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feat: add solutions to lc problems: No.3151~3153 (doocs#2838)
* No.3151.Special Array I * No.3152.Special Array II * No.3153.Sum of Digit Differences of All Pairs
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‎solution/3100-3199/3151.Special Array I/README.md

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@@ -71,32 +71,77 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3151.Sp
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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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我们从左到右遍历数组，对于每一对相邻元素，如果它们的奇偶性相同，那么数组就不是特殊数组，返回 `false`；否则，数组是特殊数组，返回 `true`。
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时间复杂度 $O(n)$，其中 $n$ 是数组的长度。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 isArraySpecial(self, nums: List[int]) -> bool:
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return all(a % 2 != b % 2 for a, b in pairwise(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 boolean isArraySpecial(int[] nums) {
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for (int i = 1; i < nums.length; ++i) {
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if (nums[i] % 2 == nums[i - 1] % 2) {
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return false;
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}
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}
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return true;
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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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bool isArraySpecial(vector<int>& nums) {
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for (int i = 1; i < nums.size(); ++i) {
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if (nums[i] % 2 == nums[i - 1] % 2) {
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return false;
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}
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}
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return true;
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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 isArraySpecial(nums []int) bool {
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for i, x := range nums[1:] {
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if x%2 == nums[i]%2 {
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return false
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}
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}
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return true
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}
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```
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#### TypeScript
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```ts
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function isArraySpecial(nums: number[]): boolean {
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for (let i = 1; i < nums.length; ++i) {
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if (nums[i] % 2 === nums[i - 1] % 2) {
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return false;
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}
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}
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return true;
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}
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```
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<!-- tabs:end -->

‎solution/3100-3199/3151.Special Array I/README_EN.md

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@@ -69,32 +69,77 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3151.Sp
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Single Pass
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We traverse the array from left to right. For each pair of adjacent elements, if their parity is the same, then the array is not a special array, return `false`; otherwise, the array is a special array, return `true`.
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The time complexity is $O(n)$, where $n$ is the length of the 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 isArraySpecial(self, nums: List[int]) -> bool:
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return all(a % 2 != b % 2 for a, b in pairwise(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 boolean isArraySpecial(int[] nums) {
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for (int i = 1; i < nums.length; ++i) {
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if (nums[i] % 2 == nums[i - 1] % 2) {
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return false;
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}
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}
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return true;
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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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bool isArraySpecial(vector<int>& nums) {
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for (int i = 1; i < nums.size(); ++i) {
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if (nums[i] % 2 == nums[i - 1] % 2) {
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return false;
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}
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}
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return true;
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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 isArraySpecial(nums []int) bool {
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for i, x := range nums[1:] {
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if x%2 == nums[i]%2 {
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return false
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}
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}
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return true
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}
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```
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#### TypeScript
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```ts
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function isArraySpecial(nums: number[]): boolean {
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for (let i = 1; i < nums.length; ++i) {
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if (nums[i] % 2 === nums[i - 1] % 2) {
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return false;
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}
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}
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return true;
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}
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```
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<!-- tabs:end -->

‎solution/3100-3199/3151.Special Array I/Solution.cpp

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class Solution {
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public:
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bool isArraySpecial(vector<int>& nums) {
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for (int i = 1; i < nums.size(); ++i) {
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if (nums[i] % 2 == nums[i - 1] % 2) {
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return false;
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}
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}
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return true;
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}
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};

‎solution/3100-3199/3151.Special Array I/Solution.go

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func isArraySpecial(nums []int) bool {
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for i, x := range nums[1:] {
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if x%2 == nums[i]%2 {
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return false
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}
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}
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return true
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}

‎solution/3100-3199/3151.Special Array I/Solution.java

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class Solution {
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public boolean isArraySpecial(int[] nums) {
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for (int i = 1; i < nums.length; ++i) {
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if (nums[i] % 2 == nums[i - 1] % 2) {
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return false;
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}
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}
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return true;
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}
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}

‎solution/3100-3199/3151.Special Array I/Solution.py

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class Solution:
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def isArraySpecial(self, nums: List[int]) -> bool:
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return all(a % 2 != b % 2 for a, b in pairwise(nums))

‎solution/3100-3199/3151.Special Array I/Solution.ts

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function isArraySpecial(nums: number[]): boolean {
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for (let i = 1; i < nums.length; ++i) {
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if (nums[i] % 2 === nums[i - 1] % 2) {
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return false;
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}
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}
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return true;
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}

‎solution/3100-3199/3152.Special Array II/README.md

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@@ -67,32 +67,110 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3152.Sp
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<!-- solution:start -->
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### 方法一
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### 方法一：记录每个位置的最左特殊数组位置
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我们可以定义一个数组 $d$ 来记录每个位置的最左特殊数组位置，初始时 $d[i] = i$。然后我们从左到右遍历数组 $nums$，如果 $nums[i]$ 和 $nums[i - 1]$ 的奇偶性不同，那么 $d[i] = d[i - 1]$。
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最后我们遍历每个查询，判断 $d[to] <= from$ 是否成立即可。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是数组的长度。
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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 isArraySpecial(self, nums: List[int], queries: List[List[int]]) -> List[bool]:
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n = len(nums)
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d = list(range(n))
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for i in range(1, n):
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if nums[i] % 2 != nums[i - 1] % 2:
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d[i] = d[i - 1]
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return [d[t] <= f for f, t in queries]
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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 boolean[] isArraySpecial(int[] nums, int[][] queries) {
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int n = nums.length;
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int[] d = new int[n];
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for (int i = 1; i < n; ++i) {
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if (nums[i] % 2 != nums[i - 1] % 2) {
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d[i] = d[i - 1];
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} else {
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d[i] = i;
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}
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}
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int m = queries.length;
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boolean[] ans = new boolean[m];
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for (int i = 0; i < m; ++i) {
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ans[i] = d[queries[i][1]] <= queries[i][0];
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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<bool> isArraySpecial(vector<int>& nums, vector<vector<int>>& queries) {
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int n = nums.size();
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vector<int> d(n);
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iota(d.begin(), d.end(), 0);
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for (int i = 1; i < n; ++i) {
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if (nums[i] % 2 != nums[i - 1] % 2) {
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d[i] = d[i - 1];
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}
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}
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vector<bool> ans;
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for (auto& q : queries) {
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ans.push_back(d[q[1]] <= q[0]);
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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 isArraySpecial(nums []int, queries [][]int) (ans []bool) {
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n := len(nums)
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d := make([]int, n)
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for i := range d {
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d[i] = i
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}
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for i := 1; i < len(nums); i++ {
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if nums[i]%2 != nums[i-1]%2 {
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d[i] = d[i-1]
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}
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}
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for _, q := range queries {
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ans = append(ans, d[q[1]] <= q[0])
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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 isArraySpecial(nums: number[], queries: number[][]): boolean[] {
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const n = nums.length;
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const d: number[] = Array.from({ length: n }, (_, i) => i);
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for (let i = 1; i < n; ++i) {
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if (nums[i] % 2 !== nums[i - 1] % 2) {
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d[i] = d[i - 1];
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}
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}
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return queries.map(([from, to]) => d[to] <= from);
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}
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```
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<!-- tabs:end -->

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