Number of Ways to Reach a Position After Exactly k Steps

Number of Ways to Reach a Position After Exactly k Steps

You are given two positive integers startPos and endPos . Initially, you are standing at position startPos on an infinite number line. With one step, you can move either one position to the left, or one position to the right.

Given a positive integer $k$, return the number of different ways to reach the position endPos starting from startPos , such that you perform exactly $k$ steps. Since the answer may be very large, return it modulo ${10}^{9} + 7$.

Two ways are considered different if the order of the steps made is not exactly the same.

Note that the number line includes negative integers.

Example 1:

Input: startPos = 1, endPos = 2, k = 3
Output: 3
Explanation: We can reach position 2 from 1 in exactly 3 steps in three ways:
- 1 -> 2 -> 3 -> 2.
- 1 -> 2 -> 1 -> 2.
- 1 -> 0 -> 1 -> 2.
It can be proven that no other way is possible, so we return 3.

Example 2:

Input: startPos = 2, endPos = 5, k = 10
Output: 0
Explanation: It is impossible to reach position 5 from position 2 in exactly 10 steps.

Constraints:

$1 \leq startPos, endPos, k \leq 1000$

 

解题思路

　　比赛的时候是用动态规划写的，没想到还有数学的解法。

　　先算一下从起点到终点要走的步数，假设是$d$，即从起点到终点要净走$d$步。假设起点在终点的左边，往右走的步数记为$r$，往左走的步数记为$l$，那么要满足

$$\begin{cases} r - l = d \\ r + l = k \end{cases}$$

解得$r = \frac{d + k}{2}$，相当于要从$k$步中选出$r$步向右走，即$C_{k}^{r}$。算这个组合数可以直接用公式加快速幂求逆元。

　　AC代码如下，时间复杂度为$O(nlogn)$：

class Solution {
public:
    int mod = 1e9 + 7;

    int qmi(int a, int k, int p) {
        int ret = 1;
        while (k) {
            if (k & 1) ret = 1ll * ret * a % p;
            a = 1ll * a * a % p;
            k >>= 1;
        }
        return ret;
    }

    int numberOfWays(int startPos, int endPos, int k) {
        int d = abs(endPos - startPos);
        if (d + k & 1 || d > k) return 0;   // 特判掉d和k奇偶不同与无法到达终点的情况
        int ret = 1, r = d + k >> 1;
        for (int i = 1, j = k; i <= r; i++, j--) {
            ret = (1ll * ret * j) % mod * qmi(i, mod - 2, mod) % mod;
        }
        return ret;
    }
};

　　下面再讲讲用dp的做法。

　　定义状态$f(i,j)$表示所有从起点花$i$步走到位置$j$的方案的集合，属性就是计数。根据最后一步的不同来划分集合，因此状态转移方程为 $$f(i,j) = f(i-1,j-1) + f(i-1,j+1)$$

　　还需要注意下标可能会出现负数的情况，因此需要加上一个偏移量。可以发现如果起点在$1$这个位置，那么最多向左$500$步，因此左边界最远为$-500$。同理如果起点在$1000$这个位置，那么最多向右走$500$步，因此右边界最远为$1500$。因此下标的最大长度为$2000$，一开始可以给起点和终点加上偏移量$500$。

　　AC代码如下，时间复杂度为$O(n^2)$：

class Solution {
public:
    int mod = 1e9 + 7, N = 2010, B = 500;

    int numberOfWays(int startPos, int endPos, int k) {
        startPos += B, endPos += B; // 加上偏移量
        vector<vector<int>> f(k + 1, vector<int>(N));
        f[0][startPos] = 1;
        for (int i = 1; i <= k; i++) {
            for (int j = 0; j < N; j++) {   // 直接枚举加上偏移量后的位置
                if (j) f[i][j] = f[i - 1][j - 1];   // 可以从左边走过来
                if (j + 1 < N) f[i][j] = (f[i][j] + f[i - 1][j + 1]) % mod; // 可以从右边走过来
            }
        }
        return f[k][endPos];
    }
};

 

参考资料

　　力扣第309场周赛：https://www.bilibili.com/video/BV1DB4y1G7gX