diff --git a/solution/3200-3299/3215.Count Triplets with Even XOR Set Bits II/README.md b/solution/3200-3299/3215.Count Triplets with Even XOR Set Bits II/README.md
new file mode 100644
index 0000000000000..ebb4b15a1117e
--- /dev/null
+++ b/solution/3200-3299/3215.Count Triplets with Even XOR Set Bits II/README.md	
@@ -0,0 +1,233 @@
+---
+comments: true
+difficulty: 中等
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3215.Count%20Triplets%20with%20Even%20XOR%20Set%20Bits%20II/README.md
+---
+
+<!-- problem:start -->
+
+# [3215. Count Triplets with Even XOR Set Bits II 🔒](https://leetcode.cn/problems/count-triplets-with-even-xor-set-bits-ii)
+
+[English Version](/solution/3200-3299/3215.Count%20Triplets%20with%20Even%20XOR%20Set%20Bits%20II/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+Given three integer arrays <code>a</code>, <code>b</code>, and <code>c</code>, return the number of triplets <code>(a[i], b[j], c[k])</code>, such that the bitwise <code>XOR</code> between the elements of each triplet has an <strong>even</strong> number of <span data-keyword="set-bit">set bits</span>.
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">a = [1], b = [2], c = [3]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">1</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The only triplet is <code>(a[0], b[0], c[0])</code> and their <code>XOR</code> is: <code>1 XOR 2 XOR 3 = 00<sub>2</sub></code>.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">a = [1,1], b = [2,3], c = [1,5]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">4</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>Consider these four triplets:</p>
+
+<ul>
+	<li><code>(a[0], b[1], c[0])</code>: <code>1 XOR 3 XOR 1 = 011<sub>2</sub></code></li>
+	<li><code>(a[1], b[1], c[0])</code>: <code>1 XOR 3 XOR 1 = 011<sub>2</sub></code></li>
+	<li><code>(a[0], b[0], c[1])</code>: <code>1 XOR 2 XOR 5 = 110<sub>2</sub></code></li>
+	<li><code>(a[1], b[0], c[1])</code>: <code>1 XOR 2 XOR 5 = 110<sub>2</sub></code></li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= a.length, b.length, c.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>0 &lt;= a[i], b[i], c[i] &lt;= 10<sup>9</sup></code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：位运算
+
+对于两个整数，异或结果中 $1$ 的个数的奇偶性，取决于两个整数的二进制表示中 $1$ 的个数的奇偶性。
+
+我们可以用三个数组 `cnt1`、`cnt2`、`cnt3` 分别记录数组 `a`、`b`、`c` 中每个数的二进制表示中 $1$ 的个数的奇偶性。
+
+然后我们在 $[0, 1]$ 的范围内枚举三个数组中的每个数的二进制表示中 $1$ 的个数的奇偶性，如果三个数的二进制表示中 $1$ 的个数的奇偶性之和为偶数，那么这三个数的异或结果中 $1$ 的个数也为偶数，此时我们将这三个数的组合数相乘累加到答案中。
+
+最后返回答案即可。
+
+时间复杂度 $O(n)$，其中 $n$ 为数组 `a`、`b`、`c` 的长度。空间复杂度 $O(1)$。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def tripletCount(self, a: List[int], b: List[int], c: List[int]) -> int:
+        cnt1 = Counter(x.bit_count() & 1 for x in a)
+        cnt2 = Counter(x.bit_count() & 1 for x in b)
+        cnt3 = Counter(x.bit_count() & 1 for x in c)
+        ans = 0
+        for i in range(2):
+            for j in range(2):
+                for k in range(2):
+                    if (i + j + k) & 1 ^ 1:
+                        ans += cnt1[i] * cnt2[j] * cnt3[k]
+        return ans
+```
+
+#### Java
+
+```java
+class Solution {
+    public long tripletCount(int[] a, int[] b, int[] c) {
+        int[] cnt1 = new int[2];
+        int[] cnt2 = new int[2];
+        int[] cnt3 = new int[2];
+        for (int x : a) {
+            ++cnt1[Integer.bitCount(x) & 1];
+        }
+        for (int x : b) {
+            ++cnt2[Integer.bitCount(x) & 1];
+        }
+        for (int x : c) {
+            ++cnt3[Integer.bitCount(x) & 1];
+        }
+        long ans = 0;
+        for (int i = 0; i < 2; ++i) {
+            for (int j = 0; j < 2; ++j) {
+                for (int k = 0; k < 2; ++k) {
+                    if ((i + j + k) % 2 == 0) {
+                        ans += 1L * cnt1[i] * cnt2[j] * cnt3[k];
+                    }
+                }
+            }
+        }
+        return ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    long long tripletCount(vector<int>& a, vector<int>& b, vector<int>& c) {
+        int cnt1[2]{};
+        int cnt2[2]{};
+        int cnt3[2]{};
+        for (int x : a) {
+            ++cnt1[__builtin_popcount(x) & 1];
+        }
+        for (int x : b) {
+            ++cnt2[__builtin_popcount(x) & 1];
+        }
+        for (int x : c) {
+            ++cnt3[__builtin_popcount(x) & 1];
+        }
+        long long ans = 0;
+        for (int i = 0; i < 2; ++i) {
+            for (int j = 0; j < 2; ++j) {
+                for (int k = 0; k < 2; ++k) {
+                    if ((i + j + k) % 2 == 0) {
+                        ans += 1LL * cnt1[i] * cnt2[j] * cnt3[k];
+                    }
+                }
+            }
+        }
+        return ans;
+    }
+};
+```
+
+#### Go
+
+```go
+func tripletCount(a []int, b []int, c []int) (ans int64) {
+	cnt1 := [2]int{}
+	cnt2 := [2]int{}
+	cnt3 := [2]int{}
+	for _, x := range a {
+		cnt1[bits.OnesCount(uint(x))%2]++
+	}
+	for _, x := range b {
+		cnt2[bits.OnesCount(uint(x))%2]++
+	}
+	for _, x := range c {
+		cnt3[bits.OnesCount(uint(x))%2]++
+	}
+	for i := 0; i < 2; i++ {
+		for j := 0; j < 2; j++ {
+			for k := 0; k < 2; k++ {
+				if (i+j+k)%2 == 0 {
+					ans += int64(cnt1[i] * cnt2[j] * cnt3[k])
+				}
+			}
+		}
+	}
+	return
+}
+```
+
+#### TypeScript
+
+```ts
+function tripletCount(a: number[], b: number[], c: number[]): number {
+    const cnt1: [number, number] = [0, 0];
+    const cnt2: [number, number] = [0, 0];
+    const cnt3: [number, number] = [0, 0];
+    for (const x of a) {
+        ++cnt1[bitCount(x) & 1];
+    }
+    for (const x of b) {
+        ++cnt2[bitCount(x) & 1];
+    }
+    for (const x of c) {
+        ++cnt3[bitCount(x) & 1];
+    }
+    let ans = 0;
+    for (let i = 0; i < 2; ++i) {
+        for (let j = 0; j < 2; ++j) {
+            for (let k = 0; k < 2; ++k) {
+                if ((i + j + k) % 2 === 0) {
+                    ans += cnt1[i] * cnt2[j] * cnt3[k];
+                }
+            }
+        }
+    }
+    return ans;
+}
+
+function bitCount(i: number): number {
+    i = i - ((i >>> 1) & 0x55555555);
+    i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
+    i = (i + (i >>> 4)) & 0x0f0f0f0f;
+    i = i + (i >>> 8);
+    i = i + (i >>> 16);
+    return i & 0x3f;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->
diff --git a/solution/3200-3299/3215.Count Triplets with Even XOR Set Bits II/README_EN.md b/solution/3200-3299/3215.Count Triplets with Even XOR Set Bits II/README_EN.md
new file mode 100644
index 0000000000000..ad7e28c68dbb9
--- /dev/null
+++ b/solution/3200-3299/3215.Count Triplets with Even XOR Set Bits II/README_EN.md	
@@ -0,0 +1,233 @@
+---
+comments: true
+difficulty: Medium
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3215.Count%20Triplets%20with%20Even%20XOR%20Set%20Bits%20II/README_EN.md
+---
+
+<!-- problem:start -->
+
+# [3215. Count Triplets with Even XOR Set Bits II 🔒](https://leetcode.com/problems/count-triplets-with-even-xor-set-bits-ii)
+
+[中文文档](/solution/3200-3299/3215.Count%20Triplets%20with%20Even%20XOR%20Set%20Bits%20II/README.md)
+
+## Description
+
+<!-- description:start -->
+
+Given three integer arrays <code>a</code>, <code>b</code>, and <code>c</code>, return the number of triplets <code>(a[i], b[j], c[k])</code>, such that the bitwise <code>XOR</code> between the elements of each triplet has an <strong>even</strong> number of <span data-keyword="set-bit">set bits</span>.
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">a = [1], b = [2], c = [3]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">1</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The only triplet is <code>(a[0], b[0], c[0])</code> and their <code>XOR</code> is: <code>1 XOR 2 XOR 3 = 00<sub>2</sub></code>.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">a = [1,1], b = [2,3], c = [1,5]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">4</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>Consider these four triplets:</p>
+
+<ul>
+	<li><code>(a[0], b[1], c[0])</code>: <code>1 XOR 3 XOR 1 = 011<sub>2</sub></code></li>
+	<li><code>(a[1], b[1], c[0])</code>: <code>1 XOR 3 XOR 1 = 011<sub>2</sub></code></li>
+	<li><code>(a[0], b[0], c[1])</code>: <code>1 XOR 2 XOR 5 = 110<sub>2</sub></code></li>
+	<li><code>(a[1], b[0], c[1])</code>: <code>1 XOR 2 XOR 5 = 110<sub>2</sub></code></li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= a.length, b.length, c.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>0 &lt;= a[i], b[i], c[i] &lt;= 10<sup>9</sup></code></li>
+</ul>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1: Bit Manipulation
+
+For two integers, the parity of the number of $1$s in the XOR result depends on the parity of the number of $1$s in the binary representations of the two integers.
+
+We can use three arrays `cnt1`, `cnt2`, `cnt3` to record the parity of the number of $1$s in the binary representations of each number in arrays `a`, `b`, `c`, respectively.
+
+Then, we enumerate the parity of the number of $1$s in the binary representations of each number in the three arrays within the range $[0, 1]$. If the sum of the parity of the number of $1$s in the binary representations of three numbers is even, then the number of $1$s in the XOR result of these three numbers is also even. At this time, we multiply the combination of these three numbers and accumulate it into the answer.
+
+Finally, return the answer.
+
+The time complexity is $O(n)$, where $n$ is the length of arrays `a`, `b`, `c`. The space complexity is $O(1)$.
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def tripletCount(self, a: List[int], b: List[int], c: List[int]) -> int:
+        cnt1 = Counter(x.bit_count() & 1 for x in a)
+        cnt2 = Counter(x.bit_count() & 1 for x in b)
+        cnt3 = Counter(x.bit_count() & 1 for x in c)
+        ans = 0
+        for i in range(2):
+            for j in range(2):
+                for k in range(2):
+                    if (i + j + k) & 1 ^ 1:
+                        ans += cnt1[i] * cnt2[j] * cnt3[k]
+        return ans
+```
+
+#### Java
+
+```java
+class Solution {
+    public long tripletCount(int[] a, int[] b, int[] c) {
+        int[] cnt1 = new int[2];
+        int[] cnt2 = new int[2];
+        int[] cnt3 = new int[2];
+        for (int x : a) {
+            ++cnt1[Integer.bitCount(x) & 1];
+        }
+        for (int x : b) {
+            ++cnt2[Integer.bitCount(x) & 1];
+        }
+        for (int x : c) {
+            ++cnt3[Integer.bitCount(x) & 1];
+        }
+        long ans = 0;
+        for (int i = 0; i < 2; ++i) {
+            for (int j = 0; j < 2; ++j) {
+                for (int k = 0; k < 2; ++k) {
+                    if ((i + j + k) % 2 == 0) {
+                        ans += 1L * cnt1[i] * cnt2[j] * cnt3[k];
+                    }
+                }
+            }
+        }
+        return ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    long long tripletCount(vector<int>& a, vector<int>& b, vector<int>& c) {
+        int cnt1[2]{};
+        int cnt2[2]{};
+        int cnt3[2]{};
+        for (int x : a) {
+            ++cnt1[__builtin_popcount(x) & 1];
+        }
+        for (int x : b) {
+            ++cnt2[__builtin_popcount(x) & 1];
+        }
+        for (int x : c) {
+            ++cnt3[__builtin_popcount(x) & 1];
+        }
+        long long ans = 0;
+        for (int i = 0; i < 2; ++i) {
+            for (int j = 0; j < 2; ++j) {
+                for (int k = 0; k < 2; ++k) {
+                    if ((i + j + k) % 2 == 0) {
+                        ans += 1LL * cnt1[i] * cnt2[j] * cnt3[k];
+                    }
+                }
+            }
+        }
+        return ans;
+    }
+};
+```
+
+#### Go
+
+```go
+func tripletCount(a []int, b []int, c []int) (ans int64) {
+	cnt1 := [2]int{}
+	cnt2 := [2]int{}
+	cnt3 := [2]int{}
+	for _, x := range a {
+		cnt1[bits.OnesCount(uint(x))%2]++
+	}
+	for _, x := range b {
+		cnt2[bits.OnesCount(uint(x))%2]++
+	}
+	for _, x := range c {
+		cnt3[bits.OnesCount(uint(x))%2]++
+	}
+	for i := 0; i < 2; i++ {
+		for j := 0; j < 2; j++ {
+			for k := 0; k < 2; k++ {
+				if (i+j+k)%2 == 0 {
+					ans += int64(cnt1[i] * cnt2[j] * cnt3[k])
+				}
+			}
+		}
+	}
+	return
+}
+```
+
+#### TypeScript
+
+```ts
+function tripletCount(a: number[], b: number[], c: number[]): number {
+    const cnt1: [number, number] = [0, 0];
+    const cnt2: [number, number] = [0, 0];
+    const cnt3: [number, number] = [0, 0];
+    for (const x of a) {
+        ++cnt1[bitCount(x) & 1];
+    }
+    for (const x of b) {
+        ++cnt2[bitCount(x) & 1];
+    }
+    for (const x of c) {
+        ++cnt3[bitCount(x) & 1];
+    }
+    let ans = 0;
+    for (let i = 0; i < 2; ++i) {
+        for (let j = 0; j < 2; ++j) {
+            for (let k = 0; k < 2; ++k) {
+                if ((i + j + k) % 2 === 0) {
+                    ans += cnt1[i] * cnt2[j] * cnt3[k];
+                }
+            }
+        }
+    }
+    return ans;
+}
+
+function bitCount(i: number): number {
+    i = i - ((i >>> 1) & 0x55555555);
+    i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
+    i = (i + (i >>> 4)) & 0x0f0f0f0f;
+    i = i + (i >>> 8);
+    i = i + (i >>> 16);
+    return i & 0x3f;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->
diff --git a/solution/3200-3299/3215.Count Triplets with Even XOR Set Bits II/Solution.cpp b/solution/3200-3299/3215.Count Triplets with Even XOR Set Bits II/Solution.cpp
new file mode 100644
index 0000000000000..10effdfaf1976
--- /dev/null
+++ b/solution/3200-3299/3215.Count Triplets with Even XOR Set Bits II/Solution.cpp	
@@ -0,0 +1,28 @@
+class Solution {
+public:
+    long long tripletCount(vector<int>& a, vector<int>& b, vector<int>& c) {
+        int cnt1[2]{};
+        int cnt2[2]{};
+        int cnt3[2]{};
+        for (int x : a) {
+            ++cnt1[__builtin_popcount(x) & 1];
+        }
+        for (int x : b) {
+            ++cnt2[__builtin_popcount(x) & 1];
+        }
+        for (int x : c) {
+            ++cnt3[__builtin_popcount(x) & 1];
+        }
+        long long ans = 0;
+        for (int i = 0; i < 2; ++i) {
+            for (int j = 0; j < 2; ++j) {
+                for (int k = 0; k < 2; ++k) {
+                    if ((i + j + k) % 2 == 0) {
+                        ans += 1LL * cnt1[i] * cnt2[j] * cnt3[k];
+                    }
+                }
+            }
+        }
+        return ans;
+    }
+};
\ No newline at end of file
diff --git a/solution/3200-3299/3215.Count Triplets with Even XOR Set Bits II/Solution.go b/solution/3200-3299/3215.Count Triplets with Even XOR Set Bits II/Solution.go
new file mode 100644
index 0000000000000..4efa959aad086
--- /dev/null
+++ b/solution/3200-3299/3215.Count Triplets with Even XOR Set Bits II/Solution.go	
@@ -0,0 +1,24 @@
+func tripletCount(a []int, b []int, c []int) (ans int64) {
+	cnt1 := [2]int{}
+	cnt2 := [2]int{}
+	cnt3 := [2]int{}
+	for _, x := range a {
+		cnt1[bits.OnesCount(uint(x))%2]++
+	}
+	for _, x := range b {
+		cnt2[bits.OnesCount(uint(x))%2]++
+	}
+	for _, x := range c {
+		cnt3[bits.OnesCount(uint(x))%2]++
+	}
+	for i := 0; i < 2; i++ {
+		for j := 0; j < 2; j++ {
+			for k := 0; k < 2; k++ {
+				if (i+j+k)%2 == 0 {
+					ans += int64(cnt1[i] * cnt2[j] * cnt3[k])
+				}
+			}
+		}
+	}
+	return
+}
\ No newline at end of file
diff --git a/solution/3200-3299/3215.Count Triplets with Even XOR Set Bits II/Solution.java b/solution/3200-3299/3215.Count Triplets with Even XOR Set Bits II/Solution.java
new file mode 100644
index 0000000000000..1dfb5b152c3e4
--- /dev/null
+++ b/solution/3200-3299/3215.Count Triplets with Even XOR Set Bits II/Solution.java	
@@ -0,0 +1,27 @@
+class Solution {
+    public long tripletCount(int[] a, int[] b, int[] c) {
+        int[] cnt1 = new int[2];
+        int[] cnt2 = new int[2];
+        int[] cnt3 = new int[2];
+        for (int x : a) {
+            ++cnt1[Integer.bitCount(x) & 1];
+        }
+        for (int x : b) {
+            ++cnt2[Integer.bitCount(x) & 1];
+        }
+        for (int x : c) {
+            ++cnt3[Integer.bitCount(x) & 1];
+        }
+        long ans = 0;
+        for (int i = 0; i < 2; ++i) {
+            for (int j = 0; j < 2; ++j) {
+                for (int k = 0; k < 2; ++k) {
+                    if ((i + j + k) % 2 == 0) {
+                        ans += 1L * cnt1[i] * cnt2[j] * cnt3[k];
+                    }
+                }
+            }
+        }
+        return ans;
+    }
+}
\ No newline at end of file
diff --git a/solution/3200-3299/3215.Count Triplets with Even XOR Set Bits II/Solution.py b/solution/3200-3299/3215.Count Triplets with Even XOR Set Bits II/Solution.py
new file mode 100644
index 0000000000000..f9dd29bb53e3f
--- /dev/null
+++ b/solution/3200-3299/3215.Count Triplets with Even XOR Set Bits II/Solution.py	
@@ -0,0 +1,12 @@
+class Solution:
+    def tripletCount(self, a: List[int], b: List[int], c: List[int]) -> int:
+        cnt1 = Counter(x.bit_count() & 1 for x in a)
+        cnt2 = Counter(x.bit_count() & 1 for x in b)
+        cnt3 = Counter(x.bit_count() & 1 for x in c)
+        ans = 0
+        for i in range(2):
+            for j in range(2):
+                for k in range(2):
+                    if (i + j + k) & 1 ^ 1:
+                        ans += cnt1[i] * cnt2[j] * cnt3[k]
+        return ans
diff --git a/solution/3200-3299/3215.Count Triplets with Even XOR Set Bits II/Solution.ts b/solution/3200-3299/3215.Count Triplets with Even XOR Set Bits II/Solution.ts
new file mode 100644
index 0000000000000..05376ee4c92be
--- /dev/null
+++ b/solution/3200-3299/3215.Count Triplets with Even XOR Set Bits II/Solution.ts	
@@ -0,0 +1,34 @@
+function tripletCount(a: number[], b: number[], c: number[]): number {
+    const cnt1: [number, number] = [0, 0];
+    const cnt2: [number, number] = [0, 0];
+    const cnt3: [number, number] = [0, 0];
+    for (const x of a) {
+        ++cnt1[bitCount(x) & 1];
+    }
+    for (const x of b) {
+        ++cnt2[bitCount(x) & 1];
+    }
+    for (const x of c) {
+        ++cnt3[bitCount(x) & 1];
+    }
+    let ans = 0;
+    for (let i = 0; i < 2; ++i) {
+        for (let j = 0; j < 2; ++j) {
+            for (let k = 0; k < 2; ++k) {
+                if ((i + j + k) % 2 === 0) {
+                    ans += cnt1[i] * cnt2[j] * cnt3[k];
+                }
+            }
+        }
+    }
+    return ans;
+}
+
+function bitCount(i: number): number {
+    i = i - ((i >>> 1) & 0x55555555);
+    i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
+    i = (i + (i >>> 4)) & 0x0f0f0f0f;
+    i = i + (i >>> 8);
+    i = i + (i >>> 16);
+    return i & 0x3f;
+}
diff --git a/solution/README.md b/solution/README.md
index d8b5ab68000ab..d5698ff7fc79c 100644
--- a/solution/README.md
+++ b/solution/README.md
@@ -3225,6 +3225,7 @@
 |  3212  |  [统计 X 和 Y 频数相等的子矩阵数量](/solution/3200-3299/3212.Count%20Submatrices%20With%20Equal%20Frequency%20of%20X%20and%20Y/README.md)  |  `数组`,`矩阵`,`前缀和`  |  中等  |  第 405 场周赛  |
 |  3213  |  [最小代价构造字符串](/solution/3200-3299/3213.Construct%20String%20with%20Minimum%20Cost/README.md)  |  `数组`,`字符串`,`动态规划`,`后缀数组`  |  困难  |  第 405 场周赛  |
 |  3214  |  [同比增长率](/solution/3200-3299/3214.Year%20on%20Year%20Growth%20Rate/README.md)  |  `数据库`  |  困难  |  🔒  |
+|  3215  |  [Count Triplets with Even XOR Set Bits II](/solution/3200-3299/3215.Count%20Triplets%20with%20Even%20XOR%20Set%20Bits%20II/README.md)  |    |  中等  |  🔒  |
 
 ## 版权
 
diff --git a/solution/README_EN.md b/solution/README_EN.md
index b4e4e0f9e97cb..82525662631a8 100644
--- a/solution/README_EN.md
+++ b/solution/README_EN.md
@@ -3223,6 +3223,7 @@ Press <kbd>Control</kbd> + <kbd>F</kbd>(or <kbd>Command</kbd> + <kbd>F</kbd> on
 |  3212  |  [Count Submatrices With Equal Frequency of X and Y](/solution/3200-3299/3212.Count%20Submatrices%20With%20Equal%20Frequency%20of%20X%20and%20Y/README_EN.md)  |  `Array`,`Matrix`,`Prefix Sum`  |  Medium  |  Weekly Contest 405  |
 |  3213  |  [Construct String with Minimum Cost](/solution/3200-3299/3213.Construct%20String%20with%20Minimum%20Cost/README_EN.md)  |  `Array`,`String`,`Dynamic Programming`,`Suffix Array`  |  Hard  |  Weekly Contest 405  |
 |  3214  |  [Year on Year Growth Rate](/solution/3200-3299/3214.Year%20on%20Year%20Growth%20Rate/README_EN.md)  |  `Database`  |  Hard  |  🔒  |
+|  3215  |  [Count Triplets with Even XOR Set Bits II](/solution/3200-3299/3215.Count%20Triplets%20with%20Even%20XOR%20Set%20Bits%20II/README_EN.md)  |    |  Medium  |  🔒  |
 
 ## Copyright