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Commit f7cf05e

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feat: add solutions to lc problem: No.2317
No.2317.Maximum XOR After Operations
1 parent ca38f01 commit f7cf05e

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7 files changed

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-38
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7 files changed

+59
-38
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‎solution/2300-2399/2317.Maximum XOR After Operations/README.md‎

Lines changed: 25 additions & 13 deletions
Original file line numberDiff line numberDiff line change
@@ -44,6 +44,14 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:位运算**
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在一次操作中,我们可以把 $nums[i]$ 更新为 $nums[i] \& (nums[i] \oplus x),ドル其中 $\&$ 表示逐位与运算,$\oplus$ 表示逐位异或运算。由于 $x$ 是任意非负整数,因此 $nums[i] \oplus x$ 的结果是一个任意值,再与 $nums[i]$ 逐位与运算,可以把 $nums[i]$ 的二进制表示中的若干位 1ドル$ 变为 0ドル$。
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而题目中要获取的是 `nums` 所有元素的最大逐位异或和,对于一个二进制位,只要在 `nums` 中存在一个元素对应的二进制位为 1ドル,ドル那么这个二进制位对于最大逐位异或和的贡献就是 1ドル$。因此答案就是 `nums` 中所有元素的逐位或运算的结果。
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时间复杂度 $O(n),ドル空间复杂度 $O(1)$。其中 $n$ 为 `nums` 的长度。
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<!-- tabs:start -->
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### **Python3**
@@ -53,10 +61,7 @@
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```python
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class Solution:
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def maximumXOR(self, nums: List[int]) -> int:
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ans = 0
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for v in nums:
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ans |= v
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return ans
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return reduce(or_, nums)
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```
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### **Java**
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class Solution {
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public int maximumXOR(int[] nums) {
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int ans = 0;
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for (int v : nums) {
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ans |= v;
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for (int x : nums) {
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ans |= x;
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}
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return ans;
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}
@@ -82,7 +87,9 @@ class Solution {
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public:
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int maximumXOR(vector<int>& nums) {
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int ans = 0;
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for (int& v : nums) ans |= v;
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for (int& x : nums) {
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ans |= x;
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}
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return ans;
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}
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};
@@ -91,19 +98,24 @@ public:
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### **Go**
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```go
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func maximumXOR(nums []int) int {
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ans := 0
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for _, v := range nums {
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ans |= v
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func maximumXOR(nums []int) (ans int) {
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for _, x := range nums {
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ans |= x
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}
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return ans
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return
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}
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```
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### **TypeScript**
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```ts
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function maximumXOR(nums: number[]): number {
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let ans = 0;
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for (const x of nums) {
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ans |= x;
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}
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return ans;
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}
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```
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### **...**

‎solution/2300-2399/2317.Maximum XOR After Operations/README_EN.md‎

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Original file line numberDiff line numberDiff line change
@@ -47,10 +47,7 @@ It can be shown that 11 is the maximum possible bitwise XOR.</pre>
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```python
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class Solution:
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def maximumXOR(self, nums: List[int]) -> int:
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ans = 0
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for v in nums:
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ans |= v
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return ans
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return reduce(or_, nums)
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```
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### **Java**
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class Solution {
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public int maximumXOR(int[] nums) {
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int ans = 0;
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for (int v : nums) {
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ans |= v;
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for (int x : nums) {
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ans |= x;
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}
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return ans;
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}
@@ -74,7 +71,9 @@ class Solution {
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public:
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int maximumXOR(vector<int>& nums) {
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int ans = 0;
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for (int& v : nums) ans |= v;
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for (int& x : nums) {
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ans |= x;
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}
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return ans;
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}
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};
@@ -83,19 +82,24 @@ public:
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### **Go**
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```go
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func maximumXOR(nums []int) int {
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ans := 0
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for _, v := range nums {
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ans |= v
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func maximumXOR(nums []int) (ans int) {
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for _, x := range nums {
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ans |= x
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}
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return ans
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return
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}
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```
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### **TypeScript**
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```ts
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function maximumXOR(nums: number[]): number {
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let ans = 0;
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for (const x of nums) {
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ans |= x;
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}
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return ans;
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}
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```
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### **...**

‎solution/2300-2399/2317.Maximum XOR After Operations/Solution.cpp‎

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Original file line numberDiff line numberDiff line change
@@ -2,7 +2,9 @@ class Solution {
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public:
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int maximumXOR(vector<int>& nums) {
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int ans = 0;
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for (int& v : nums) ans |= v;
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for (int& x : nums) {
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ans |= x;
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}
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return ans;
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}
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};
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Original file line numberDiff line numberDiff line change
@@ -1,7 +1,6 @@
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func maximumXOR(nums []int) int {
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ans := 0
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for _, v := range nums {
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ans |= v
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func maximumXOR(nums []int) (ans int) {
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for _, x := range nums {
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ans |= x
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}
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returnans
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return
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}

‎solution/2300-2399/2317.Maximum XOR After Operations/Solution.java‎

Lines changed: 2 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -1,8 +1,8 @@
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class Solution {
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public int maximumXOR(int[] nums) {
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int ans = 0;
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for (int v : nums) {
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ans |= v;
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for (int x : nums) {
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ans |= x;
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}
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return ans;
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}
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Original file line numberDiff line numberDiff line change
@@ -1,6 +1,3 @@
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class Solution:
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def maximumXOR(self, nums: List[int]) -> int:
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ans = 0
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for v in nums:
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ans |= v
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return ans
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return reduce(or_, nums)
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,7 @@
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function maximumXOR(nums: number[]): number {
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let ans = 0;
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for (const x of nums) {
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ans |= x;
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}
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return ans;
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}

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