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Commit 5e3d4a0

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feat: add solutions to lc problems: No.3674,3675 (#4702)
1 parent 29183b6 commit 5e3d4a0

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14 files changed

+285
-16
lines changed

14 files changed

+285
-16
lines changed

‎solution/3600-3699/3674.Minimum Operations to Equalize Array/README.md

Lines changed: 49 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -63,32 +63,77 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3674.Mi
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<!-- solution:start -->
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### 方法一
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### 方法一:一次遍历
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如果 $\textit{nums}$ 中所有元素都相等,则不需要任何操作;否则,选择整个数组作为子数组进行一次操作即可。
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时间复杂度 $O(n),ドル其中 $n$ 是数组 $\textit{nums}$ 的长度。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 minOperations(self, nums: List[int]) -> int:
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return int(any(x != nums[0] for x in nums))
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```
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#### Java
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```java
79-
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class Solution {
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public int minOperations(int[] nums) {
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for (int x : nums) {
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if (x != nums[0]) {
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return 1;
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}
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}
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return 0;
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}
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}
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```
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#### C++
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```cpp
85-
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class Solution {
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public:
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int minOperations(vector<int>& nums) {
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for (int x : nums) {
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if (x != nums[0]) {
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return 1;
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}
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}
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return 0;
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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 minOperations(nums []int) int {
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for _, x := range nums {
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if x != nums[0] {
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return 1
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}
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}
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return 0
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}
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```
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#### TypeScript
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```ts
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function minOperations(nums: number[]): number {
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for (const x of nums) {
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if (x !== nums[0]) {
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return 1;
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}
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}
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return 0;
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}
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```
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<!-- tabs:end -->

‎solution/3600-3699/3674.Minimum Operations to Equalize Array/README_EN.md

Lines changed: 49 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -59,32 +59,77 @@ A <strong>subarray</strong> is a contiguous <b>non-empty</b> sequence of element
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Single Pass
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If all elements in $\textit{nums}$ are equal, no operations are needed; otherwise, we can select the entire array as a subarray and perform one operation.
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The time complexity is $O(n),ドル where $n$ is the length of the array $\textit{nums}$. 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 minOperations(self, nums: List[int]) -> int:
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return int(any(x != nums[0] for x in nums))
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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 minOperations(int[] nums) {
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for (int x : nums) {
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if (x != nums[0]) {
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return 1;
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}
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}
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return 0;
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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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int minOperations(vector<int>& nums) {
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for (int x : nums) {
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if (x != nums[0]) {
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return 1;
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}
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}
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return 0;
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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 minOperations(nums []int) int {
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for _, x := range nums {
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if x != nums[0] {
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return 1
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}
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}
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return 0
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}
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```
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#### TypeScript
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```ts
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function minOperations(nums: number[]): number {
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for (const x of nums) {
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if (x !== nums[0]) {
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return 1;
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}
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}
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return 0;
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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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int minOperations(vector<int>& nums) {
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for (int x : nums) {
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if (x != nums[0]) {
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return 1;
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}
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}
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return 0;
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}
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};
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func minOperations(nums []int) int {
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for _, x := range nums {
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if x != nums[0] {
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return 1
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}
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}
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return 0
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}
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Original file line numberDiff line numberDiff line change
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class Solution {
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public int minOperations(int[] nums) {
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for (int x : nums) {
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if (x != nums[0]) {
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return 1;
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}
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}
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return 0;
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}
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}
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class Solution:
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def minOperations(self, nums: List[int]) -> int:
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return int(any(x != nums[0] for x in nums))
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function minOperations(nums: number[]): number {
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for (const x of nums) {
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if (x !== nums[0]) {
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return 1;
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}
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}
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return 0;
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}

‎solution/3600-3699/3675.Minimum Operations to Transform String/README.md

Lines changed: 52 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -76,32 +76,80 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3675.Mi
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<!-- solution:start -->
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### 方法一
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### 方法一:一次遍历
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根据题目描述,我们一定是先从字符 'b' 开始,依次将每个字符变为下一个字符,直到变为 'a'。因此,我们只需要统计字符串中距离 'a' 最远的字符与 'a' 的距离,即可得到答案。
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时间复杂度 $O(n),ドル其中 $n$ 是字符串 $s$ 的长度。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 minOperations(self, s: str) -> int:
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return max((26 - (ord(c) - 97) for c in s if c != "a"), default=0)
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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 minOperations(String s) {
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int ans = 0;
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for (char c : s.toCharArray()) {
102+
if (c != 'a') {
103+
ans = Math.max(ans, 26 - (c - 'a'));
104+
}
105+
}
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return ans;
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}
108+
}
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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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int minOperations(string s) {
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int ans = 0;
118+
for (char c : s) {
119+
if (c != 'a') {
120+
ans = max(ans, 26 - (c - 'a'));
121+
}
122+
}
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return ans;
124+
}
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};
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```
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#### Go
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```go
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func minOperations(s string) (ans int) {
132+
for _, c := range s {
133+
if c != 'a' {
134+
ans = max(ans, 26-int(c-'a'))
135+
}
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}
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return
138+
}
139+
```
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#### TypeScript
142+
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```ts
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function minOperations(s: string): number {
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let ans = 0;
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for (const c of s) {
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if (c !== 'a') {
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ans = Math.max(ans, 26 - (c.charCodeAt(0) - 97));
149+
}
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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/3600-3699/3675.Minimum Operations to Transform String/README_EN.md

Lines changed: 52 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -74,32 +74,80 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3675.Mi
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Single Pass
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According to the problem description, we always start from the character 'b' and successively change each character to the next one until it becomes 'a'. Therefore, we only need to find the character in the string that is farthest from 'a' and calculate its distance to 'a' to get the answer.
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The time complexity is $O(n),ドル where $n$ is the length of the string $s$. 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:
89+
def minOperations(self, s: str) -> int:
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return max((26 - (ord(c) - 97) for c in s if c != "a"), default=0)
8591
```
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#### Java
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```java
90-
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class Solution {
97+
public int minOperations(String s) {
98+
int ans = 0;
99+
for (char c : s.toCharArray()) {
100+
if (c != 'a') {
101+
ans = Math.max(ans, 26 - (c - 'a'));
102+
}
103+
}
104+
return ans;
105+
}
106+
}
91107
```
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#### C++
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```cpp
96-
112+
class Solution {
113+
public:
114+
int minOperations(string s) {
115+
int ans = 0;
116+
for (char c : s) {
117+
if (c != 'a') {
118+
ans = max(ans, 26 - (c - 'a'));
119+
}
120+
}
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return ans;
122+
}
123+
};
97124
```
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#### Go
100127
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```go
129+
func minOperations(s string) (ans int) {
130+
for _, c := range s {
131+
if c != 'a' {
132+
ans = max(ans, 26-int(c-'a'))
133+
}
134+
}
135+
return
136+
}
137+
```
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139+
#### TypeScript
140+
141+
```ts
142+
function minOperations(s: string): number {
143+
let ans = 0;
144+
for (const c of s) {
145+
if (c !== 'a') {
146+
ans = Math.max(ans, 26 - (c.charCodeAt(0) - 97));
147+
}
148+
}
149+
return ans;
150+
}
103151
```
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<!-- tabs:end -->
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Original file line numberDiff line numberDiff line change
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class Solution {
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public:
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int minOperations(string s) {
4+
int ans = 0;
5+
for (char c : s) {
6+
if (c != 'a') {
7+
ans = max(ans, 26 - (c - 'a'));
8+
}
9+
}
10+
return ans;
11+
}
12+
};

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