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yanglbme merged 1 commit into from
Dec 7, 2025
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feat: add weekly contest 479 #4885

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68 changes: 34 additions & 34 deletions solution/3700-3799/3766.Minimum Operations to Make Binary Palindrome/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -157,9 +157,9 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3766.Mi

### 方法一：预处理 + 二分查找

我们注意到，题目中给定的数字范围仅为 $[1, 5000]$，因此，我们直接预处理 $[0, 2^{14})$ 范围内的所有二进制回文数，并将其存储在一个数组中，记为 $\textit{primes}$。
我们注意到，题目中给定的数字范围仅为 $[1, 5000]$，因此，我们直接预处理 $[0, 2^{14})$ 范围内的所有二进制回文数，并将其存储在一个数组中，记为 $\textit{p}$。

接下来，对于每个数字 $x$，我们使用二分查找在数组 $\textit{primes}$ 中找到第一个大于等于 $x$ 的回文数 $\textit{primes}[i]$，以及第一个小于 $x$ 的回文数 $\textit{primes}[i - 1]$。然后，我们计算将 $x$ 转换为这两个回文数所需的操作次数，并取其中的最小值作为答案。
接下来，对于每个数字 $x$，我们使用二分查找在数组 $\textit{p}$ 中找到第一个大于等于 $x$ 的回文数 $\textit{p}[i]$，以及第一个小于 $x$ 的回文数 $\textit{p}[i - 1]$。然后，我们计算将 $x$ 转换为这两个回文数所需的操作次数，并取其中的最小值作为答案。

时间复杂度 $O(n \times \log M)$，空间复杂度 $O(M)$。其中 $n$ 是数组 $\textit{nums}$ 的长度，而 $M$ 是预处理的二进制回文数的数量。

Expand All @@ -168,23 +168,23 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3766.Mi
#### Python3

```python
primes = []
p = []
for i in range(1 << 14):
s = bin(i)[2:]
if s == s[::-1]:
primes.append(i)
p.append(i)


class Solution:
def minOperations(self, nums: List[int]) -> List[int]:
ans = []
for x in nums:
i = bisect_left(primes, x)
i = bisect_left(p, x)
times = inf
if i < len(primes):
times = min(times, primes[i] - x)
if i < len(p):
times = min(times, p[i] - x)
if i >= 1:
times = min(times, x - primes[i - 1])
times = min(times, x - p[i - 1])
ans.append(times)
return ans
```
Expand All @@ -193,15 +193,15 @@ class Solution:

```java
class Solution {
private static final List<Integer> primes = new ArrayList<>();
private static final List<Integer> p = new ArrayList<>();

static {
int N = 1 << 14;
for (int i = 0; i < N; i++) {
String s = Integer.toBinaryString(i);
String rs = new StringBuilder(s).reverse().toString();
if (s.equals(rs)) {
primes.add(i);
p.add(i);
}
}
}
Expand All @@ -212,23 +212,23 @@ class Solution {
Arrays.fill(ans, Integer.MAX_VALUE);
for (int k = 0; k < n; ++k) {
int x = nums[k];
int i = binarySearch(primes, x);
if (i < primes.size()) {
ans[k] = Math.min(ans[k], primes.get(i) - x);
int i = binarySearch(p, x);
if (i < p.size()) {
ans[k] = Math.min(ans[k], p.get(i) - x);
}
if (i >= 1) {
ans[k] = Math.min(ans[k], x - primes.get(i - 1));
ans[k] = Math.min(ans[k], x - p.get(i - 1));
}
}

return ans;
}

private int binarySearch(List<Integer> primes, int x) {
int l = 0, r = primes.size();
private int binarySearch(List<Integer> p, int x) {
int l = 0, r = p.size();
while (l < r) {
int mid = (l + r) >>> 1;
if (primes.get(mid) >= x) {
if (p.get(mid) >= x) {
r = mid;
} else {
l = mid + 1;
Expand All @@ -242,7 +242,7 @@ class Solution {
#### C++

```cpp
vector<int> primes;
vector<int> p;

auto init = [] {
int N = 1 << 14;
Expand All @@ -252,7 +252,7 @@ auto init = [] {
string rs = s;
reverse(rs.begin(), rs.end());
if (s == rs) {
primes.push_back(i);
p.push_back(i);
}
}
return 0;
Expand All @@ -265,12 +265,12 @@ public:
vector<int> ans(n, INT_MAX);
for (int k = 0; k < n; ++k) {
int x = nums[k];
int i = lower_bound(primes.begin(), primes.end(), x) - primes.begin();
if (i < (int) primes.size()) {
ans[k] = min(ans[k], primes[i] - x);
int i = lower_bound(p.begin(), p.end(), x) - p.begin();
if (i < (int) p.size()) {
ans[k] = min(ans[k], p[i] - x);
}
if (i >= 1) {
ans[k] = min(ans[k], x - primes[i - 1]);
ans[k] = min(ans[k], x - p[i - 1]);
}
}
return ans;
Expand All @@ -281,14 +281,14 @@ public:
#### Go

```go
var primes []int
var p []int

func init() {
N := 1 << 14
for i := 0; i < N; i++ {
s := strconv.FormatInt(int64(i), 2)
if isPalindrome(s) {
primes = append(primes, i)
p = append(p, i)
}
}
}
Expand All @@ -306,13 +306,13 @@ func isPalindrome(s string) bool {
func minOperations(nums []int) []int {
ans := make([]int, len(nums))
for k, x := range nums {
i := sort.SearchInts(primes, x)
i := sort.SearchInts(p, x)
t := math.MaxInt32
if i < len(primes) {
t = primes[i] - x
if i < len(p) {
t = p[i] - x
}
if i >= 1 {
t = min(t, x-primes[i-1])
t = min(t, x-p[i-1])
}
ans[k] = t
}
Expand All @@ -323,7 +323,7 @@ func minOperations(nums []int) []int {
#### TypeScript

```ts
const primes: number[] = (() => {
const p: number[] = (() => {
const res: number[] = [];
const N = 1 << 14;
for (let i = 0; i < N; i++) {
Expand All @@ -340,12 +340,12 @@ function minOperations(nums: number[]): number[] {

for (let k = 0; k < nums.length; k++) {
const x = nums[k];
const i = _.sortedIndex(primes, x);
if (i < primes.length) {
ans[k] = primes[i] - x;
const i = _.sortedIndex(p, x);
if (i < p.length) {
ans[k] = p[i] - x;
}
if (i >= 1) {
ans[k] = Math.min(ans[k], x - primes[i - 1]);
ans[k] = Math.min(ans[k], x - p[i - 1]);
}
}

Expand Down
68 changes: 34 additions & 34 deletions solution/3700-3799/3766.Minimum Operations to Make Binary Palindrome/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -155,9 +155,9 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3766.Mi

### Solution 1: Preprocessing + Binary Search

We observe that the range of numbers given in the problem is only $[1, 5000]$. Therefore, we directly preprocess all binary palindromic numbers in the range $[0, 2^{14})$ and store them in an array, denoted as $\textit{primes}$.
We observe that the range of numbers given in the problem is only $[1, 5000]$. Therefore, we directly preprocess all binary palindromic numbers in the range $[0, 2^{14})$ and store them in an array, denoted as $\textit{p}$.

Next, for each number $x$, we use binary search to find the first palindromic number greater than or equal to $x$ in the array $\textit{primes}$, denoted as $\textit{primes}[i]$, as well as the first palindromic number less than $x$, denoted as $\textit{primes}[i - 1]$. Then, we calculate the number of operations required to convert $x$ to these two palindromic numbers and take the minimum value as the answer.
Next, for each number $x$, we use binary search to find the first palindromic number greater than or equal to $x$ in the array $\textit{p}$, denoted as $\textit{p}[i]$, as well as the first palindromic number less than $x$, denoted as $\textit{p}[i - 1]$. Then, we calculate the number of operations required to convert $x$ to these two palindromic numbers and take the minimum value as the answer.

The time complexity is $O(n \times \log M)$, and the space complexity is $O(M)$. Where $n$ is the length of the array $\textit{nums}$, and $M$ is the number of preprocessed binary palindromic numbers.

Expand All @@ -166,23 +166,23 @@ The time complexity is $O(n \times \log M)$, and the space complexity is $O(M)$.
#### Python3

```python
primes = []
p = []
for i in range(1 << 14):
s = bin(i)[2:]
if s == s[::-1]:
primes.append(i)
p.append(i)


class Solution:
def minOperations(self, nums: List[int]) -> List[int]:
ans = []
for x in nums:
i = bisect_left(primes, x)
i = bisect_left(p, x)
times = inf
if i < len(primes):
times = min(times, primes[i] - x)
if i < len(p):
times = min(times, p[i] - x)
if i >= 1:
times = min(times, x - primes[i - 1])
times = min(times, x - p[i - 1])
ans.append(times)
return ans
```
Expand All @@ -191,15 +191,15 @@ class Solution:

```java
class Solution {
private static final List<Integer> primes = new ArrayList<>();
private static final List<Integer> p = new ArrayList<>();

static {
int N = 1 << 14;
for (int i = 0; i < N; i++) {
String s = Integer.toBinaryString(i);
String rs = new StringBuilder(s).reverse().toString();
if (s.equals(rs)) {
primes.add(i);
p.add(i);
}
}
}
Expand All @@ -210,23 +210,23 @@ class Solution {
Arrays.fill(ans, Integer.MAX_VALUE);
for (int k = 0; k < n; ++k) {
int x = nums[k];
int i = binarySearch(primes, x);
if (i < primes.size()) {
ans[k] = Math.min(ans[k], primes.get(i) - x);
int i = binarySearch(p, x);
if (i < p.size()) {
ans[k] = Math.min(ans[k], p.get(i) - x);
}
if (i >= 1) {
ans[k] = Math.min(ans[k], x - primes.get(i - 1));
ans[k] = Math.min(ans[k], x - p.get(i - 1));
}
}

return ans;
}

private int binarySearch(List<Integer> primes, int x) {
int l = 0, r = primes.size();
private int binarySearch(List<Integer> p, int x) {
int l = 0, r = p.size();
while (l < r) {
int mid = (l + r) >>> 1;
if (primes.get(mid) >= x) {
if (p.get(mid) >= x) {
r = mid;
} else {
l = mid + 1;
Expand All @@ -240,7 +240,7 @@ class Solution {
#### C++

```cpp
vector<int> primes;
vector<int> p;

auto init = [] {
int N = 1 << 14;
Expand All @@ -250,7 +250,7 @@ auto init = [] {
string rs = s;
reverse(rs.begin(), rs.end());
if (s == rs) {
primes.push_back(i);
p.push_back(i);
}
}
return 0;
Expand All @@ -263,12 +263,12 @@ public:
vector<int> ans(n, INT_MAX);
for (int k = 0; k < n; ++k) {
int x = nums[k];
int i = lower_bound(primes.begin(), primes.end(), x) - primes.begin();
if (i < (int) primes.size()) {
ans[k] = min(ans[k], primes[i] - x);
int i = lower_bound(p.begin(), p.end(), x) - p.begin();
if (i < (int) p.size()) {
ans[k] = min(ans[k], p[i] - x);
}
if (i >= 1) {
ans[k] = min(ans[k], x - primes[i - 1]);
ans[k] = min(ans[k], x - p[i - 1]);
}
}
return ans;
Expand All @@ -279,14 +279,14 @@ public:
#### Go

```go
var primes []int
var p []int

func init() {
N := 1 << 14
for i := 0; i < N; i++ {
s := strconv.FormatInt(int64(i), 2)
if isPalindrome(s) {
primes = append(primes, i)
p = append(p, i)
}
}
}
Expand All @@ -304,13 +304,13 @@ func isPalindrome(s string) bool {
func minOperations(nums []int) []int {
ans := make([]int, len(nums))
for k, x := range nums {
i := sort.SearchInts(primes, x)
i := sort.SearchInts(p, x)
t := math.MaxInt32
if i < len(primes) {
t = primes[i] - x
if i < len(p) {
t = p[i] - x
}
if i >= 1 {
t = min(t, x-primes[i-1])
t = min(t, x-p[i-1])
}
ans[k] = t
}
Expand All @@ -321,7 +321,7 @@ func minOperations(nums []int) []int {
#### TypeScript

```ts
const primes: number[] = (() => {
const p: number[] = (() => {
const res: number[] = [];
const N = 1 << 14;
for (let i = 0; i < N; i++) {
Expand All @@ -338,12 +338,12 @@ function minOperations(nums: number[]): number[] {

for (let k = 0; k < nums.length; k++) {
const x = nums[k];
const i = _.sortedIndex(primes, x);
if (i < primes.length) {
ans[k] = primes[i] - x;
const i = _.sortedIndex(p, x);
if (i < p.length) {
ans[k] = p[i] - x;
}
if (i >= 1) {
ans[k] = Math.min(ans[k], x - primes[i - 1]);
ans[k] = Math.min(ans[k], x - p[i - 1]);
}
}

Expand Down
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