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Commit 9f3a109

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feat: add solutions to lc problems: No.3314~3316 (doocs#3662)
* No.3314.Construct the Minimum Bitwise Array I * No.3315.Construct the Minimum Bitwise Array II * No.3316.Find Maximum Removals From Source String
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‎solution/3300-3399/3314.Construct the Minimum Bitwise Array I/README.md‎

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@@ -80,32 +80,123 @@ tags:
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<!-- solution:start -->
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### 方法一
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### 方法一:位运算
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对于一个整数 $a,ドル满足 $a \lor (a + 1)$ 的结果一定为奇数,因此,如果 $\text{nums[i]}$ 是偶数,那么 $\text{ans}[i]$ 一定不存在,直接返回 $-1$。本题中 $\textit{nums}[i]$ 是质数,判断是否是偶数,只需要判断是否等于 2ドル$ 即可。
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如果 $\text{nums[i]}$ 是奇数,假设 $\text{nums[i]} = \text{0b1101101},ドル由于 $a \lor (a + 1) = \text{nums[i]},ドル等价于将 $a$ 的最后一个为 0ドル$ 的二进制位变为 1ドル$。那么求解 $a,ドル就等价于将 $\text{nums[i]}$ 的最后一个 0ドル$ 的下一位 1ドル$ 变为 0ドル$。我们只需要从低位(下标为 1ドル$)开始遍历,找到第一个为 0ドル$ 的二进制位,如果是第 $i$ 位,那么我们就将 $\text{nums[i]}$ 的第 $i - 1$ 位变为 1ドル,ドル即 $\text{ans}[i] = \text{nums[i]} \oplus 2^{i - 1}$。
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遍历所有的 $\text{nums[i]},ドル即可得到答案。
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时间复杂度 $O(n \times \log M),ドル其中 $n$ 和 $M$ 分别是数组 $\text{nums}$ 的长度和数组中的最大值。忽略答案数组的空间消耗,空间复杂度 $O(1)$。
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def minBitwiseArray(self, nums: List[int]) -> List[int]:
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ans = []
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for x in nums:
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if x == 2:
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ans.append(-1)
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else:
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for i in range(1, 32):
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if x >> i & 1 ^ 1:
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ans.append(x ^ 1 << (i - 1))
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break
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return ans
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```
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#### Java
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```java
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class Solution {
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public int[] minBitwiseArray(List<Integer> nums) {
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int n = nums.size();
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int[] ans = new int[n];
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for (int i = 0; i < n; ++i) {
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int x = nums.get(i);
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if (x == 2) {
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ans[i] = -1;
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} else {
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for (int j = 1; j < 32; ++j) {
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if ((x >> j & 1) == 0) {
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ans[i] = x ^ 1 << (j - 1);
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break;
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}
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}
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}
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}
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return ans;
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}
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}
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```
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#### C++
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```cpp
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class Solution {
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public:
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vector<int> minBitwiseArray(vector<int>& nums) {
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vector<int> ans;
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for (int x : nums) {
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if (x == 2) {
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ans.push_back(-1);
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} else {
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for (int i = 1; i < 32; ++i) {
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if (x >> i & 1 ^ 1) {
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ans.push_back(x ^ 1 << (i - 1));
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break;
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}
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}
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}
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}
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return ans;
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}
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};
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```
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#### Go
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```go
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func minBitwiseArray(nums []int) (ans []int) {
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for _, x := range nums {
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if x == 2 {
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ans = append(ans, -1)
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} else {
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for i := 1; i < 32; i++ {
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if x>>i&1 == 0 {
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ans = append(ans, x^1<<(i-1))
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break
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}
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}
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}
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}
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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 minBitwiseArray(nums: number[]): number[] {
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const ans: number[] = [];
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for (const x of nums) {
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if (x === 2) {
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ans.push(-1);
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} else {
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for (let i = 1; i < 32; ++i) {
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if (((x >> i) & 1) ^ 1) {
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ans.push(x ^ (1 << (i - 1)));
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break;
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}
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}
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}
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}
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return ans;
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}
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```
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<!-- tabs:end -->

‎solution/3300-3399/3314.Construct the Minimum Bitwise Array I/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Bit Manipulation
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For an integer $a,ドル the result of $a \lor (a + 1)$ is always odd. Therefore, if $\text{nums[i]}$ is even, then $\text{ans}[i]$ does not exist, and we directly return $-1$. In this problem, $\textit{nums}[i]$ is a prime number, so to check if it is even, we only need to check if it equals 2ドル$.
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If $\text{nums[i]}$ is odd, suppose $\text{nums[i]} = \text{0b1101101}$. Since $a \lor (a + 1) = \text{nums[i]},ドル this is equivalent to changing the last 0ドル$ bit of $a$ to 1ドル$. To solve for $a,ドル we need to change the bit after the last 0ドル$ in $\text{nums[i]}$ to 0ドル$. We start traversing from the least significant bit (index 1ドル$) and find the first 0ドル$ bit. If it is at position $i,ドル we change the $(i - 1)$-th bit of $\text{nums[i]}$ to 1ドル,ドル i.e., $\text{ans}[i] = \text{nums[i]} \oplus 2^{i - 1}$.
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By traversing all elements in $\text{nums},ドル we can obtain the answer.
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The time complexity is $O(n \times \log M),ドル where $n$ and $M$ are the length of the array $\text{nums}$ and the maximum value in the array, respectively. Ignoring the space consumption of the answer array, the space complexity is $O(1)$.
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def minBitwiseArray(self, nums: List[int]) -> List[int]:
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ans = []
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for x in nums:
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if x == 2:
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ans.append(-1)
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else:
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for i in range(1, 32):
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if x >> i & 1 ^ 1:
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ans.append(x ^ 1 << (i - 1))
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break
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return ans
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```
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#### Java
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```java
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class Solution {
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public int[] minBitwiseArray(List<Integer> nums) {
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int n = nums.size();
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int[] ans = new int[n];
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for (int i = 0; i < n; ++i) {
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int x = nums.get(i);
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if (x == 2) {
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ans[i] = -1;
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} else {
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for (int j = 1; j < 32; ++j) {
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if ((x >> j & 1) == 0) {
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ans[i] = x ^ 1 << (j - 1);
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break;
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}
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}
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}
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}
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return ans;
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}
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}
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```
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#### C++
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```cpp
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class Solution {
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public:
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vector<int> minBitwiseArray(vector<int>& nums) {
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vector<int> ans;
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for (int x : nums) {
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if (x == 2) {
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ans.push_back(-1);
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} else {
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for (int i = 1; i < 32; ++i) {
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if (x >> i & 1 ^ 1) {
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ans.push_back(x ^ 1 << (i - 1));
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break;
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}
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}
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}
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}
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return ans;
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}
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};
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```
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#### Go
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```go
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func minBitwiseArray(nums []int) (ans []int) {
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for _, x := range nums {
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if x == 2 {
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ans = append(ans, -1)
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} else {
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for i := 1; i < 32; i++ {
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if x>>i&1 == 0 {
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ans = append(ans, x^1<<(i-1))
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break
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}
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}
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}
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}
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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 minBitwiseArray(nums: number[]): number[] {
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const ans: number[] = [];
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for (const x of nums) {
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if (x === 2) {
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ans.push(-1);
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} else {
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for (let i = 1; i < 32; ++i) {
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if (((x >> i) & 1) ^ 1) {
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ans.push(x ^ (1 << (i - 1)));
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break;
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}
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}
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}
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}
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return ans;
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}
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```
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<!-- tabs:end -->
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class Solution {
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public:
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vector<int> minBitwiseArray(vector<int>& nums) {
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vector<int> ans;
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for (int x : nums) {
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if (x == 2) {
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ans.push_back(-1);
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} else {
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for (int i = 1; i < 32; ++i) {
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if (x >> i & 1 ^ 1) {
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ans.push_back(x ^ 1 << (i - 1));
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break;
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}
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}
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}
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}
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return ans;
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}
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};
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func minBitwiseArray(nums []int) (ans []int) {
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for _, x := range nums {
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if x == 2 {
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ans = append(ans, -1)
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} else {
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for i := 1; i < 32; i++ {
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if x>>i&1 == 0 {
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ans = append(ans, x^1<<(i-1))
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break
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}
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}
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}
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}
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return
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}
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class Solution {
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public int[] minBitwiseArray(List<Integer> nums) {
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int n = nums.size();
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int[] ans = new int[n];
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for (int i = 0; i < n; ++i) {
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int x = nums.get(i);
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if (x == 2) {
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ans[i] = -1;
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} else {
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for (int j = 1; j < 32; ++j) {
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if ((x >> j & 1) == 0) {
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ans[i] = x ^ 1 << (j - 1);
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break;
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}
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}
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}
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}
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return ans;
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}
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}
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class Solution:
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def minBitwiseArray(self, nums: List[int]) -> List[int]:
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ans = []
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for x in nums:
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if x == 2:
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ans.append(-1)
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else:
8+
for i in range(1, 32):
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if x >> i & 1 ^ 1:
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ans.append(x ^ 1 << (i - 1))
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break
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return ans
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function minBitwiseArray(nums: number[]): number[] {
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const ans: number[] = [];
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for (const x of nums) {
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if (x === 2) {
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ans.push(-1);
6+
} else {
7+
for (let i = 1; i < 32; ++i) {
8+
if (((x >> i) & 1) ^ 1) {
9+
ans.push(x ^ (1 << (i - 1)));
10+
break;
11+
}
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}
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}
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}
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return ans;
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}

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