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Commit b91312c

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feat: add solutions to lc problem: No.3696 (doocs#4759)
1 parent 2d95ff9 commit b91312c

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7 files changed

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‎solution/3600-3699/3696.Maximum Distance Between Unequal Words in Array I/README.md‎

Lines changed: 78 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -76,32 +76,106 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3696.Ma
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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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我们可以发现,最大距离的两个单词中至少有一个单词在数组的两端(即下标为 0ドル$ 或 $n - 1$)。否则,假设最大距离的两个单词分别在下标 $i$ 和 $j$ 处,即 0ドル < i < j < n - 1,ドル那么单词 $\textit{words}[0]$ 和 $\textit{words}[j]$ 相同,而单词 $\textit{words}[n - 1]$ 和 $\textit{words}[i]$ 也相同(否则距离会更大),因此单词 $\textit{words}[0]$ 和 $\textit{words}[n - 1]$ 不同,且它们的距离 $n - 1 - 0 + 1 = n$ 一定大于 $j - i + 1,ドル与假设矛盾。因此,最大距离的两个单词中至少有一个单词在数组的两端。
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所以,我们只需要遍历数组,计算每个单词与数组两端单词的距离,并更新最大距离。
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时间复杂度 $O(n),ドル其中 $n$ 是数组 $\textit{words}$ 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 maxDistance(self, words: List[str]) -> int:
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n = len(words)
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ans = 0
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for i in range(n):
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if words[i] != words[0]:
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ans = max(ans, i + 1)
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if words[i] != words[-1]:
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ans = max(ans, n - i)
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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 maxDistance(String[] words) {
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int n = words.length;
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int ans = 0;
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for (int i = 0; i < n; ++i) {
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if (!words[i].equals(words[0])) {
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ans = Math.max(ans, i + 1);
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}
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if (!words[i].equals(words[n - 1])) {
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ans = Math.max(ans, n - i);
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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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int maxDistance(vector<string>& words) {
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int n = words.size();
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int ans = 0;
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for (int i = 0; i < n; ++i) {
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if (words[i] != words[0]) {
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ans = max(ans, i + 1);
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}
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if (words[i] != words[n - 1]) {
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ans = max(ans, n - i);
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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 maxDistance(words []string) int {
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n := len(words)
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ans := 0
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for i := 0; i < n; i++ {
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if words[i] != words[0] {
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ans = max(ans, i+1)
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}
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if words[i] != words[n-1] {
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ans = max(ans, n-i)
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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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#### TypeScript
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```ts
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function maxDistance(words: string[]): number {
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const n = words.length;
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let ans = 0;
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for (let i = 0; i < n; i++) {
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if (words[i] !== words[0]) {
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ans = Math.max(ans, i + 1);
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}
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if (words[i] !== words[n - 1]) {
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ans = Math.max(ans, n - i);
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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/3600-3699/3696.Maximum Distance Between Unequal Words in Array I/README_EN.md‎

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Original file line numberDiff line numberDiff line change
@@ -76,32 +76,106 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3600-3699/3696.Ma
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<!-- solution:start -->
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### Solution 1
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### Solution: Single Pass
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We can observe that at least one of the two words with maximum distance must be at either end of the array (i.e., at index 0ドル$ or $n - 1$). Otherwise, suppose the two words with maximum distance are at indices $i$ and $j$ where 0ドル < i < j < n - 1$. Then $\textit{words}[0]$ must be the same as $\textit{words}[j],ドル and $\textit{words}[n - 1]$ must be the same as $\textit{words}[i]$ (otherwise the distance would be greater). This means $\textit{words}[0]$ and $\textit{words}[n - 1]$ are different, and their distance $n - 1 - 0 + 1 = n$ is definitely greater than $j - i + 1,ドル which contradicts our assumption. Therefore, at least one of the two words with maximum distance must be at either end of the array.
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So, we only need to traverse the array, calculate the distance between each word and the words at both ends of the array, and update the maximum distance.
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The time complexity is $O(n),ドル where $n$ is the length of array $\textit{words}$. 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 maxDistance(self, words: List[str]) -> int:
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n = len(words)
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ans = 0
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for i in range(n):
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if words[i] != words[0]:
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ans = max(ans, i + 1)
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if words[i] != words[-1]:
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ans = max(ans, n - i)
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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 maxDistance(String[] words) {
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int n = words.length;
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int ans = 0;
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for (int i = 0; i < n; ++i) {
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if (!words[i].equals(words[0])) {
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ans = Math.max(ans, i + 1);
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}
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if (!words[i].equals(words[n - 1])) {
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ans = Math.max(ans, n - i);
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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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int maxDistance(vector<string>& words) {
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int n = words.size();
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int ans = 0;
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for (int i = 0; i < n; ++i) {
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if (words[i] != words[0]) {
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ans = max(ans, i + 1);
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}
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if (words[i] != words[n - 1]) {
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ans = max(ans, n - i);
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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 maxDistance(words []string) int {
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n := len(words)
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ans := 0
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for i := 0; i < n; i++ {
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if words[i] != words[0] {
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ans = max(ans, i+1)
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}
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if words[i] != words[n-1] {
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ans = max(ans, n-i)
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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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#### TypeScript
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```ts
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function maxDistance(words: string[]): number {
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const n = words.length;
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let ans = 0;
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for (let i = 0; i < n; i++) {
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if (words[i] !== words[0]) {
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ans = Math.max(ans, i + 1);
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}
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if (words[i] !== words[n - 1]) {
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ans = Math.max(ans, n - i);
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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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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,16 @@
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class Solution {
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public:
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int maxDistance(vector<string>& words) {
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int n = words.size();
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int ans = 0;
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for (int i = 0; i < n; ++i) {
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if (words[i] != words[0]) {
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ans = max(ans, i + 1);
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}
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if (words[i] != words[n - 1]) {
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ans = max(ans, n - i);
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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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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,13 @@
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func maxDistance(words []string) int {
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n := len(words)
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ans := 0
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for i := 0; i < n; i++ {
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if words[i] != words[0] {
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ans = max(ans, i+1)
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}
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if words[i] != words[n-1] {
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ans = max(ans, n-i)
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}
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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
@@ -0,0 +1,15 @@
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class Solution {
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public int maxDistance(String[] words) {
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int n = words.length;
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int ans = 0;
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for (int i = 0; i < n; ++i) {
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if (!words[i].equals(words[0])) {
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ans = Math.max(ans, i + 1);
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}
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if (!words[i].equals(words[n - 1])) {
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ans = Math.max(ans, n - i);
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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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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,10 @@
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class Solution:
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def maxDistance(self, words: List[str]) -> int:
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n = len(words)
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ans = 0
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for i in range(n):
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if words[i] != words[0]:
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ans = max(ans, i + 1)
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if words[i] != words[-1]:
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ans = max(ans, n - i)
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return ans
Lines changed: 13 additions & 0 deletions
Original file line numberDiff line numberDiff line change
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function maxDistance(words: string[]): number {
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const n = words.length;
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let ans = 0;
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for (let i = 0; i < n; i++) {
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if (words[i] !== words[0]) {
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ans = Math.max(ans, i + 1);
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}
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if (words[i] !== words[n - 1]) {
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ans = Math.max(ans, n - i);
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}
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}
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return ans;
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}

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