feat: add solutions to lc problem: No.3215 (doocs#3253)

No.3215.Count Triplets with Even XOR Set Bits II

Author: yanglbme

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 -->