added solutions to 2851 String Transformation

solution/2800-2899/2851.String Transformation/Solution.cpp

@@ -0,0 +1,78 @@
+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;
+    }
+};

solution/2800-2899/2851.String Transformation/Solution.java

@@ -0,0 +1,83 @@
+class 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;
+    }
+}

solution/2800-2899/2851.String Transformation/Solution.py

@@ -0,0 +1,100 @@
+"""
+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 result