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Commit 6b8a81d

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feat: add solutions to lc problem: No.2851 (doocs#1676)
No.2851.String Transformation
1 parent f82e843 commit 6b8a81d

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class Solution {
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const int M = 1000000007;
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int add(int x, int y) {
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if ((x += y) >= M) {
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x -= M;
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}
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return x;
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}
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int mul(long long x, long long y) {
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return x * y % M;
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}
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vector<int> getz(const string& s) {
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const int n = s.length();
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vector<int> z(n);
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for (int i = 1, left = 0, right = 0; i < n; ++i) {
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if (i <= right && z[i - left] <= right - i) {
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z[i] = z[i - left];
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} else {
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for (z[i] = max(0, right - i + 1); i + z[i] < n && s[i + z[i]] == s[z[i]]; ++z[i])
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;
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}
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if (i + z[i] - 1 > right) {
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left = i;
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right = i + z[i] - 1;
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}
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}
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return z;
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}
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vector<vector<int>> mul(const vector<vector<int>>& a, const vector<vector<int>>& b) {
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const int m = a.size(), n = a[0].size(), p = b[0].size();
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vector<vector<int>> r(m, vector<int>(p));
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for (int i = 0; i < m; ++i) {
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for (int j = 0; j < n; ++j) {
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for (int k = 0; k < p; ++k) {
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r[i][k] = add(r[i][k], mul(a[i][j], b[j][k]));
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}
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}
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}
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return r;
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}
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vector<vector<int>> pow(const vector<vector<int>>& a, long long y) {
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const int n = a.size();
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vector<vector<int>> r(n, vector<int>(n));
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for (int i = 0; i < n; ++i) {
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r[i][i] = 1;
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}
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auto x = a;
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for (; y; y >>= 1) {
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if (y & 1) {
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r = mul(r, x);
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}
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x = mul(x, x);
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}
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return r;
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}
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public:
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int numberOfWays(string s, string t, long long k) {
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const int n = s.length();
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const auto dp = pow({{0, 1}, {n - 1, n - 2}}, k)[0];
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s.append(t);
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s.append(t);
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const auto z = getz(s);
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const int m = n + n;
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int r = 0;
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for (int i = n; i < m; ++i) {
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if (z[i] >= n) {
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r = add(r, dp[!!(i - n)]);
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}
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}
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return r;
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}
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};
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class Solution {
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private static final int M = 1000000007;
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private int add(int x, int y) {
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if ((x += y) >= M) {
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x -= M;
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}
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return x;
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}
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private int mul(long x, long y) {
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return (int) (x * y % M);
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}
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private int[] getZ(String s) {
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int n = s.length();
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int[] z = new int[n];
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for (int i = 1, left = 0, right = 0; i < n; ++i) {
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if (i <= right && z[i - left] <= right - i) {
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z[i] = z[i - left];
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} else {
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int z_i = Math.max(0, right - i + 1);
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while (i + z_i < n && s.charAt(i + z_i) == s.charAt(z_i)) {
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z_i++;
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}
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z[i] = z_i;
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}
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if (i + z[i] - 1 > right) {
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left = i;
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right = i + z[i] - 1;
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}
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}
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return z;
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}
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private int[][] matrixMultiply(int[][] a, int[][] b) {
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int m = a.length, n = a[0].length, p = b[0].length;
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int[][] r = new int[m][p];
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for (int i = 0; i < m; ++i) {
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for (int j = 0; j < p; ++j) {
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for (int k = 0; k < n; ++k) {
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r[i][j] = add(r[i][j], mul(a[i][k], b[k][j]));
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}
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}
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}
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return r;
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}
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private int[][] matrixPower(int[][] a, long y) {
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int n = a.length;
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int[][] r = new int[n][n];
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for (int i = 0; i < n; ++i) {
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r[i][i] = 1;
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}
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int[][] x = new int[n][n];
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for (int i = 0; i < n; ++i) {
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System.arraycopy(a[i], 0, x[i], 0, n);
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}
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while (y > 0) {
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if ((y & 1) == 1) {
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r = matrixMultiply(r, x);
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}
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x = matrixMultiply(x, x);
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y >>= 1;
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}
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return r;
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}
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public int numberOfWays(String s, String t, long k) {
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int n = s.length();
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int[] dp = matrixPower(new int[][] {{0, 1}, {n - 1, n - 2}}, k)[0];
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s += t + t;
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int[] z = getZ(s);
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int m = n + n;
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int result = 0;
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for (int i = n; i < m; ++i) {
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if (z[i] >= n) {
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result = add(result, dp[i - n == 0 ? 0 : 1]);
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}
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}
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return result;
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}
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}
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"""
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DP, Z-algorithm, Fast mod.
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Approach
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How to represent a string?
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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].
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How to find the integer(s) that can represent string t?
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Create a new string s + t + t (length = 3 * n).
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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.
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How to get the result?
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It's a very obvious DP.
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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.
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So let dp[t][i = 0/1] be the number of ways to get the zero/nonzero string
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after excatly t steps.
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Then
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dp[t][0] = dp[t - 1][1] * (n - 1).
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All the non zero strings can make it.
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dp[t][1] = dp[t - 1][0] + dp[t - 1] * (n - 2).
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For a particular non zero string, all the other non zero strings and zero string can make it.
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We have dp[0][0] = 1 and dp[0][1] = 0
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Use matrix multiplication.
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How to calculate dp[k][x = 0, 1] faster?
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Use matrix multiplication
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vector (dp[t - 1][0], dp[t - 1][1])
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multiplies matrix
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[0 1]
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[n - 1 n - 2]
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== vector (dp[t][0], dp[t - 1][1]).
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So we just need to calculate the kth power of the matrix which can be done by fast power algorith.
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Complexity
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Time complexity:
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O(n + logk)
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Space complexity:
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O(n)
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"""
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class Solution:
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M: int = 1000000007
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def add(self, x: int, y: int) -> int:
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x += y
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if x >= self.M:
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x -= self.M
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return x
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def mul(self, x: int, y: int) -> int:
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return int(x * y % self.M)
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def getZ(self, s: str) -> List[int]:
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n = len(s)
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z = [0] * n
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left = right = 0
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for i in range(1, n):
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if i <= right and z[i - left] <= right - i:
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z[i] = z[i - left]
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else:
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z_i = max(0, right - i + 1)
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while i + z_i < n and s[i + z_i] == s[z_i]:
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z_i += 1
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z[i] = z_i
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if i + z[i] - 1 > right:
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left = i
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right = i + z[i] - 1
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return z
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def matrixMultiply(self, a: List[List[int]], b: List[List[int]]) -> List[List[int]]:
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m = len(a)
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n = len(a[0])
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p = len(b[0])
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r = [[0] * p for _ in range(m)]
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for i in range(m):
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for j in range(p):
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for k in range(n):
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r[i][j] = self.add(r[i][j], self.mul(a[i][k], b[k][j]))
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return r
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def matrixPower(self, a: List[List[int]], y: int) -> List[List[int]]:
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n = len(a)
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r = [[0] * n for _ in range(n)]
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for i in range(n):
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r[i][i] = 1
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x = [a[i][:] for i in range(n)]
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while y > 0:
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if y & 1:
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r = self.matrixMultiply(r, x)
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x = self.matrixMultiply(x, x)
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y >>= 1
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return r
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def numberOfWays(self, s: str, t: str, k: int) -> int:
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n = len(s)
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dp = self.matrixPower([[0, 1], [n - 1, n - 2]], k)[0]
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s += t + t
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z = self.getZ(s)
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m = n + n
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result = 0
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for i in range(n, m):
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if z[i] >= n:
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result = self.add(result, dp[0] if i - n == 0 else dp[1])
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return result

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