| comments | difficulty | edit_url | rating | source | tags | ||||
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true |
困难 |
2857 |
第 362 场周赛 Q4 |
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给你两个长度都为 n 的字符串 s 和 t 。你可以对字符串 s 执行以下操作：
- 将
s长度为l（0 < l < n）的 后缀字符串 删除，并将它添加在s的开头。
比方说，s = 'abcd'，那么一次操作中，你可以删除后缀'cd'，并将它添加到s的开头，得到s = 'cdab'。
给你一个整数 k ，请你返回 恰好 k 次操作将 s 变为 t 的方案数。
由于答案可能很大，返回答案对 109 + 7 取余 后的结果。
示例 1：
输入：s = "abcd", t = "cdab", k = 2 输出：2 解释： 第一种方案： 第一次操作，选择 index = 3 开始的后缀，得到 s = "dabc" 。 第二次操作，选择 index = 3 开始的后缀，得到 s = "cdab" 。 第二种方案： 第一次操作，选择 index = 1 开始的后缀，得到 s = "bcda" 。 第二次操作，选择 index = 1 开始的后缀，得到 s = "cdab" 。
示例 2：
输入：s = "ababab", t = "ababab", k = 1 输出：2 解释： 第一种方案： 选择 index = 2 开始的后缀，得到 s = "ababab" 。 第二种方案： 选择 index = 4 开始的后缀，得到 s = "ababab" 。
提示：
2 <= s.length <= 5 * 1051 <= k <= 1015s.length == t.lengths和t都只包含小写英文字母。
"""
DP, Z-algorithm, Fast mod.
Approach
How to represent a string?
Each operation is just a rotation. Each result string can be represented by an integer from 0 to n - 1. Namely, it's just the new index of s[0].
How to find the integer(s) that can represent string t?
Create a new string s + t + t (length = 3 * n).
Use Z-algorithm (or KMP), for each n <= index < 2 * n, calculate the maximum prefix length that each substring starts from index can match, if the length >= n, then (index - n) is a valid integer representation.
How to get the result?
It's a very obvious DP.
If we use an integer to represent a string, we only need to consider the transition from zero to non-zero and from non-zero to zero. In other words, all the non-zero strings should have the same result.
So let dp[t][i = 0/1] be the number of ways to get the zero/nonzero string
after excatly t steps.
Then
dp[t][0] = dp[t - 1][1] * (n - 1).
All the non zero strings can make it.
dp[t][1] = dp[t - 1][0] + dp[t - 1] * (n - 2).
For a particular non zero string, all the other non zero strings and zero string can make it.
We have dp[0][0] = 1 and dp[0][1] = 0
Use matrix multiplication.
How to calculate dp[k][x = 0, 1] faster?
Use matrix multiplication
vector (dp[t - 1][0], dp[t - 1][1])
multiplies matrix
[0 1]
[n - 1 n - 2]
== vector (dp[t][0], dp[t - 1][1]).
So we just need to calculate the kth power of the matrix which can be done by fast power algorith.
Complexity
Time complexity:
O(n + logk)
Space complexity:
O(n)
"""
class Solution:
M: int = 1000000007
def add(self, x: int, y: int) -> int:
x += y
if x >= self.M:
x -= self.M
return x
def mul(self, x: int, y: int) -> int:
return int(x * y % self.M)
def getZ(self, s: str) -> List[int]:
n = len(s)
z = [0] * n
left = right = 0
for i in range(1, n):
if i <= right and z[i - left] <= right - i:
z[i] = z[i - left]
else:
z_i = max(0, right - i + 1)
while i + z_i < n and s[i + z_i] == s[z_i]:
z_i += 1
z[i] = z_i
if i + z[i] - 1 > right:
left = i
right = i + z[i] - 1
return z
def matrixMultiply(self, a: List[List[int]], b: List[List[int]]) -> List[List[int]]:
m = len(a)
n = len(a[0])
p = len(b[0])
r = [[0] * p for _ in range(m)]
for i in range(m):
for j in range(p):
for k in range(n):
r[i][j] = self.add(r[i][j], self.mul(a[i][k], b[k][j]))
return r
def matrixPower(self, a: List[List[int]], y: int) -> List[List[int]]:
n = len(a)
r = [[0] * n for _ in range(n)]
for i in range(n):
r[i][i] = 1
x = [a[i][:] for i in range(n)]
while y > 0:
if y & 1:
r = self.matrixMultiply(r, x)
x = self.matrixMultiply(x, x)
y >>= 1
return r
def numberOfWays(self, s: str, t: str, k: int) -> int:
n = len(s)
dp = self.matrixPower([[0, 1], [n - 1, n - 2]], k)[0]
s += t + t
z = self.getZ(s)
m = n + n
result = 0
for i in range(n, m):
if z[i] >= n:
result = self.add(result, dp[0] if i - n == 0 else dp[1])
return resultclass Solution {
private static final int M = 1000000007;
private int add(int x, int y) {
if ((x += y) >= M) {
x -= M;
}
return x;
}
private int mul(long x, long y) {
return (int) (x * y % M);
}
private int[] getZ(String s) {
int n = s.length();
int[] z = new int[n];
for (int i = 1, left = 0, right = 0; i < n; ++i) {
if (i <= right && z[i - left] <= right - i) {
z[i] = z[i - left];
} else {
int z_i = Math.max(0, right - i + 1);
while (i + z_i < n && s.charAt(i + z_i) == s.charAt(z_i)) {
z_i++;
}
z[i] = z_i;
}
if (i + z[i] - 1 > right) {
left = i;
right = i + z[i] - 1;
}
}
return z;
}
private int[][] matrixMultiply(int[][] a, int[][] b) {
int m = a.length, n = a[0].length, p = b[0].length;
int[][] r = new int[m][p];
for (int i = 0; i < m; ++i) {
for (int j = 0; j < p; ++j) {
for (int k = 0; k < n; ++k) {
r[i][j] = add(r[i][j], mul(a[i][k], b[k][j]));
}
}
}
return r;
}
private int[][] matrixPower(int[][] a, long y) {
int n = a.length;
int[][] r = new int[n][n];
for (int i = 0; i < n; ++i) {
r[i][i] = 1;
}
int[][] x = new int[n][n];
for (int i = 0; i < n; ++i) {
System.arraycopy(a[i], 0, x[i], 0, n);
}
while (y > 0) {
if ((y & 1) == 1) {
r = matrixMultiply(r, x);
}
x = matrixMultiply(x, x);
y >>= 1;
}
return r;
}
public int numberOfWays(String s, String t, long k) {
int n = s.length();
int[] dp = matrixPower(new int[][] {{0, 1}, {n - 1, n - 2}}, k)[0];
s += t + t;
int[] z = getZ(s);
int m = n + n;
int result = 0;
for (int i = n; i < m; ++i) {
if (z[i] >= n) {
result = add(result, dp[i - n == 0 ? 0 : 1]);
}
}
return result;
}
}class Solution {
const int M = 1000000007;
int add(int x, int y) {
if ((x += y) >= M) {
x -= M;
}
return x;
}
int mul(long long x, long long y) {
return x * y % M;
}
vector<int> getz(const string& s) {
const int n = s.length();
vector<int> z(n);
for (int i = 1, left = 0, right = 0; i < n; ++i) {
if (i <= right && z[i - left] <= right - i) {
z[i] = z[i - left];
} else {
for (z[i] = max(0, right - i + 1); i + z[i] < n && s[i + z[i]] == s[z[i]]; ++z[i])
;
}
if (i + z[i] - 1 > right) {
left = i;
right = i + z[i] - 1;
}
}
return z;
}
vector<vector<int>> mul(const vector<vector<int>>& a, const vector<vector<int>>& b) {
const int m = a.size(), n = a[0].size(), p = b[0].size();
vector<vector<int>> r(m, vector<int>(p));
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
for (int k = 0; k < p; ++k) {
r[i][k] = add(r[i][k], mul(a[i][j], b[j][k]));
}
}
}
return r;
}
vector<vector<int>> pow(const vector<vector<int>>& a, long long y) {
const int n = a.size();
vector<vector<int>> r(n, vector<int>(n));
for (int i = 0; i < n; ++i) {
r[i][i] = 1;
}
auto x = a;
for (; y; y >>= 1) {
if (y & 1) {
r = mul(r, x);
}
x = mul(x, x);
}
return r;
}
public:
int numberOfWays(string s, string t, long long k) {
const int n = s.length();
const auto dp = pow({{0, 1}, {n - 1, n - 2}}, k)[0];
s.append(t);
s.append(t);
const auto z = getz(s);
const int m = n + n;
int r = 0;
for (int i = n; i < m; ++i) {
if (z[i] >= n) {
r = add(r, dp[!!(i - n)]);
}
}
return r;
}
};