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‎MinimumTimeToCollectAllApplesInATree/minimum_time_to_collect_all_apples_in_a_tree.dart

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/*
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-* 1443. Minimum Time to Collect All Apples in a Tree *-
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Given an undirected tree consisting of n vertices numbered from 0 to n-1, which has some apples in their vertices. You spend 1 second to walk over one edge of the tree. Return the minimum time in seconds you have to spend to collect all apples in the tree, starting at vertex 0 and coming back to this vertex.
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The edges of the undirected tree are given in the array edges, where edges[i] = [ai, bi] means that exists an edge connecting the vertices ai and bi. Additionally, there is a boolean array hasApple, where hasApple[i] = true means that vertex i has an apple; otherwise, it does not have any apple.
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Example 1:
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Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,true,false,true,true,false]
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Output: 8
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Explanation: The figure above represents the given tree where red vertices have an apple. One optimal path to collect all apples is shown by the green arrows.
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Example 2:
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Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,true,false,false,true,false]
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Output: 6
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Explanation: The figure above represents the given tree where red vertices have an apple. One optimal path to collect all apples is shown by the green arrows.
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Example 3:
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Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,false,false,false,false,false]
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Output: 0
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Constraints:
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1 <= n <= 105
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edges.length == n - 1
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edges[i].length == 2
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0 <= ai < bi <= n - 1
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from-i < toi
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hasApple.length == n
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*/
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import 'dart:collection';
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class A {
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late int ans;
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int minTime(int n, List<List<int>> edges, List<bool> hasApple) {
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ans = 0;
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List<List<int>> a = List.filled(n, 0, growable: true)
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.map((e) => List.filled(n, 0, growable: true))
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.toList();
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for (int i = 0; i < n; i++) {
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a[i] = [];
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}
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for (List<int> arr in edges) {
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a[arr[0]].add(arr[1]);
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a[arr[1]].add(arr[0]);
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}
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List<bool> visited = List.filled(n, false);
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visited[0] = true;
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rec(n, a, hasApple, visited, 0);
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return ans;
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}
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bool rec(int n, List<List<int>> a, List<bool> hasApple, List<bool> visited,
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int ind) {
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bool flag = false;
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for (int i = 0; i < a[ind].length; i++) {
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if (!visited[a[ind].elementAt(i)]) {
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visited[a[ind].elementAt(i)] = true;
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bool val = rec(n, a, hasApple, visited, a[ind].elementAt(i));
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if (val) {
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ans += 2;
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}
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flag = flag | val;
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}
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}
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flag = flag | hasApple.elementAt(ind);
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return flag;
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}
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}
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class B {
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int res = 0;
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int minTime(int n, List<List<int>> edges, List<bool> hasApple) {
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HashMap<int, HashSet<int>> map = HashMap();
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for (List<int> edge in edges) {
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HashSet<int> list1 = map[edge[0]] ?? HashSet();
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HashSet<int> list2 = map[edge[1]] ?? HashSet();
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list1.add(edge[1]);
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list2.add(edge[0]);
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map.putIfAbsent(edge[0], () => list1);
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map.putIfAbsent(edge[1], () => list2);
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}
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List<bool> visited = List.filled(n, false);
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visited[0] = true;
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dfs(hasApple, map, 0, visited);
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return res;
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}
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void dfs(List<bool> hasApple, HashMap<int, HashSet<int>> map, int index,
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List<bool> visited) {
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HashSet<int> list1 = map[index] ?? HashSet();
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for (int neighbor in list1) {
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if (!visited[neighbor]) {
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int temp = res;
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visited[neighbor] = true;
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if (hasApple[neighbor]) res += 2;
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dfs(hasApple, map, neighbor, visited);
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if (!hasApple[neighbor] && temp < res) {
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res += 2;
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}
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}
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}
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}
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}
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class C {
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int minTime(int n, List<List<int>> edges, List<bool> hasApple) {
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HashMap<int, List<int>> map = HashMap();
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buildTree(edges, map);
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HashSet<int> visited = HashSet();
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return helper(0, map, hasApple, visited);
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}
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void buildTree(List<List<int>> edges, HashMap<int, List<int>> map) {
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for (List<int> edge in edges) {
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int a = edge[0], b = edge[1];
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map.putIfAbsent(a, () => []);
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map.putIfAbsent(b, () => []);
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map[a]?.add(b);
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map[b]?.add(a);
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}
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}
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int helper(int node, HashMap<int, List<int>> map, List<bool> hasApple,
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HashSet<int> visited) {
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visited.add(node);
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int res = 0;
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for (int child in map[node] ?? []) {
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if (visited.contains(child)) continue;
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res += helper(child, map, hasApple, visited);
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}
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if ((res > 0 || hasApple[node]) && node != 0) res += 2;
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return res;
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}
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}

‎MinimumTimeToCollectAllApplesInATree/minimum_time_to_collect_all_apples_in_a_tree.go

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package main
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func minTime(n int, edges [][]int, hasApple []bool) int {
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graph := make([][]int, n)
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for _, edge := range edges {
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graph[edge[0]] = append(graph[edge[0]], edge[1])
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graph[edge[1]] = append(graph[edge[1]], edge[0])
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}
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out := 0
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visited := make([]bool, n)
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visited[0] = true
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for _, node := range graph[0] {
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out += traverse(graph, node, hasApple, visited)
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}
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return out
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}
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func traverse(graph [][]int, node int, hasApple []bool, visited []bool) int {
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out := 0
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visited[node] = true
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for _, n := range graph[node] {
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if !visited[n] {
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out += traverse(graph, n, hasApple, visited)
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}
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}
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if hasApple[node] || out != 0 {
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out += 2
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}
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return out
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}

‎MinimumTimeToCollectAllApplesInATree/minimum_time_to_collect_all_apples_in_a_tree.md

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# 🔥 3 Approaches 🔥 || Simple Fast and Easy || with Explanation
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## Approach
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Still need bi-direction graph and use visited set. Example test case for that:
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4
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[[0,2],[0,3],[1,2]]
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[false,true,false,false]
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Expected Answer: 4
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Key Notes:
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You need to consume 2 seconds to simply collect an apple node (come and go)
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Consider a node:
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If none of descendant (including itself) has an apple, we don't need to waste time on this node
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If any of descendant has an apple (no matter if it-self has an apple or not), we need to consume 2 seconds on this node anyway
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Collect node 0 does not need to consume any time
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Then, we can have a helper dfs function meaning: time needs to waste on this node to collect all apples. (0 or > 0).
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## Solution - 1 Classic Depth First Search
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```dart
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class Solution {
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int res = 0;
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int minTime(int n, List<List<int>> edges, List<bool> hasApple) {
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HashMap<int, HashSet<int>> map = HashMap();
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for (List<int> edge in edges) {
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HashSet<int> list1 = map[edge[0]] ?? HashSet();
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HashSet<int> list2 = map[edge[1]] ?? HashSet();
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list1.add(edge[1]);
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list2.add(edge[0]);
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map.putIfAbsent(edge[0], () => list1);
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map.putIfAbsent(edge[1], () => list2);
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}
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List<bool> visited = List.filled(n, false);
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visited[0] = true;
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dfs(hasApple, map, 0, visited);
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return res;
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}
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void dfs(List<bool> hasApple, HashMap<int, HashSet<int>> map, int index,
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List<bool> visited) {
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HashSet<int> list1 = map[index] ?? HashSet();
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for (int neighbor in list1) {
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if (!visited[neighbor]) {
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int temp = res;
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visited[neighbor] = true;
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if (hasApple[neighbor]) res += 2;
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dfs(hasApple, map, neighbor, visited);
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if (!hasApple[neighbor] && temp < res) {
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res += 2;
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}
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}
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}
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}
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}
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```
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## Solution - 2 Depth First Search - Via Building Tree
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```dart
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class Solution {
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int minTime(int n, List<List<int>> edges, List<bool> hasApple) {
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HashMap<int, List<int>> map = HashMap();
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buildTree(edges, map);
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HashSet<int> visited = HashSet();
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return helper(0, map, hasApple, visited);
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}
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void buildTree(List<List<int>> edges, HashMap<int, List<int>> map) {
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for (List<int> edge in edges) {
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int a = edge[0], b = edge[1];
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map.putIfAbsent(a, () => []);
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map.putIfAbsent(b, () => []);
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map[a]?.add(b);
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map[b]?.add(a);
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}
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}
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int helper(int node, HashMap<int, List<int>> map, List<bool> hasApple,
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HashSet<int> visited) {
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visited.add(node);
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int res = 0;
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for (int child in map[node] ?? []) {
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if (visited.contains(child)) continue;
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res += helper(child, map, hasApple, visited);
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}
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if ((res > 0 || hasApple[node]) && node != 0) res += 2;
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return res;
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}
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}
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```
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## Solution - 3 For FUN
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```dart
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class Solution {
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late int ans;
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int minTime(int n, List<List<int>> edges, List<bool> hasApple) {
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ans = 0;
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List<List<int>> a = List.filled(n, 0, growable: true)
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.map((e) => List.filled(n, 0, growable: true))
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.toList();
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for (int i = 0; i < n; i++) {
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a[i] = [];
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}
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for (List<int> arr in edges) {
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a[arr[0]].add(arr[1]);
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a[arr[1]].add(arr[0]);
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}
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List<bool> visited = List.filled(n, false);
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visited[0] = true;
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rec(n, a, hasApple, visited, 0);
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return ans;
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}
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bool rec(int n, List<List<int>> a, List<bool> hasApple, List<bool> visited,
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int ind) {
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bool flag = false;
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for (int i = 0; i < a[ind].length; i++) {
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if (!visited[a[ind].elementAt(i)]) {
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visited[a[ind].elementAt(i)] = true;
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bool val = rec(n, a, hasApple, visited, a[ind].elementAt(i));
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if (val) {
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ans += 2;
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}
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flag = flag | val;
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}
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}
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flag = flag | hasApple.elementAt(ind);
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return flag;
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}
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}
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```

‎README.md

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Original file line numberDiff line numberDiff line change
@@ -170,7 +170,7 @@ This repo contain leetcode solution using DART and GO programming language. Most
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- [**1143.** Longest Common Subsequence](LongestCommonSubsequence/longest_common_subsequence.dart)
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- [**150.** Evaluate Reverse Polish Notation](EvaluateReversePolishNotation/evaluate_reverse_polish_notation.dart)
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- [**739.** Daily Temperatures](DailyTemperatures/daily_temperatures.dart)
173-
- [**1971.* Find if Path Exists in Graph](FindIfPathExistsInGraph/find_if_path_exists_in_graph.dart)
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- [**1971.** Find if Path Exists in Graph](FindIfPathExistsInGraph/find_if_path_exists_in_graph.dart)
174174
- [**841.** Keys and Rooms](KeysAndRooms/keys_and_rooms.dart)
175175
- [**886.** Possible Bipartition](PossibleBipartition/possible_bipartition.dart)
176176
- [**980.** Unique Paths III](UniquePathsIII/unique_paths_iii.dart)
@@ -180,6 +180,7 @@ This repo contain leetcode solution using DART and GO programming language. Most
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- [**452.** Minimum Number of Arrows to Burst Balloons](MinimumNumberOfArrowsToBurstBalloons/minimum_number_of_arrows_to_burst_balloons.dart)
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- [**1833.** Maximum Ice Cream Bars](MaximumIceCreamBars/maximum_ice_cream_bars.dart)
182182
- [**149.** Max Points on a Line](MaxPointsOnALine/max_points_on_a_line.dart)
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- [**1443.** Minimum Time to Collect All Apples in a Tree](MinimumTimeToCollectAllApplesInATree/minimum_time_to_collect_all_apples_in_a_tree.dart)
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## Reach me via
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