Solution For Minimum Time To Collect All Apples In A Tree

Problem Statement:

Given a tree in which each vertex represents a fruit-bearing apple tree, you need to collect all the apples without skipping your instructions.

The apple tree is represented by an integer array where the value of the ith element is the location of the ith apple tree. You can start at any vertex of the tree and visit each vertex in any order.

You can collect apples at any vertex as long as you’ve reached the equipment that’s installed there.

Return the minimum time in seconds you need to collect all apples in the tree starting at vertex 0.

Solution:

The problem statement asks us to find the minimum time required to collect all apples in the tree. We can solve this problem using the Breadth First Search algorithm.

We start from the root node (node 0) and traverse the tree to collect the apples. We maintain a variable called distance, which represents the distance traveled so far, and a boolean array called visited, which represents whether a particular node has been visited or not.

We use a priority queue that stores the next node that we should visit based on the distance, with the minimum distance always on top.

During the traversal, we check if the current node has an apple tree. If yes, we mark it as visited and add the distance to the total distance.

We continue the traversal until we visit all the nodes with apple trees and return the total distance traveled.

Algorithm:

1. Initialize distance = 0 and visited[0] = true.

2. Create a priority queue pq and push the root node 0 with distance 0.

3. While pq is not empty:

   a. Pop the node with minimum distance.

   b. If node i has an apple tree and has not been visited yet, add the distance to the total distance and mark visited[i] as true.

   c. For each neighbor j of node i:

   i. If the neighbor j has not been visited yet, push it to pq with the distance traveled so far plus the distance between i and j.
   

4. Return the total distance traveled.

Time Complexity: O(nlogn), where n is the number of nodes in the tree.

Space Complexity: O(n), where n is the number of nodes in the tree.

Implementation:

Here’s the implementation of the above algorithm in Python:

import heapq

class Solution:
def minTime(self, n: int, edges: List[List[int]], hasApple: List[bool]) -> int:
visited = [False] * n
distance = 0
graph = [[] for _ in range(n)]

    for i in range(len(edges)):
        u, v = edges[i]
        graph[u].append(v)
        graph[v].append(u)

    pq = [(0, 0)]
    visited[0] = True

    while pq:
        dist, node = heapq.heappop(pq)

        if hasApple[node] and not visited[node]:
            visited[node] = True
            distance += dist

        for neighbor in graph[node]:
            if not visited[neighbor]:
                heapq.heappush(pq, (dist + 1, neighbor))

    return distance * 2

Step by Step Implementation For Minimum Time To Collect All Apples In A Tree

This problem can be solved using a breadth-first search algorithm. The idea is to start from the root node and explore all of the nodes at the same level before moving on to the next level. The time required to collect all of the apples will be the sum of the times required to collect all of the apples at each level.

Here is some pseudocode for the algorithm:

1. Create a queue.
2. Add the root node to the queue.
3. While the queue is not empty:
4.     Remove a node from the queue.
5.     If the node has any children:
6.         Add the children to the queue.
7.         Increment the time required to collect all of the apples.
8. return the time required to collect all of the apples.

This is a breadth-first search problem.

from collections import deque

def minimumTime(numNodes, edges, hasApple):
    # Create a graph represented by an adjacency list
    graph = {i: [] for i in range(numNodes)}
    for u, v in edges:
        graph[u].append(v)
        graph[v].append(u)

    # Perform a breadth-first search, keeping track of the number of steps
    # needed to reach each node
    visited = [False] * numNodes
    steps = [0] * numNodes
    queue = deque([0])
    visited[0] = True
    while queue:
        u = queue.popleft()
        for v in graph[u]:
            if not visited[v]:
                steps[v] = steps[u] + 1
                visited[v] = True
                queue.append(v)

    # The total number of steps needed is the sum of the steps needed to reach
    # each node that has an apple
    return sum(steps[i] for i in range(numNodes) if hasApple[i])

/**
 * @param {number[][]} edges
 * @param {number[]} apples
 * @return {number}
 */
var minimumTime = function(edges, apples) {
    // create a map of all the edges
    // key is the node, value is an array of all the nodes it's connected to
    const map = new Map();
    for (let i = 0; i < edges.length; i++) {
        const [node1, node2] = edges[i];
        if (!map.has(node1)) {
            map.set(node1, [node2]);
        } else {
            map.get(node1).push(node2);
        }
        if (!map.has(node2)) {
            map.set(node2, [node1]);
        } else {
            map.get(node2).push(node1);
        }
    }
    
    // create a set of all the apples
    // this will be used to check if a node has an apple or not
    const appleSet = new Set(apples);
    
    // create a set of all the nodes that have been visited
    // this will be used to keep track of which nodes have been visited
    const visited = new Set();
    
    // create a variable to keep track of the total time it takes to collect all the apples
    let time = 0;
    
    // create a queue to keep track of all the nodes that need to be visited
    // we will use a breadth first search to find the shortest path from the root node to all the apples
    const queue = [0];
    
    // while the queue is not empty, keep searching
    while (queue.length > 0) {
        // the number of nodes at the current level
        // we need to keep track of this so we know when we have visited all the nodes at the current level
        let size = queue.length;
        
        // iterate through all the nodes at the current level
        for (let i = 0; i < size; i++) {
            // get the node from the queue
            const node = queue.shift();
            
            // if the node has not been visited yet
            if (!visited.has(node)) {
                // mark the node as visited
                visited.add(node);
                
                // if the node has an apple, increment the time
                if (appleSet.has(node)) {
                    time++;
                }
                
                // get all the nodes that the current node is connected to
                const connectedNodes = map.get(node);
                
                // add all the connected nodes to the queue
                for (let j = 0; j < connectedNodes.length; j++) {
                    queue.push(connectedNodes[j]);
                }
            }
        }
        
        // after visiting all the nodes at the current level
        // we need to increment the time again
        // this is because it takes one unit of time to move from one level to the next
        time++;
    }
    
    return time;
};

There are n apples in a tree, and each apple has a unique value. You are given an integer array tree of size n, where tree[i] is the value of the ith apple in the tree.

You start at the root of the tree and collect apples as you move down the tree. The value of an apple at each node is given by tree[node]. You can only move to adjacent nodes (children of the current node) if tree[node] is equal to tree[parent], where parent is the index of the current node's parent.

Return the minimum number of moves required to collect all apples in the tree.



int minTime(vector& tree) {
        // Base case
        if (tree.empty()) {
            return 0;
        }
        // DFS
        int res = 0;
        stack> stk; // {node, parent}
        stk.push({0, -1});
        while (!stk.empty()) {
            auto [node, parent] = stk.top();
            stk.pop();
            // Check if current node has been visited
            if (tree[node] != tree[parent]) {
                res++;
            }
            // Visit children
            for (int child : {node * 2 + 1, node * 2 + 2}) {
                if (child < tree.size() && tree[child] == tree[node]) {
                    stk.push({child, node});
                }
            }
        }
        return res;
    }

There are a few different ways to solve this problem. One way would be to use a breadth first search algorithm. Another way would be to use a depth first search algorithm.

Here is one possible solution using a breadth first search algorithm:

1. Create a queue.
2. Enqueue the root node.
3. While the queue is not empty:
4. Dequeue a node.
5. If the node is an apple, add its value to a total.
6. Enqueue the node's children.
7. Return the total.