with math. move toward downside m - 1 steps, right n - 1 steps. res = C((m - 1 + n - 1), m - 1) = (m + n - 2)! / (m - 1)!(n - 1)! = (m * m+1 * m+n-2) / (1 * 2 * n-1)

tips: res * (m - 1 + i) / i would overflow if res is int. res / i * (m - 1 + i) would reduce the precision.

time is O(n) and memory is O(1)

int uniquePaths(int m, int n) {
        if (m == 0 || n == 0) {
            return 0;
        }
        
        long res = 1;
        for (int i = 1; i <= n - 1; i++) {
            res = res * (m - 1 + i) / i;
        }
        return res;
    }
⚠️ **GitHub.com Fallback** ⚠️