diff --git a/solution/2900-2999/2930.Number of Strings Which Can Be Rearranged to Contain Substring/README.md b/solution/2900-2999/2930.Number of Strings Which Can Be Rearranged to Contain Substring/README.md
index c1254911711bb..fd313967385b4 100644
--- a/solution/2900-2999/2930.Number of Strings Which Can Be Rearranged to Contain Substring/README.md	
+++ b/solution/2900-2999/2930.Number of Strings Which Can Be Rearranged to Contain Substring/README.md	
@@ -53,6 +53,42 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：记忆化搜索**
+
+我们设计一个函数 $dfs(i, l, e, t)$，表示当前剩余字符串长度为 $i$，且已至少有 $l$ 个字符 `'l'`, $e$ 个字符 `'e'` 和 $t$ 个字符 `'t'`，构成的字符串是一个好字符串的方案数。那么答案为 $dfs(n, 0, 0, 0)$。
+
+函数 $dfs(i, l, e, t)$ 的执行逻辑如下：
+
+如果 $i = 0$，说明当前字符串已经构造完毕，如果 $l = 1$, $e = 2$ 且 $t = 1$，说明当前字符串是一个好字符串，返回 $1$，否则返回 $0$。
+
+否则，我们可以考虑在当前位置添加除 `'l'`, `'e'`, `'t'` 以外的任意一个小写字母，一共有 $23$ 种，那么此时得到的方案数为 $dfs(i - 1, l, e, t) \times 23$。
+
+我们也可以考虑在当前位置添加 `'l'`，此时得到的方案数为 $dfs(i - 1, \min(1, l + 1), e, t)$。同理，添加 `'e'` 和 `'t'` 的方案数分别为 $dfs(i - 1, l, \min(2, e + 1), t)$ 和 $dfs(i - 1, l, e, \min(1, t + 1))$。累加起来，并对 $10^9 + 7$ 取模，即可得到 $dfs(i, l, e, t)$ 的值。
+
+为了避免重复计算，我们可以使用记忆化搜索。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为字符串长度。
+
+**方法二：逆向思维 + 容斥原理**
+
+我们可以考虑逆向思维，即计算不包含子字符串 `"leet"` 的字符串数目，然后用总数减去该数目即可。
+
+我们分成以下几种情况：
+
+-   情况 $a$：表示字符串中不包含字符 `'l'` 的方案数，那么有 $a = 25^n$。
+-   情况 $b$：与 $a$ 类似，表示字符串中不包含字符 `'t'` 的方案数，那么有 $b = 25^n$。
+-   情况 $c$：表示字符串中不包含字符 `'e'` 或者只包含一个字符 `'e'` 的方案数，那么有 $c = 25^n + n \times 25^{n - 1}$。
+-   情况 $ab$：表示字符串中不包含字符 `'l'` 和 `'t'` 的方案数，那么有 $ab = 24^n$。
+-   情况 $ac$：表示字符串中不包含字符 `'l'` 和 `'e'` 或者只包含一个字符 `'e'` 的方案数，那么有 $ac = 24^n + n \times 24^{n - 1}$。
+-   情况 $bc$：与 $ac$ 类似，表示字符串中不包含字符 `'t'` 和 `'e'` 或者只包含一个字符 `'e'` 的方案数，那么有 $bc = 24^n + n \times 24^{n - 1}$。
+-   情况 $abc$：表示字符串中不包含字符 `'l'`、`'t'` 和 `'e'` 或者只包含一个字符 `'e'` 的方案数，那么有 $abc = 23^n + n \times 23^{n - 1}$。
+
+那么根据容斥原理，可以得到 $a + b + c - ab - ac - bc + abc$，就是不包含子字符串 `"leet"` 的字符串数目。
+
+而总数 $tot = 26^n$，所以答案为 $tot - (a + b + c - ab - ac - bc + abc)$，注意要对 $10^9 + 7$ 取模。
+
+时间复杂度 $O(\log n)$，其中 $n$ 为字符串长度。空间复杂度 $O(1)$。
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -60,7 +96,33 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
+class Solution:
+    def stringCount(self, n: int) -> int:
+        @cache
+        def dfs(i: int, l: int, e: int, t: int) -> int:
+            if i == 0:
+                return int(l == 1 and e == 2 and t == 1)
+            a = dfs(i - 1, l, e, t) * 23 % mod
+            b = dfs(i - 1, min(1, l + 1), e, t)
+            c = dfs(i - 1, l, min(2, e + 1), t)
+            d = dfs(i - 1, l, e, min(1, t + 1))
+            return (a + b + c + d) % mod
+
+        mod = 10**9 + 7
+        return dfs(n, 0, 0, 0)
+```
 
+```python
+class Solution:
+    def stringCount(self, n: int) -> int:
+        mod = 10**9 + 7
+        a = b = pow(25, n, mod)
+        c = pow(25, n, mod) + n * pow(25, n - 1, mod)
+        ab = pow(24, n, mod)
+        ac = bc = (pow(24, n, mod) + n * pow(24, n - 1, mod)) % mod
+        abc = (pow(23, n, mod) + n * pow(23, n - 1, mod)) % mod
+        tot = pow(26, n, mod)
+        return (tot - (a + b + c - ab - ac - bc + abc)) % mod
 ```
 
 ### **Java**
@@ -68,19 +130,228 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    private final int mod = (int) 1e9 + 7;
+    private Long[][][][] f;
+
+    public int stringCount(int n) {
+        f = new Long[n + 1][2][3][2];
+        return (int) dfs(n, 0, 0, 0);
+    }
+
+    private long dfs(int i, int l, int e, int t) {
+        if (i == 0) {
+            return l == 1 && e == 2 && t == 1 ? 1 : 0;
+        }
+        if (f[i][l][e][t] != null) {
+            return f[i][l][e][t];
+        }
+        long a = dfs(i - 1, l, e, t) * 23 % mod;
+        long b = dfs(i - 1, Math.min(1, l + 1), e, t);
+        long c = dfs(i - 1, l, Math.min(2, e + 1), t);
+        long d = dfs(i - 1, l, e, Math.min(1, t + 1));
+        return f[i][l][e][t] = (a + b + c + d) % mod;
+    }
+}
+```
 
+```java
+class Solution {
+    private final int mod = (int) 1e9 + 7;
+
+    public int stringCount(int n) {
+        long a = qpow(25, n);
+        long b = a;
+        long c = (qpow(25, n) + n * qpow(25, n - 1) % mod) % mod;
+        long ab = qpow(24, n);
+        long ac = (qpow(24, n) + n * qpow(24, n - 1) % mod) % mod;
+        long bc = ac;
+        long abc = (qpow(23, n) + n * qpow(23, n - 1) % mod) % mod;
+        long tot = qpow(26, n);
+        return (int) ((tot - (a + b + c - ab - ac - bc + abc)) % mod + mod) % mod;
+    }
+
+    private long qpow(long a, int n) {
+        long ans = 1;
+        for (; n > 0; n >>= 1) {
+            if ((n & 1) == 1) {
+                ans = ans * a % mod;
+            }
+            a = a * a % mod;
+        }
+        return ans;
+    }
+}
 ```
 
 ### **C++**
 
 ```cpp
+class Solution {
+public:
+    int stringCount(int n) {
+        const int mod = 1e9 + 7;
+        using ll = long long;
+        ll f[n + 1][2][3][2];
+        memset(f, -1, sizeof(f));
+        function<ll(int, int, int, int)> dfs = [&](int i, int l, int e, int t) -> ll {
+            if (i == 0) {
+                return l == 1 && e == 2 && t == 1 ? 1 : 0;
+            }
+            if (f[i][l][e][t] != -1) {
+                return f[i][l][e][t];
+            }
+            ll a = dfs(i - 1, l, e, t) * 23 % mod;
+            ll b = dfs(i - 1, min(1, l + 1), e, t) % mod;
+            ll c = dfs(i - 1, l, min(2, e + 1), t) % mod;
+            ll d = dfs(i - 1, l, e, min(1, t + 1)) % mod;
+            return f[i][l][e][t] = (a + b + c + d) % mod;
+        };
+        return dfs(n, 0, 0, 0);
+    }
+};
+```
 
+```cpp
+class Solution {
+public:
+    int stringCount(int n) {
+        const int mod = 1e9 + 7;
+        using ll = long long;
+        auto qpow = [&](ll a, int n) {
+            ll ans = 1;
+            for (; n; n >>= 1) {
+                if (n & 1) {
+                    ans = ans * a % mod;
+                }
+                a = a * a % mod;
+            }
+            return ans;
+        };
+        ll a = qpow(25, n);
+        ll b = a;
+        ll c = (qpow(25, n) + n * qpow(25, n - 1) % mod) % mod;
+        ll ab = qpow(24, n);
+        ll ac = (qpow(24, n) + n * qpow(24, n - 1) % mod) % mod;
+        ll bc = ac;
+        ll abc = (qpow(23, n) + n * qpow(23, n - 1) % mod) % mod;
+        ll tot = qpow(26, n);
+        return ((tot - (a + b + c - ab - ac - bc + abc)) % mod + mod) % mod;
+    }
+};
 ```
 
 ### **Go**
 
 ```go
+func stringCount(n int) int {
+	const mod int = 1e9 + 7
+	f := make([][2][3][2]int, n+1)
+	for i := range f {
+		for j := range f[i] {
+			for k := range f[i][j] {
+				for l := range f[i][j][k] {
+					f[i][j][k][l] = -1
+				}
+			}
+		}
+	}
+	var dfs func(i, l, e, t int) int
+	dfs = func(i, l, e, t int) int {
+		if i == 0 {
+			if l == 1 && e == 2 && t == 1 {
+				return 1
+			}
+			return 0
+		}
+		if f[i][l][e][t] == -1 {
+			a := dfs(i-1, l, e, t) * 23 % mod
+			b := dfs(i-1, min(1, l+1), e, t)
+			c := dfs(i-1, l, min(2, e+1), t)
+			d := dfs(i-1, l, e, min(1, t+1))
+			f[i][l][e][t] = (a + b + c + d) % mod
+		}
+		return f[i][l][e][t]
+	}
+	return dfs(n, 0, 0, 0)
+}
+```
+
+```go
+func stringCount(n int) int {
+	const mod int = 1e9 + 7
+	qpow := func(a, n int) int {
+		ans := 1
+		for ; n > 0; n >>= 1 {
+			if n&1 == 1 {
+				ans = ans * a % mod
+			}
+			a = a * a % mod
+		}
+		return ans
+	}
+	a := qpow(25, n)
+	b := a
+	c := qpow(25, n) + n*qpow(25, n-1)
+	ab := qpow(24, n)
+	ac := (qpow(24, n) + n*qpow(24, n-1)) % mod
+	bc := ac
+	abc := (qpow(23, n) + n*qpow(23, n-1)) % mod
+	tot := qpow(26, n)
+	return ((tot-(a+b+c-ab-ac-bc+abc))%mod + mod) % mod
+}
+```
+
+### **TypeScript**
+
+```ts
+function stringCount(n: number): number {
+    const mod = 10 ** 9 + 7;
+    const f: number[][][][] = Array.from({ length: n + 1 }, () =>
+        Array.from({ length: 2 }, () =>
+            Array.from({ length: 3 }, () => Array.from({ length: 2 }, () => -1)),
+        ),
+    );
+    const dfs = (i: number, l: number, e: number, t: number): number => {
+        if (i === 0) {
+            return l === 1 && e === 2 && t === 1 ? 1 : 0;
+        }
+        if (f[i][l][e][t] !== -1) {
+            return f[i][l][e][t];
+        }
+        const a = (dfs(i - 1, l, e, t) * 23) % mod;
+        const b = dfs(i - 1, Math.min(1, l + 1), e, t);
+        const c = dfs(i - 1, l, Math.min(2, e + 1), t);
+        const d = dfs(i - 1, l, e, Math.min(1, t + 1));
+        return (f[i][l][e][t] = (a + b + c + d) % mod);
+    };
+    return dfs(n, 0, 0, 0);
+}
+```
 
+```ts
+function stringCount(n: number): number {
+    const mod = BigInt(10 ** 9 + 7);
+    const qpow = (a: bigint, n: number): bigint => {
+        let ans = 1n;
+        for (; n; n >>>= 1) {
+            if (n & 1) {
+                ans = (ans * a) % mod;
+            }
+            a = (a * a) % mod;
+        }
+        return ans;
+    };
+    const a = qpow(25n, n);
+    const b = a;
+    const c = (qpow(25n, n) + ((BigInt(n) * qpow(25n, n - 1)) % mod)) % mod;
+    const ab = qpow(24n, n);
+    const ac = (qpow(24n, n) + ((BigInt(n) * qpow(24n, n - 1)) % mod)) % mod;
+    const bc = ac;
+    const abc = (qpow(23n, n) + ((BigInt(n) * qpow(23n, n - 1)) % mod)) % mod;
+    const tot = qpow(26n, n);
+    return Number((((tot - (a + b + c - ab - ac - bc + abc)) % mod) + mod) % mod);
+}
 ```
 
 ### **...**
diff --git a/solution/2900-2999/2930.Number of Strings Which Can Be Rearranged to Contain Substring/README_EN.md b/solution/2900-2999/2930.Number of Strings Which Can Be Rearranged to Contain Substring/README_EN.md
index 12167841de55d..c484ff3b672e4 100644
--- a/solution/2900-2999/2930.Number of Strings Which Can Be Rearranged to Contain Substring/README_EN.md	
+++ b/solution/2900-2999/2930.Number of Strings Which Can Be Rearranged to Contain Substring/README_EN.md	
@@ -49,30 +49,301 @@
 
 ## Solutions
 
+**Solution 1: Memorization Search**
+
+We design a function $dfs(i, l, e, t)$, which represents the number of good strings that can be formed when the remaining string length is $i$, and there are at least $l$ characters 'l', $e$ characters 'e' and $t$ characters 't'. The answer is $dfs(n, 0, 0, 0)$.
+
+The execution logic of the function $dfs(i, l, e, t)$ is as follows:
+
+If $i = 0$, it means that the current string has been constructed. If $l = 1$, $e = 2$ and $t = 1$, it means that the current string is a good string, return $1$, otherwise return $0$.
+
+Otherwise, we can consider adding any lowercase letter other than 'l', 'e', 't' at the current position, there are 23 in total, so the number of schemes obtained at this time is $dfs(i - 1, l, e, t) \times 23$.
+
+We can also consider adding 'l' at the current position, and the number of schemes obtained at this time is $dfs(i - 1, \min(1, l + 1), e, t)$. Similarly, the number of schemes for adding 'e' and 't' are $dfs(i - 1, l, \min(2, e + 1), t)$ and $dfs(i - 1, l, e, \min(1, t + 1))$ respectively. Add them up and take the modulus of $10^9 + 7$ to get the value of $dfs(i, l, e, t)$.
+
+To avoid repeated calculations, we can use memorization search.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string.
+
+**Solution 2: Reverse Thinking + Inclusion-Exclusion Principle**
+
+We can consider reverse thinking, that is, calculate the number of strings that do not contain the substring "leet", and then subtract this number from the total.
+
+We divide it into the following cases:
+
+-   Case $a$: represents the number of schemes where the string does not contain the character 'l', so we have $a = 25^n$.
+-   Case $b$: similar to $a$, represents the number of schemes where the string does not contain the character 't', so we have $b = 25^n$.
+-   Case $c$: represents the number of schemes where the string does not contain the character 'e' or only contains one character 'e', so we have $c = 25^n + n \times 25^{n - 1}$.
+-   Case $ab$: represents the number of schemes where the string does not contain the characters 'l' and 't', so we have $ab = 24^n$.
+-   Case $ac$: represents the number of schemes where the string does not contain the characters 'l' and 'e' or only contains one character 'e', so we have $ac = 24^n + n \times 24^{n - 1}$.
+-   Case $bc$: similar to $ac$, represents the number of schemes where the string does not contain the characters 't' and 'e' or only contains one character 'e', so we have $bc = 24^n + n \times 24^{n - 1}$.
+-   Case $abc$: represents the number of schemes where the string does not contain the characters 'l', 't' and 'e' or only contains one character 'e', so we have $abc = 23^n + n \times 23^{n - 1}$.
+
+Then according to the inclusion-exclusion principle, $a + b + c - ab - ac - bc + abc$ is the number of strings that do not contain the substring "leet".
+
+The total number $tot = 26^n$, so the answer is $tot - (a + b + c - ab - ac - bc + abc)$, remember to take the modulus of $10^9 + 7$.
+
+The time complexity is $O(\log n)$, and the space complexity is $O(1)$. Here, $n$ is the length of the string.
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 ```python
+class Solution:
+    def stringCount(self, n: int) -> int:
+        @cache
+        def dfs(i: int, l: int, e: int, t: int) -> int:
+            if i == 0:
+                return int(l == 1 and e == 2 and t == 1)
+            a = dfs(i - 1, l, e, t) * 23 % mod
+            b = dfs(i - 1, min(1, l + 1), e, t)
+            c = dfs(i - 1, l, min(2, e + 1), t)
+            d = dfs(i - 1, l, e, min(1, t + 1))
+            return (a + b + c + d) % mod
+
+        mod = 10**9 + 7
+        return dfs(n, 0, 0, 0)
+```
 
+```python
+class Solution:
+    def stringCount(self, n: int) -> int:
+        mod = 10**9 + 7
+        a = b = pow(25, n, mod)
+        c = pow(25, n, mod) + n * pow(25, n - 1, mod)
+        ab = pow(24, n, mod)
+        ac = bc = (pow(24, n, mod) + n * pow(24, n - 1, mod)) % mod
+        abc = (pow(23, n, mod) + n * pow(23, n - 1, mod)) % mod
+        tot = pow(26, n, mod)
+        return (tot - (a + b + c - ab - ac - bc + abc)) % mod
 ```
 
 ### **Java**
 
 ```java
+class Solution {
+    private final int mod = (int) 1e9 + 7;
+    private Long[][][][] f;
+
+    public int stringCount(int n) {
+        f = new Long[n + 1][2][3][2];
+        return (int) dfs(n, 0, 0, 0);
+    }
+
+    private long dfs(int i, int l, int e, int t) {
+        if (i == 0) {
+            return l == 1 && e == 2 && t == 1 ? 1 : 0;
+        }
+        if (f[i][l][e][t] != null) {
+            return f[i][l][e][t];
+        }
+        long a = dfs(i - 1, l, e, t) * 23 % mod;
+        long b = dfs(i - 1, Math.min(1, l + 1), e, t);
+        long c = dfs(i - 1, l, Math.min(2, e + 1), t);
+        long d = dfs(i - 1, l, e, Math.min(1, t + 1));
+        return f[i][l][e][t] = (a + b + c + d) % mod;
+    }
+}
+```
 
+```java
+class Solution {
+    private final int mod = (int) 1e9 + 7;
+
+    public int stringCount(int n) {
+        long a = qpow(25, n);
+        long b = a;
+        long c = (qpow(25, n) + n * qpow(25, n - 1) % mod) % mod;
+        long ab = qpow(24, n);
+        long ac = (qpow(24, n) + n * qpow(24, n - 1) % mod) % mod;
+        long bc = ac;
+        long abc = (qpow(23, n) + n * qpow(23, n - 1) % mod) % mod;
+        long tot = qpow(26, n);
+        return (int) ((tot - (a + b + c - ab - ac - bc + abc)) % mod + mod) % mod;
+    }
+
+    private long qpow(long a, int n) {
+        long ans = 1;
+        for (; n > 0; n >>= 1) {
+            if ((n & 1) == 1) {
+                ans = ans * a % mod;
+            }
+            a = a * a % mod;
+        }
+        return ans;
+    }
+}
 ```
 
 ### **C++**
 
 ```cpp
+class Solution {
+public:
+    int stringCount(int n) {
+        const int mod = 1e9 + 7;
+        using ll = long long;
+        ll f[n + 1][2][3][2];
+        memset(f, -1, sizeof(f));
+        function<ll(int, int, int, int)> dfs = [&](int i, int l, int e, int t) -> ll {
+            if (i == 0) {
+                return l == 1 && e == 2 && t == 1 ? 1 : 0;
+            }
+            if (f[i][l][e][t] != -1) {
+                return f[i][l][e][t];
+            }
+            ll a = dfs(i - 1, l, e, t) * 23 % mod;
+            ll b = dfs(i - 1, min(1, l + 1), e, t) % mod;
+            ll c = dfs(i - 1, l, min(2, e + 1), t) % mod;
+            ll d = dfs(i - 1, l, e, min(1, t + 1)) % mod;
+            return f[i][l][e][t] = (a + b + c + d) % mod;
+        };
+        return dfs(n, 0, 0, 0);
+    }
+};
+```
 
+```cpp
+class Solution {
+public:
+    int stringCount(int n) {
+        const int mod = 1e9 + 7;
+        using ll = long long;
+        auto qpow = [&](ll a, int n) {
+            ll ans = 1;
+            for (; n; n >>= 1) {
+                if (n & 1) {
+                    ans = ans * a % mod;
+                }
+                a = a * a % mod;
+            }
+            return ans;
+        };
+        ll a = qpow(25, n);
+        ll b = a;
+        ll c = (qpow(25, n) + n * qpow(25, n - 1) % mod) % mod;
+        ll ab = qpow(24, n);
+        ll ac = (qpow(24, n) + n * qpow(24, n - 1) % mod) % mod;
+        ll bc = ac;
+        ll abc = (qpow(23, n) + n * qpow(23, n - 1) % mod) % mod;
+        ll tot = qpow(26, n);
+        return ((tot - (a + b + c - ab - ac - bc + abc)) % mod + mod) % mod;
+    }
+};
 ```
 
 ### **Go**
 
 ```go
+func stringCount(n int) int {
+	const mod int = 1e9 + 7
+	f := make([][2][3][2]int, n+1)
+	for i := range f {
+		for j := range f[i] {
+			for k := range f[i][j] {
+				for l := range f[i][j][k] {
+					f[i][j][k][l] = -1
+				}
+			}
+		}
+	}
+	var dfs func(i, l, e, t int) int
+	dfs = func(i, l, e, t int) int {
+		if i == 0 {
+			if l == 1 && e == 2 && t == 1 {
+				return 1
+			}
+			return 0
+		}
+		if f[i][l][e][t] == -1 {
+			a := dfs(i-1, l, e, t) * 23 % mod
+			b := dfs(i-1, min(1, l+1), e, t)
+			c := dfs(i-1, l, min(2, e+1), t)
+			d := dfs(i-1, l, e, min(1, t+1))
+			f[i][l][e][t] = (a + b + c + d) % mod
+		}
+		return f[i][l][e][t]
+	}
+	return dfs(n, 0, 0, 0)
+}
+```
+
+```go
+func stringCount(n int) int {
+	const mod int = 1e9 + 7
+	qpow := func(a, n int) int {
+		ans := 1
+		for ; n > 0; n >>= 1 {
+			if n&1 == 1 {
+				ans = ans * a % mod
+			}
+			a = a * a % mod
+		}
+		return ans
+	}
+	a := qpow(25, n)
+	b := a
+	c := qpow(25, n) + n*qpow(25, n-1)
+	ab := qpow(24, n)
+	ac := (qpow(24, n) + n*qpow(24, n-1)) % mod
+	bc := ac
+	abc := (qpow(23, n) + n*qpow(23, n-1)) % mod
+	tot := qpow(26, n)
+	return ((tot-(a+b+c-ab-ac-bc+abc))%mod + mod) % mod
+}
+```
+
+### **TypeScript**
+
+```ts
+function stringCount(n: number): number {
+    const mod = 10 ** 9 + 7;
+    const f: number[][][][] = Array.from({ length: n + 1 }, () =>
+        Array.from({ length: 2 }, () =>
+            Array.from({ length: 3 }, () => Array.from({ length: 2 }, () => -1)),
+        ),
+    );
+    const dfs = (i: number, l: number, e: number, t: number): number => {
+        if (i === 0) {
+            return l === 1 && e === 2 && t === 1 ? 1 : 0;
+        }
+        if (f[i][l][e][t] !== -1) {
+            return f[i][l][e][t];
+        }
+        const a = (dfs(i - 1, l, e, t) * 23) % mod;
+        const b = dfs(i - 1, Math.min(1, l + 1), e, t);
+        const c = dfs(i - 1, l, Math.min(2, e + 1), t);
+        const d = dfs(i - 1, l, e, Math.min(1, t + 1));
+        return (f[i][l][e][t] = (a + b + c + d) % mod);
+    };
+    return dfs(n, 0, 0, 0);
+}
+```
 
+```ts
+function stringCount(n: number): number {
+    const mod = BigInt(10 ** 9 + 7);
+    const qpow = (a: bigint, n: number): bigint => {
+        let ans = 1n;
+        for (; n; n >>>= 1) {
+            if (n & 1) {
+                ans = (ans * a) % mod;
+            }
+            a = (a * a) % mod;
+        }
+        return ans;
+    };
+    const a = qpow(25n, n);
+    const b = a;
+    const c = (qpow(25n, n) + ((BigInt(n) * qpow(25n, n - 1)) % mod)) % mod;
+    const ab = qpow(24n, n);
+    const ac = (qpow(24n, n) + ((BigInt(n) * qpow(24n, n - 1)) % mod)) % mod;
+    const bc = ac;
+    const abc = (qpow(23n, n) + ((BigInt(n) * qpow(23n, n - 1)) % mod)) % mod;
+    const tot = qpow(26n, n);
+    return Number((((tot - (a + b + c - ab - ac - bc + abc)) % mod) + mod) % mod);
+}
 ```
 
 ### **...**
diff --git a/solution/2900-2999/2930.Number of Strings Which Can Be Rearranged to Contain Substring/Solution.cpp b/solution/2900-2999/2930.Number of Strings Which Can Be Rearranged to Contain Substring/Solution.cpp
new file mode 100644
index 0000000000000..111d7c1c1ff77
--- /dev/null
+++ b/solution/2900-2999/2930.Number of Strings Which Can Be Rearranged to Contain Substring/Solution.cpp	
@@ -0,0 +1,26 @@
+class Solution {
+public:
+    int stringCount(int n) {
+        const int mod = 1e9 + 7;
+        using ll = long long;
+        auto qpow = [&](ll a, int n) {
+            ll ans = 1;
+            for (; n; n >>= 1) {
+                if (n & 1) {
+                    ans = ans * a % mod;
+                }
+                a = a * a % mod;
+            }
+            return ans;
+        };
+        ll a = qpow(25, n);
+        ll b = a;
+        ll c = (qpow(25, n) + n * qpow(25, n - 1) % mod) % mod;
+        ll ab = qpow(24, n);
+        ll ac = (qpow(24, n) + n * qpow(24, n - 1) % mod) % mod;
+        ll bc = ac;
+        ll abc = (qpow(23, n) + n * qpow(23, n - 1) % mod) % mod;
+        ll tot = qpow(26, n);
+        return ((tot - (a + b + c - ab - ac - bc + abc)) % mod + mod) % mod;
+    }
+};
\ No newline at end of file
diff --git a/solution/2900-2999/2930.Number of Strings Which Can Be Rearranged to Contain Substring/Solution.go b/solution/2900-2999/2930.Number of Strings Which Can Be Rearranged to Contain Substring/Solution.go
new file mode 100644
index 0000000000000..ded6388eea146
--- /dev/null
+++ b/solution/2900-2999/2930.Number of Strings Which Can Be Rearranged to Contain Substring/Solution.go	
@@ -0,0 +1,22 @@
+func stringCount(n int) int {
+	const mod int = 1e9 + 7
+	qpow := func(a, n int) int {
+		ans := 1
+		for ; n > 0; n >>= 1 {
+			if n&1 == 1 {
+				ans = ans * a % mod
+			}
+			a = a * a % mod
+		}
+		return ans
+	}
+	a := qpow(25, n)
+	b := a
+	c := qpow(25, n) + n*qpow(25, n-1)
+	ab := qpow(24, n)
+	ac := (qpow(24, n) + n*qpow(24, n-1)) % mod
+	bc := ac
+	abc := (qpow(23, n) + n*qpow(23, n-1)) % mod
+	tot := qpow(26, n)
+	return ((tot-(a+b+c-ab-ac-bc+abc))%mod + mod) % mod
+}
\ No newline at end of file
diff --git a/solution/2900-2999/2930.Number of Strings Which Can Be Rearranged to Contain Substring/Solution.java b/solution/2900-2999/2930.Number of Strings Which Can Be Rearranged to Contain Substring/Solution.java
new file mode 100644
index 0000000000000..f38e4381dffcf
--- /dev/null
+++ b/solution/2900-2999/2930.Number of Strings Which Can Be Rearranged to Contain Substring/Solution.java	
@@ -0,0 +1,26 @@
+class Solution {
+    private final int mod = (int) 1e9 + 7;
+
+    public int stringCount(int n) {
+        long a = qpow(25, n);
+        long b = a;
+        long c = (qpow(25, n) + n * qpow(25, n - 1) % mod) % mod;
+        long ab = qpow(24, n);
+        long ac = (qpow(24, n) + n * qpow(24, n - 1) % mod) % mod;
+        long bc = ac;
+        long abc = (qpow(23, n) + n * qpow(23, n - 1) % mod) % mod;
+        long tot = qpow(26, n);
+        return (int) ((tot - (a + b + c - ab - ac - bc + abc)) % mod + mod) % mod;
+    }
+
+    private long qpow(long a, int n) {
+        long ans = 1;
+        for (; n > 0; n >>= 1) {
+            if ((n & 1) == 1) {
+                ans = ans * a % mod;
+            }
+            a = a * a % mod;
+        }
+        return ans;
+    }
+}
\ No newline at end of file
diff --git a/solution/2900-2999/2930.Number of Strings Which Can Be Rearranged to Contain Substring/Solution.py b/solution/2900-2999/2930.Number of Strings Which Can Be Rearranged to Contain Substring/Solution.py
new file mode 100644
index 0000000000000..32ba3853603f6
--- /dev/null
+++ b/solution/2900-2999/2930.Number of Strings Which Can Be Rearranged to Contain Substring/Solution.py	
@@ -0,0 +1,10 @@
+class Solution:
+    def stringCount(self, n: int) -> int:
+        mod = 10**9 + 7
+        a = b = pow(25, n, mod)
+        c = pow(25, n, mod) + n * pow(25, n - 1, mod)
+        ab = pow(24, n, mod)
+        ac = bc = (pow(24, n, mod) + n * pow(24, n - 1, mod)) % mod
+        abc = (pow(23, n, mod) + n * pow(23, n - 1, mod)) % mod
+        tot = pow(26, n, mod)
+        return (tot - (a + b + c - ab - ac - bc + abc)) % mod
diff --git a/solution/2900-2999/2930.Number of Strings Which Can Be Rearranged to Contain Substring/Solution.ts b/solution/2900-2999/2930.Number of Strings Which Can Be Rearranged to Contain Substring/Solution.ts
new file mode 100644
index 0000000000000..745655b6c5a77
--- /dev/null
+++ b/solution/2900-2999/2930.Number of Strings Which Can Be Rearranged to Contain Substring/Solution.ts	
@@ -0,0 +1,22 @@
+function stringCount(n: number): number {
+    const mod = BigInt(10 ** 9 + 7);
+    const qpow = (a: bigint, n: number): bigint => {
+        let ans = 1n;
+        for (; n; n >>>= 1) {
+            if (n & 1) {
+                ans = (ans * a) % mod;
+            }
+            a = (a * a) % mod;
+        }
+        return ans;
+    };
+    const a = qpow(25n, n);
+    const b = a;
+    const c = (qpow(25n, n) + ((BigInt(n) * qpow(25n, n - 1)) % mod)) % mod;
+    const ab = qpow(24n, n);
+    const ac = (qpow(24n, n) + ((BigInt(n) * qpow(24n, n - 1)) % mod)) % mod;
+    const bc = ac;
+    const abc = (qpow(23n, n) + ((BigInt(n) * qpow(23n, n - 1)) % mod)) % mod;
+    const tot = qpow(26n, n);
+    return Number((((tot - (a + b + c - ab - ac - bc + abc)) % mod) + mod) % mod);
+}