2023-08-08: Give you a tree of n nodes (connected undirected and acyclic graph)
Nodes are numbered from 0 to n - 1 and have exactly n - 1 edges
You are given an integer array vals of length n indexed from 0
Represent the value of each node separately
At the same time give you a two-dimensional integer array edges
Where edges[i] = [ai, bi] means there is an undirected edge between nodes ai and bi
A good path needs to meet the following conditions:
The start node and end node have the same value.
All node values between the start node and the end node are less than or equal to the value of the start node.
(That is to say, the value of the start node should be the maximum value of all nodes on the path).
Please return the number of different good paths.
Note that a path and its inverse path count as the same path.
For example, 0 -> 1 is considered the same path as 1 -> 0. A single node is also considered a legal path.
Input: vals = [1,1,2,2,3], edges = [[0,1],[1,2],[2,3],[2,4]].
Output: 7.
from Google.
from Zuo Shen
Answer 2023-08-08:
The general steps are as follows:
1. Create a graph (tree) data structure, and initialize the values and connections of nodes.
2. Sort the values of the nodes, and process the nodes in order of value.
3. Initialize and check the set, which is used to manage the connectivity of nodes.
4. Create an array to record the index of the node with the largest value in each connected component.
5. Create an array to record the number of nodes in the connected component where the node with the largest value in each connected component is located.
6. Initialize the answer as the total number of nodes.
7. Traverse the sorted list of nodes and process each node in turn:
7.1. Get the index and value of the current node.
7.2. Find the connected component representative node of the current node.
7.3. Find the index of the maximum node representing the node of the current connected component.
7.4. Traverse the neighbor nodes of the current node, and compare the value of the neighbor node with the value of the current node.
7.5. If the value of the neighbor node is less than or equal to the value of the current node, and the connected component where the neighbor node is located is different from the current connected component, then perform the following operations:
7.5.1. Find the index of the maximum value node of the representative node of the connected component of the neighbor node.
7.5.2. If the value of the neighbor node is equal to the value of the maximum value node, then update the answer and accumulate the number of connected components where the maximum value node is located.
7.5.3. Merge the connected components of the current node and neighbor nodes.
7.5.4. Update the index of the node representing the current connected component.
8. Return the answer.
The time complexity is O(nlogn).
The space complexity is O(n).
The complete code of go is as follows:
package main
import (
"fmt"
"sort"
)
func numberOfGoodPaths(vals []int, edges [][]int) int {
n := len(vals)
graph := make([][]int, n)
for i := 0; i < n; i++ {
graph[i] = make([]int, 0)
}
for _, e := range edges {
graph[e[0]] = append(graph[e[0]], e[1])
graph[e[1]] = append(graph[e[1]], e[0])
}
nodes := make([][]int, n)
for i := 0; i < n; i++ {
nodes[i] = make([]int, 2)
nodes[i][0] = i
nodes[i][1] = vals[i]
}
sort.Slice(nodes, func(i, j int) bool {
return nodes[i][1] < nodes[j][1]
})
uf := NewUnionFind(n)
maxIndex := make([]int, n)
for i := 0; i < n; i++ {
maxIndex[i] = i
}
maxCnts := make([]int, n)
for i := 0; i < n; i++ {
maxCnts[i] = 1
}
ans := n
for _, node := range nodes {
curi := node[0]
curv := vals[curi]
curCandidate := uf.Find(curi)
curMaxIndex := maxIndex[curCandidate]
for _, nexti := range graph[curi] {
nextv := vals[nexti]
nextCandidate := uf.Find(nexti)
if curCandidate != nextCandidate && curv >= nextv {
nextMaxIndex := maxIndex[nextCandidate]
if curv == vals[nextMaxIndex] {
ans += maxCnts[curMaxIndex] * maxCnts[nextMaxIndex]
maxCnts[curMaxIndex] += maxCnts[nextMaxIndex]
}
candidate := uf.Union(curi, nexti)
maxIndex[candidate] = curMaxIndex
}
}
}
return ans
}
type UnionFind struct {
parent []int
size []int
help []int
}
func NewUnionFind(n int) *UnionFind {
uf := &UnionFind{
parent: make([]int, n),
size: make([]int, n),
help: make([]int, n),
}
for i := 0; i < n; i++ {
uf.parent[i] = i
uf.size[i] = 1
}
return uf
}
func (uf *UnionFind) Find(i int) int {
hi := 0
for i != uf.parent[i] {
uf.help[hi] = i
hi++
i = uf.parent[i]
}
for hi--; hi >= 0; hi-- {
uf.parent[uf.help[hi]] = i
}
return i
}
func (uf *UnionFind) Union(i, j int) int {
f1 := uf.Find(i)
f2 := uf.Find(j)
if f1 != f2 {
if uf.size[f1] >= uf.size[f2] {
uf.size[f1] += uf.size[f2]
uf.parent[f2] = f1
return f1
} else {
uf.size[f2] += uf.size[f1]
uf.parent[f1] = f2
return f2
}
}
return f1
}
func main() {
vals := []int{
1, 1, 2, 2, 3}
edges := [][]int{
{
0, 1}, {
1, 2}, {
2, 3}, {
2, 4}}
result := numberOfGoodPaths(vals, edges)
fmt.Println(result)
}
The complete code of rust is as follows:
use std::cmp::Ordering;
fn number_of_good_paths(vals: Vec<i32>, edges: Vec<Vec<i32>>) -> i32 {
let n = vals.len();
let mut graph: Vec<Vec<i32>> = vec![Vec::new(); n];
for edge in edges {
let a = edge[0] as i32;
let b = edge[1] as i32;
graph[a as usize].push(b);
graph[b as usize].push(a);
}
let mut nodes: Vec<[i32; 2]> = vals
.iter()
.enumerate()
.map(|(i, &v)| [i as i32, v as i32])
.collect();
nodes.sort_by(|a, b| a[1].cmp(&b[1]));
let mut uf = UnionFind::new(n as i32);
let mut max_index: Vec<i32> = (0..n as i32).collect();
let mut max_counts: Vec<i32> = vec![1; n];
let mut ans = n as i32;
for node in nodes {
let cur_i = node[0];
let cur_v = vals[cur_i as usize];
let cur_candidate = uf.find(cur_i);
let cur_max_index = max_index[cur_candidate as usize];
for &next_i in &graph[cur_i as usize] {
let next_v = vals[next_i as usize];
let next_candidate = uf.find(next_i);
if cur_candidate != next_candidate && cur_v >= next_v {
let next_max_index = max_index[next_candidate as usize];
if cur_v == vals[next_max_index as usize] {
ans += max_counts[cur_max_index as usize] * max_counts[next_max_index as usize];
max_counts[cur_max_index as usize] += max_counts[next_max_index as usize];
}
let candidate = uf.union(cur_i, next_i);
max_index[candidate as usize] = cur_max_index;
}
}
}
ans
}
struct UnionFind {
parent: Vec<i32>,
size: Vec<i32>,
help: Vec<i32>,
}
impl UnionFind {
fn new(n: i32) -> UnionFind {
UnionFind {
parent: (0..n).collect(),
size: vec![1; n as usize],
help: Vec::new(),
}
}
fn find(&mut self, i: i32) -> i32 {
let mut hi = 0;
let mut x = i;
while x != self.parent[x as usize] {
self.help.push(x);
x = self.parent[x as usize];
hi += 1;
}
for hi in (0..hi).rev() {
self.parent[self.help[hi as usize] as usize] = x;
}
self.help.clear();
x
}
fn union(&mut self, i: i32, j: i32) -> i32 {
let f1 = self.find(i);
let f2 = self.find(j);
if f1 != f2 {
if self.size[f1 as usize] >= self.size[f2 as usize] {
self.size[f1 as usize] += self.size[f2 as usize];
self.parent[f2 as usize] = f1;
return f1;
} else {
self.size[f2 as usize] += self.size[f1 as usize];
self.parent[f1 as usize] = f2;
return f2;
}
}
f1
}
}
fn main() {
let vals = vec![1, 1, 2, 2, 3];
let edges = vec![vec![0, 1], vec![1, 2], vec![2, 3], vec![2, 4]];
let result = number_of_good_paths(vals, edges);
println!("Number of good paths: {}", result);
}
The complete c++ code is as follows:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int numberOfGoodPaths(vector<int>& vals, vector<vector<int>>& edges) {
int n = vals.size();
// Build the graph
vector<vector<int>> graph(n);
for (auto& edge : edges) {
int a = edge[0];
int b = edge[1];
graph[a].push_back(b);
graph[b].push_back(a);
}
// Sort the nodes based on values
vector<pair<int, int>> nodes;
for (int i = 0; i < n; i++) {
nodes.push_back({
vals[i], i });
}
sort(nodes.begin(), nodes.end());
vector<int> parent(n);
vector<int> size(n, 1);
vector<int> maxIndex(n);
vector<int> maxCnts(n, 1);
for (int i = 0; i < n; i++) {
parent[i] = i;
maxIndex[i] = i;
}
int ans = n;
// Traverse the nodes in ascending order of values
for (int i = 0; i < n; i++) {
int curi = nodes[i].second;
int curv = vals[curi];
int curCandidate = parent[curi];
int curMaxIndex = maxIndex[curCandidate];
// Iterate over the neighbors
for (int nexti : graph[curi]) {
int nextv = vals[nexti];
int nextCandidate = parent[nexti];
if (curCandidate != nextCandidate && curv >= nextv) {
int nextMaxIndex = maxIndex[nextCandidate];
if (curv == vals[nextMaxIndex]) {
ans += maxCnts[curMaxIndex] * maxCnts[nextMaxIndex];
maxCnts[curMaxIndex] += maxCnts[nextMaxIndex];
}
int candidate = parent[curi] = parent[nexti] = curMaxIndex;
maxIndex[candidate] = curMaxIndex;
}
}
}
return ans;
}
int main() {
vector<int> vals = {
1, 1, 2, 2, 3 };
vector<vector<int>> edges = {
{
0, 1}, {
1, 2}, {
2, 3}, {
2, 4} };
int result = numberOfGoodPaths(vals, edges);
cout << "Number of Good Paths: " << result << endl;
return 0;
}
The complete code of c is as follows:
#include <stdio.h>
#include <stdlib.h>
struct UnionFind {
int* parent;
int* size;
int* help;
};
struct UnionFind* createUnionFind(int n) {
struct UnionFind* uf = (struct UnionFind*)malloc(sizeof(struct UnionFind));
uf->parent = (int*)malloc(n * sizeof(int));
uf->size = (int*)malloc(n * sizeof(int));
uf->help = (int*)malloc(n * sizeof(int));
int i;
for (i = 0; i < n; i++) {
uf->parent[i] = i;
uf->size[i] = 1;
}
return uf;
}
int find(struct UnionFind* uf, int i) {
int hi = 0;
while (i != uf->parent[i]) {
uf->help[hi++] = i;
i = uf->parent[i];
}
for (hi--; hi >= 0; hi--) {
uf->parent[uf->help[hi]] = i;
}
return i;
}
int unionSet(struct UnionFind* uf, int i, int j) {
int f1 = find(uf, i);
int f2 = find(uf, j);
if (f1 != f2) {
if (uf->size[f1] >= uf->size[f2]) {
uf->size[f1] += uf->size[f2];
uf->parent[f2] = f1;
return f1;
}
else {
uf->size[f2] += uf->size[f1];
uf->parent[f1] = f2;
return f2;
}
}
return f1;
}
// Comparison function for qsort
int compare(const void* a, const void* b) {
int* na = *(int**)a;
int* nb = *(int**)b;
return na[1] - nb[1];
}
int numberOfGoodPaths(int* vals, int valsSize, int** edges, int edgesSize, int* edgesColSize) {
int n = valsSize;
int i, j;
// 创建图
int** graph = (int**)malloc(n * sizeof(int*));
for (i = 0; i < n; i++) {
graph[i] = (int*)malloc(n * sizeof(int));
edgesColSize[i] = 0;
}
for (i = 0; i < edgesSize; i++) {
int u = edges[i][0];
int v = edges[i][1];
graph[u][edgesColSize[u]++] = v;
graph[v][edgesColSize[v]++] = u;
}
// 创建节点数组
int** nodes = (int**)malloc(n * sizeof(int*));
for (i = 0; i < n; i++) {
nodes[i] = (int*)malloc(2 * sizeof(int));
nodes[i][0] = i;
nodes[i][1] = vals[i];
}
// 根据节点值排序
qsort(nodes, n, sizeof(nodes[0]),compare);
// 创建并初始化并查集
struct UnionFind* uf = createUnionFind(n);
int* maxIndex = (int*)malloc(n * sizeof(int));
int* maxCnts = (int*)malloc(n * sizeof(int));
for (i = 0; i < n; i++) {
maxIndex[i] = i;
maxCnts[i] = 1;
}
int ans = n;
// 遍历节点
for (i = 0; i < n; i++) {
int curi = nodes[i][0];
int curv = vals[curi];
int curCandidate = find(uf, curi);
int curMaxIndex = maxIndex[curCandidate];
// 遍历邻居
for (j = 0; j < edgesColSize[curi]; j++) {
int nexti = graph[curi][j];
int nextv = vals[nexti];
int nextCandidate = find(uf, nexti);
if (curCandidate != nextCandidate && curv >= nextv) {
int nextMaxIndex = maxIndex[nextCandidate];
if (curv == vals[nextMaxIndex]) {
ans += maxCnts[curMaxIndex] * maxCnts[nextMaxIndex];
maxCnts[curMaxIndex] += maxCnts[nextMaxIndex];
}
int candidate = unionSet(uf, curi, nexti);
maxIndex[candidate] = curMaxIndex;
}
}
}
// 释放内存
for (i = 0; i < n; i++) {
free(graph[i]);
free(nodes[i]);
}
free(graph);
free(nodes);
free(maxIndex);
free(maxCnts);
free(uf->parent);
free(uf->size);
free(uf->help);
free(uf);
return ans;
}
int main() {
int vals[] = {
1, 1, 2, 2, 3 };
int** edges = (int**)malloc(4 * sizeof(int*));
edges[0] = (int*)malloc(2 * sizeof(int));
edges[0][0] = 0; edges[0][1] = 1;
edges[1] = (int*)malloc(2 * sizeof(int));
edges[1][0] = 1; edges[1][1] = 2;
edges[2] = (int*)malloc(2 * sizeof(int));
edges[2][0] = 2; edges[2][1] = 3;
edges[3] = (int*)malloc(2 * sizeof(int));
edges[3][0] = 2; edges[3][1] = 4;
//int edgesColSize[] = { 2, 2, 2, 2 };
int* edgesColSize = (int*)malloc(4 * sizeof(int));
edgesColSize[0] = 2;
edgesColSize[1] = 2;
edgesColSize[2] = 2;
edgesColSize[3] = 2;
int result = numberOfGoodPaths(vals, 5, edges, 4, edgesColSize);
printf("Number of Good Paths: %d\n", result);
for (int i = 0; i < 4; i++) {
free(edges[i]);
}
free(edges);
//free(edgesColSize);
return 0;
}
