diff --git a/solution/2900-2999/2913.Subarrays Distinct Element Sum of Squares I/README.md b/solution/2900-2999/2913.Subarrays Distinct Element Sum of Squares I/README.md
index 74df06f8ac071..809f86da4d805 100644
--- a/solution/2900-2999/2913.Subarrays Distinct Element Sum of Squares I/README.md	
+++ b/solution/2900-2999/2913.Subarrays Distinct Element Sum of Squares I/README.md	
@@ -62,6 +62,14 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：枚举**
+
+我们可以枚举子数组的左端点下标 $i$，对于每个 $i$，我们在 $[i, n)$ 的范围内枚举子数组的右端点下标 $j$，并统计 $nums[j]$ 的值，将其加入到集合 $s$ 中，记 $s$ 的大小为 $cnt$，那么 $nums[i..j]$ 的不同计数为 $cnt$，将其平方后加入到答案中。
+
+枚举结束后，返回答案即可。
+
+时间复杂度 $O(n^2)$，空间复杂度 $O(n)$。其中 $n$ 是数组 $nums$ 的长度。
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -69,7 +77,15 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def sumCounts(self, nums: List[int]) -> int:
+        ans, n = 0, len(nums)
+        for i in range(n):
+            s = set()
+            for j in range(i, n):
+                s.add(nums[j])
+                ans += len(s) * len(s)
+        return ans
 ```
 
 ### **Java**
@@ -77,19 +93,85 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public int sumCounts(List<Integer> nums) {
+        int ans = 0;
+        int n = nums.size();
+        for (int i = 0; i < n; ++i) {
+            int[] s = new int[101];
+            int cnt = 0;
+            for (int j = i; j < n; ++j) {
+                if (++s[nums.get(j)] == 1) {
+                    ++cnt;
+                }
+                ans += cnt * cnt;
+            }
+        }
+        return ans;
+    }
+}
 ```
 
 ### **C++**
 
 ```cpp
-
+class Solution {
+public:
+    int sumCounts(vector<int>& nums) {
+        int ans = 0;
+        int n = nums.size();
+        for (int i = 0; i < n; ++i) {
+            int s[101]{};
+            int cnt = 0;
+            for (int j = i; j < n; ++j) {
+                if (++s[nums[j]] == 1) {
+                    ++cnt;
+                }
+                ans += cnt * cnt;
+            }
+        }
+        return ans;
+    }
+};
 ```
 
 ### **Go**
 
 ```go
+func sumCounts(nums []int) (ans int) {
+	for i := range nums {
+		s := [101]int{}
+		cnt := 0
+		for _, x := range nums[i:] {
+			s[x]++
+			if s[x] == 1 {
+				cnt++
+			}
+			ans += cnt * cnt
+		}
+	}
+	return
+}
+```
 
+### **TypeScript**
+
+```ts
+function sumCounts(nums: number[]): number {
+    let ans = 0;
+    const n = nums.length;
+    for (let i = 0; i < n; ++i) {
+        const s: number[] = Array(101).fill(0);
+        let cnt = 0;
+        for (const x of nums.slice(i)) {
+            if (++s[x] === 1) {
+                ++cnt;
+            }
+            ans += cnt * cnt;
+        }
+    }
+    return ans;
+}
 ```
 
 ### **...**
diff --git a/solution/2900-2999/2913.Subarrays Distinct Element Sum of Squares I/README_EN.md b/solution/2900-2999/2913.Subarrays Distinct Element Sum of Squares I/README_EN.md
index db319da3a4143..6d35a45f57e73 100644
--- a/solution/2900-2999/2913.Subarrays Distinct Element Sum of Squares I/README_EN.md	
+++ b/solution/2900-2999/2913.Subarrays Distinct Element Sum of Squares I/README_EN.md	
@@ -53,30 +53,112 @@ The sum of the squares of the distinct counts in all subarrays is equal to 1<sup
 
 ## Solutions
 
+**Solution 1: Enumeration**
+
+We can enumerate the left endpoint index $i$ of the subarray, and for each $i$, we enumerate the right endpoint index $j$ in the range $[i, n)$, and calculate the distinct count of $nums[i..j]$ by adding the count of $nums[j]$ to a set $s$, and then taking the square of the size of $s$ as the contribution of $nums[i..j]$ to the answer.
+
+After the enumeration, we return the answer.
+
+The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 ```python
-
+class Solution:
+    def sumCounts(self, nums: List[int]) -> int:
+        ans, n = 0, len(nums)
+        for i in range(n):
+            s = set()
+            for j in range(i, n):
+                s.add(nums[j])
+                ans += len(s) * len(s)
+        return ans
 ```
 
 ### **Java**
 
 ```java
-
+class Solution {
+    public int sumCounts(List<Integer> nums) {
+        int ans = 0;
+        int n = nums.size();
+        for (int i = 0; i < n; ++i) {
+            int[] s = new int[101];
+            int cnt = 0;
+            for (int j = i; j < n; ++j) {
+                if (++s[nums.get(j)] == 1) {
+                    ++cnt;
+                }
+                ans += cnt * cnt;
+            }
+        }
+        return ans;
+    }
+}
 ```
 
 ### **C++**
 
 ```cpp
-
+class Solution {
+public:
+    int sumCounts(vector<int>& nums) {
+        int ans = 0;
+        int n = nums.size();
+        for (int i = 0; i < n; ++i) {
+            int s[101]{};
+            int cnt = 0;
+            for (int j = i; j < n; ++j) {
+                if (++s[nums[j]] == 1) {
+                    ++cnt;
+                }
+                ans += cnt * cnt;
+            }
+        }
+        return ans;
+    }
+};
 ```
 
 ### **Go**
 
 ```go
+func sumCounts(nums []int) (ans int) {
+	for i := range nums {
+		s := [101]int{}
+		cnt := 0
+		for _, x := range nums[i:] {
+			s[x]++
+			if s[x] == 1 {
+				cnt++
+			}
+			ans += cnt * cnt
+		}
+	}
+	return
+}
+```
 
+### **TypeScript**
+
+```ts
+function sumCounts(nums: number[]): number {
+    let ans = 0;
+    const n = nums.length;
+    for (let i = 0; i < n; ++i) {
+        const s: number[] = Array(101).fill(0);
+        let cnt = 0;
+        for (const x of nums.slice(i)) {
+            if (++s[x] === 1) {
+                ++cnt;
+            }
+            ans += cnt * cnt;
+        }
+    }
+    return ans;
+}
 ```
 
 ### **...**
diff --git a/solution/2900-2999/2913.Subarrays Distinct Element Sum of Squares I/Solution.cpp b/solution/2900-2999/2913.Subarrays Distinct Element Sum of Squares I/Solution.cpp
new file mode 100644
index 0000000000000..5e59e40435c04
--- /dev/null
+++ b/solution/2900-2999/2913.Subarrays Distinct Element Sum of Squares I/Solution.cpp	
@@ -0,0 +1,18 @@
+class Solution {
+public:
+    int sumCounts(vector<int>& nums) {
+        int ans = 0;
+        int n = nums.size();
+        for (int i = 0; i < n; ++i) {
+            int s[101]{};
+            int cnt = 0;
+            for (int j = i; j < n; ++j) {
+                if (++s[nums[j]] == 1) {
+                    ++cnt;
+                }
+                ans += cnt * cnt;
+            }
+        }
+        return ans;
+    }
+};
\ No newline at end of file
diff --git a/solution/2900-2999/2913.Subarrays Distinct Element Sum of Squares I/Solution.go b/solution/2900-2999/2913.Subarrays Distinct Element Sum of Squares I/Solution.go
new file mode 100644
index 0000000000000..263dc63058cc8
--- /dev/null
+++ b/solution/2900-2999/2913.Subarrays Distinct Element Sum of Squares I/Solution.go	
@@ -0,0 +1,14 @@
+func sumCounts(nums []int) (ans int) {
+	for i := range nums {
+		s := [101]int{}
+		cnt := 0
+		for _, x := range nums[i:] {
+			s[x]++
+			if s[x] == 1 {
+				cnt++
+			}
+			ans += cnt * cnt
+		}
+	}
+	return
+}
\ No newline at end of file
diff --git a/solution/2900-2999/2913.Subarrays Distinct Element Sum of Squares I/Solution.java b/solution/2900-2999/2913.Subarrays Distinct Element Sum of Squares I/Solution.java
new file mode 100644
index 0000000000000..c2b72d1199bfe
--- /dev/null
+++ b/solution/2900-2999/2913.Subarrays Distinct Element Sum of Squares I/Solution.java	
@@ -0,0 +1,17 @@
+class Solution {
+    public int sumCounts(List<Integer> nums) {
+        int ans = 0;
+        int n = nums.size();
+        for (int i = 0; i < n; ++i) {
+            int[] s = new int[101];
+            int cnt = 0;
+            for (int j = i; j < n; ++j) {
+                if (++s[nums.get(j)] == 1) {
+                    ++cnt;
+                }
+                ans += cnt * cnt;
+            }
+        }
+        return ans;
+    }
+}
\ No newline at end of file
diff --git a/solution/2900-2999/2913.Subarrays Distinct Element Sum of Squares I/Solution.py b/solution/2900-2999/2913.Subarrays Distinct Element Sum of Squares I/Solution.py
new file mode 100644
index 0000000000000..9c8720ab45025
--- /dev/null
+++ b/solution/2900-2999/2913.Subarrays Distinct Element Sum of Squares I/Solution.py	
@@ -0,0 +1,9 @@
+class Solution:
+    def sumCounts(self, nums: List[int]) -> int:
+        ans, n = 0, len(nums)
+        for i in range(n):
+            s = set()
+            for j in range(i, n):
+                s.add(nums[j])
+                ans += len(s) * len(s)
+        return ans
diff --git a/solution/2900-2999/2913.Subarrays Distinct Element Sum of Squares I/Solution.ts b/solution/2900-2999/2913.Subarrays Distinct Element Sum of Squares I/Solution.ts
new file mode 100644
index 0000000000000..98d79c8832372
--- /dev/null
+++ b/solution/2900-2999/2913.Subarrays Distinct Element Sum of Squares I/Solution.ts	
@@ -0,0 +1,15 @@
+function sumCounts(nums: number[]): number {
+    let ans = 0;
+    const n = nums.length;
+    for (let i = 0; i < n; ++i) {
+        const s: number[] = Array(101).fill(0);
+        let cnt = 0;
+        for (const x of nums.slice(i)) {
+            if (++s[x] === 1) {
+                ++cnt;
+            }
+            ans += cnt * cnt;
+        }
+    }
+    return ans;
+}
diff --git a/solution/2900-2999/2914.Minimum Number of Changes to Make Binary String Beautiful/README.md b/solution/2900-2999/2914.Minimum Number of Changes to Make Binary String Beautiful/README.md
index 3e456c39dcfca..7d17255da6c08 100644
--- a/solution/2900-2999/2914.Minimum Number of Changes to Make Binary String Beautiful/README.md	
+++ b/solution/2900-2999/2914.Minimum Number of Changes to Make Binary String Beautiful/README.md	
@@ -63,6 +63,14 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：计数**
+
+我们只需要遍历字符串 $s$ 的所有奇数下标 $1, 3, 5, \cdots$，如果当前奇数下标与前一个下标的字符不同，即 $s[i] \ne s[i - 1]$，那么就需要修改当前字符，使得 $s[i] = s[i - 1]$。因此，此时答案需要加 $1$。
+
+遍历结束后，返回答案即可。
+
+时间复杂度 $O(n)$，其中 $n$ 是字符串 $s$ 的长度。空间复杂度 $O(1)$。
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -70,7 +78,9 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def minChanges(self, s: str) -> int:
+        return sum(s[i] != s[i - 1] for i in range(1, len(s), 2))
 ```
 
 ### **Java**
@@ -78,19 +88,60 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public int minChanges(String s) {
+        int ans = 0;
+        for (int i = 1; i < s.length(); i += 2) {
+            if (s.charAt(i) != s.charAt(i - 1)) {
+                ++ans;
+            }
+        }
+        return ans;
+    }
+}
 ```
 
 ### **C++**
 
 ```cpp
-
+class Solution {
+public:
+    int minChanges(string s) {
+        int ans = 0;
+        int n = s.size();
+        for (int i = 1; i < n; i += 2) {
+            ans += s[i] != s[i - 1];
+        }
+        return ans;
+    }
+};
 ```
 
 ### **Go**
 
 ```go
+func minChanges(s string) (ans int) {
+	for i := 1; i < len(s); i += 2 {
+		if s[i] != s[i-1] {
+			ans++
+		}
+	}
+	return
+}
+```
 
+### **TypeScript**
+
+```ts
+function minChanges(s: string): number {
+    let ans = 0;
+    for (let i = 1; i < s.length; i += 2) {
+        if (s[i] !== s[i - 1]) {
+            ++ans;
+        }
+    }
+    return ans;
+}
 ```
 
 ### **...**
diff --git a/solution/2900-2999/2914.Minimum Number of Changes to Make Binary String Beautiful/README_EN.md b/solution/2900-2999/2914.Minimum Number of Changes to Make Binary String Beautiful/README_EN.md
index 85a086dc7af46..1e3af94b77977 100644
--- a/solution/2900-2999/2914.Minimum Number of Changes to Make Binary String Beautiful/README_EN.md	
+++ b/solution/2900-2999/2914.Minimum Number of Changes to Make Binary String Beautiful/README_EN.md	
@@ -57,30 +57,81 @@ It can be proven that 1 is the minimum number of changes needed to make the stri
 
 ## Solutions
 
+**Solution 1: Counting**
+
+We only need to traverse all odd indices $1, 3, 5, \cdots$ of the string $s$. If the current odd index is different from the previous index, i.e., $s[i] \ne s[i - 1]$, we need to modify the current character so that $s[i] = s[i - 1]$. Therefore, the answer needs to be incremented by $1$.
+
+After the traversal, we return the answer.
+
+The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 ```python
-
+class Solution:
+    def minChanges(self, s: str) -> int:
+        return sum(s[i] != s[i - 1] for i in range(1, len(s), 2))
 ```
 
 ### **Java**
 
 ```java
-
+class Solution {
+    public int minChanges(String s) {
+        int ans = 0;
+        for (int i = 1; i < s.length(); i += 2) {
+            if (s.charAt(i) != s.charAt(i - 1)) {
+                ++ans;
+            }
+        }
+        return ans;
+    }
+}
 ```
 
 ### **C++**
 
 ```cpp
-
+class Solution {
+public:
+    int minChanges(string s) {
+        int ans = 0;
+        int n = s.size();
+        for (int i = 1; i < n; i += 2) {
+            ans += s[i] != s[i - 1];
+        }
+        return ans;
+    }
+};
 ```
 
 ### **Go**
 
 ```go
+func minChanges(s string) (ans int) {
+	for i := 1; i < len(s); i += 2 {
+		if s[i] != s[i-1] {
+			ans++
+		}
+	}
+	return
+}
+```
 
+### **TypeScript**
+
+```ts
+function minChanges(s: string): number {
+    let ans = 0;
+    for (let i = 1; i < s.length; i += 2) {
+        if (s[i] !== s[i - 1]) {
+            ++ans;
+        }
+    }
+    return ans;
+}
 ```
 
 ### **...**
diff --git a/solution/2900-2999/2914.Minimum Number of Changes to Make Binary String Beautiful/Solution.cpp b/solution/2900-2999/2914.Minimum Number of Changes to Make Binary String Beautiful/Solution.cpp
new file mode 100644
index 0000000000000..2e28922e7c939
--- /dev/null
+++ b/solution/2900-2999/2914.Minimum Number of Changes to Make Binary String Beautiful/Solution.cpp	
@@ -0,0 +1,11 @@
+class Solution {
+public:
+    int minChanges(string s) {
+        int ans = 0;
+        int n = s.size();
+        for (int i = 1; i < n; i += 2) {
+            ans += s[i] != s[i - 1];
+        }
+        return ans;
+    }
+};
\ No newline at end of file
diff --git a/solution/2900-2999/2914.Minimum Number of Changes to Make Binary String Beautiful/Solution.go b/solution/2900-2999/2914.Minimum Number of Changes to Make Binary String Beautiful/Solution.go
new file mode 100644
index 0000000000000..2e5eff143b5fb
--- /dev/null
+++ b/solution/2900-2999/2914.Minimum Number of Changes to Make Binary String Beautiful/Solution.go	
@@ -0,0 +1,8 @@
+func minChanges(s string) (ans int) {
+	for i := 1; i < len(s); i += 2 {
+		if s[i] != s[i-1] {
+			ans++
+		}
+	}
+	return
+}
\ No newline at end of file
diff --git a/solution/2900-2999/2914.Minimum Number of Changes to Make Binary String Beautiful/Solution.java b/solution/2900-2999/2914.Minimum Number of Changes to Make Binary String Beautiful/Solution.java
new file mode 100644
index 0000000000000..3064ac3a0b425
--- /dev/null
+++ b/solution/2900-2999/2914.Minimum Number of Changes to Make Binary String Beautiful/Solution.java	
@@ -0,0 +1,11 @@
+class Solution {
+    public int minChanges(String s) {
+        int ans = 0;
+        for (int i = 1; i < s.length(); i += 2) {
+            if (s.charAt(i) != s.charAt(i - 1)) {
+                ++ans;
+            }
+        }
+        return ans;
+    }
+}
\ No newline at end of file
diff --git a/solution/2900-2999/2914.Minimum Number of Changes to Make Binary String Beautiful/Solution.py b/solution/2900-2999/2914.Minimum Number of Changes to Make Binary String Beautiful/Solution.py
new file mode 100644
index 0000000000000..cbf348a9a4cf2
--- /dev/null
+++ b/solution/2900-2999/2914.Minimum Number of Changes to Make Binary String Beautiful/Solution.py	
@@ -0,0 +1,3 @@
+class Solution:
+    def minChanges(self, s: str) -> int:
+        return sum(s[i] != s[i - 1] for i in range(1, len(s), 2))
diff --git a/solution/2900-2999/2914.Minimum Number of Changes to Make Binary String Beautiful/Solution.ts b/solution/2900-2999/2914.Minimum Number of Changes to Make Binary String Beautiful/Solution.ts
new file mode 100644
index 0000000000000..5d76149314451
--- /dev/null
+++ b/solution/2900-2999/2914.Minimum Number of Changes to Make Binary String Beautiful/Solution.ts	
@@ -0,0 +1,9 @@
+function minChanges(s: string): number {
+    let ans = 0;
+    for (let i = 1; i < s.length; i += 2) {
+        if (s[i] !== s[i - 1]) {
+            ++ans;
+        }
+    }
+    return ans;
+}