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feat: add solutions to lc problem: No.1786
No.1786.Number of Restricted Paths From First to Last Node
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‎solution/1700-1799/1785.Minimum Elements to Add to Form a Given Sum/README.md

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**方法一：贪心**
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先计算数组元素总和 $s$，然后计算 $s$ 与 $goal$ 的差值 $d$。
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我们先计算数组元素总和 $s$，然后计算 $s$ 与 $goal$ 的差值 $d$。
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那么需要添加的元素数量为 $d$ 的绝对值除以 $limit$ 向上取整，即 $\lceil \frac{|d|}{limit} \rceil$。
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```java
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class Solution {
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public int minElements(int[] nums, int limit, int goal) {
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// long s = Arrays.stream(nums).asLongStream().sum();
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long s = 0;
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for (int v : nums) {
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s += v;

‎solution/1700-1799/1785.Minimum Elements to Add to Form a Given Sum/README_EN.md

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```java
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class Solution {
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public int minElements(int[] nums, int limit, int goal) {
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// long s = Arrays.stream(nums).asLongStream().sum();
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long s = 0;
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for (int v : nums) {
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s += v;

‎solution/1700-1799/1785.Minimum Elements to Add to Form a Given Sum/Solution.java

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class Solution {
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public int minElements(int[] nums, int limit, int goal) {
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// long s = Arrays.stream(nums).asLongStream().sum();
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long s = 0;
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for (int v : nums) {
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s += v;

‎solution/1700-1799/1786.Number of Restricted Paths From First to Last Node/README.md

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return dfs(1)
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```
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```python
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class Solution:
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def countRestrictedPaths(self, n: int, edges: List[List[int]]) -> int:
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g = defaultdict(list)
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for u, v, w in edges:
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g[u].append((v, w))
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g[v].append((u, w))
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dist = [inf] * (n + 1)
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dist[n] = 0
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q = [(0, n)]
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mod = 10**9 + 7
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while q:
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_, u = heappop(q)
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for v, w in g[u]:
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if dist[v] > dist[u] + w:
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dist[v] = dist[u] + w
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heappush(q, (dist[v], v))
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arr = list(range(1, n + 1))
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arr.sort(key=lambda i: dist[i])
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f = [0] * (n + 1)
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f[n] = 1
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for i in arr:
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for j, _ in g[i]:
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if dist[i] > dist[j]:
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f[i] = (f[i] + f[j]) % mod
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return f[1]
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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}
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```
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```java
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class Solution {
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private static final int INF = Integer.MAX_VALUE;
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private static final int MOD = (int) 1e9 + 7;
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public int countRestrictedPaths(int n, int[][] edges) {
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List<int[]>[] g = new List[n + 1];
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Arrays.setAll(g, k -> new ArrayList<>());
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for (int[] e : edges) {
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int u = e[0], v = e[1], w = e[2];
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g[u].add(new int[] {v, w});
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g[v].add(new int[] {u, w});
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}
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PriorityQueue<int[]> q = new PriorityQueue<>((a, b) -> a[0] - b[0]);
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q.offer(new int[] {0, n});
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int[] dist = new int[n + 1];
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Arrays.fill(dist, INF);
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dist[n] = 0;
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while (!q.isEmpty()) {
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int[] p = q.poll();
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int u = p[1];
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for (int[] ne : g[u]) {
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int v = ne[0], w = ne[1];
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if (dist[v] > dist[u] + w) {
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dist[v] = dist[u] + w;
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q.offer(new int[] {dist[v], v});
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}
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}
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}
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int[] f = new int[n + 1];
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f[n] = 1;
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Integer[] arr = new Integer[n];
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for (int i = 0; i < n; ++i) {
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arr[i] = i + 1;
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}
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Arrays.sort(arr, (i, j) -> dist[i] - dist[j]);
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for (int i : arr) {
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for (int[] ne : g[i]) {
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int j = ne[0];
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if (dist[i] > dist[j]) {
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f[i] = (f[i] + f[j]) % MOD;
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}
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}
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}
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return f[1];
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}
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}
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```
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### **C++**
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```cpp

‎solution/1700-1799/1786.Number of Restricted Paths From First to Last Node/README_EN.md

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return dfs(1)
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```
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```python
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class Solution:
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def countRestrictedPaths(self, n: int, edges: List[List[int]]) -> int:
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g = defaultdict(list)
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for u, v, w in edges:
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g[u].append((v, w))
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g[v].append((u, w))
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dist = [inf] * (n + 1)
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dist[n] = 0
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q = [(0, n)]
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mod = 10**9 + 7
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while q:
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_, u = heappop(q)
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for v, w in g[u]:
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if dist[v] > dist[u] + w:
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dist[v] = dist[u] + w
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heappush(q, (dist[v], v))
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arr = list(range(1, n + 1))
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arr.sort(key=lambda i: dist[i])
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f = [0] * (n + 1)
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f[n] = 1
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for i in arr:
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for j, _ in g[i]:
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if dist[i] > dist[j]:
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f[i] = (f[i] + f[j]) % mod
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return f[1]
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```
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### **Java**
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```java
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}
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```
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```java
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class Solution {
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private static final int INF = Integer.MAX_VALUE;
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private static final int MOD = (int) 1e9 + 7;
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public int countRestrictedPaths(int n, int[][] edges) {
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List<int[]>[] g = new List[n + 1];
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Arrays.setAll(g, k -> new ArrayList<>());
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for (int[] e : edges) {
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int u = e[0], v = e[1], w = e[2];
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g[u].add(new int[] {v, w});
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g[v].add(new int[] {u, w});
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}
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PriorityQueue<int[]> q = new PriorityQueue<>((a, b) -> a[0] - b[0]);
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q.offer(new int[] {0, n});
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int[] dist = new int[n + 1];
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Arrays.fill(dist, INF);
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dist[n] = 0;
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while (!q.isEmpty()) {
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int[] p = q.poll();
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int u = p[1];
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for (int[] ne : g[u]) {
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int v = ne[0], w = ne[1];
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if (dist[v] > dist[u] + w) {
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dist[v] = dist[u] + w;
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q.offer(new int[] {dist[v], v});
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}
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}
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}
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int[] f = new int[n + 1];
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f[n] = 1;
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Integer[] arr = new Integer[n];
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for (int i = 0; i < n; ++i) {
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arr[i] = i + 1;
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}
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Arrays.sort(arr, (i, j) -> dist[i] - dist[j]);
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for (int i : arr) {
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for (int[] ne : g[i]) {
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int j = ne[0];
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if (dist[i] > dist[j]) {
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f[i] = (f[i] + f[j]) % MOD;
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}
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}
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}
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return f[1];
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}
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}
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```
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### **C++**
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```cpp

‎solution/1800-1899/1814.Count Nice Pairs in an Array/README.md

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**方法一：式子变换 + 哈希表**
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对于下标对 $(i, j)$，如果满足条件，那么有 $nums[i] + rev(nums[j]) = nums[j] + rev(nums[i])$，即 $nums[i] - rev(nums[i]) = nums[j] - rev(nums[j])$。移项后，有 $nums[i] - nums[j] = rev(nums[j]) - rev(nums[i])$。因此，我们可以将 $nums[i] - rev(nums[i])$ 作为哈希表的键，$nums[i] - nums[j]$ 作为哈希表的值，统计每个键出现的次数，然后计算每个键对应的值的组合数，最后将所有组合数相加即可。
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对于下标对 $(i, j)$，如果满足条件，那么有 $nums[i] + rev(nums[j]) = nums[j] + rev(nums[i])$，即 $nums[i] - rev(nums[i]) = nums[j] - rev(nums[j])$。移项后，有 $nums[i] - nums[j] = rev(nums[j]) - rev(nums[i])$。因此，我们可以将 $nums[i] - rev(nums[i])$ 作为哈希表的键，将 $nums[i] - nums[j]$ 作为哈希表的值，统计每个键出现的次数，然后计算每个键对应的值的组合数，最后将所有组合数相加即可。
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时间复杂度 $O(n \times \log M)$，其中 $n$ 和 $M$ 分别是数组 `nums` 的长度和数组 `nums` 中的最大值。空间复杂度 $O(n)$。
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‎solution/2100-2199/2179.Count Good Triplets in an Array/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：树状数组或线段树
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对于本题，我们先用 pos 记录每个数在 nums2 中的位置，然后依次对 nums1 中的每个元素进行处理。
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考虑**以当前数字作为三元组中间数字**的好三元组的数目。第一个数字需要是之前已经遍历过的，并且在 nums2 中的位置比当前数字更靠前的；第三个数字需要是当前还没有遍历过的，并且在 nums2 中的位置比当前数字更靠后的。

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