Find Closest Node to Given Two Nodes LeetCode Solution

Problem – Find Closest Node to Given Two Nodes LeetCode Solution

You are given a directed graph of n nodes numbered from 0 to n - 1, where each node has at most one outgoing edge.

The graph is represented with a given 0-indexed array edges of size n, indicating that there is a directed edge from node i to node edges[i]. If there is no outgoing edge from i, then edges[i] == -1.

You are also given two integers node1 and node2.

Return the index of the node that can be reached from both node1 and node2, such that the maximum between the distance from node1 to that node, and from node2 to that node is minimized. If there are multiple answers, return the node with the smallest index, and if no possible answer exists, return -1.

Note that edges may contain cycles.

Example 1:

Input: edges = [2,2,3,-1], node1 = 0, node2 = 1
Output: 2
Explanation: The distance from node 0 to node 2 is 1, and the distance from node 1 to node 2 is 1.
The maximum of those two distances is 1. It can be proven that we cannot get a node with a smaller maximum distance than 1, so we return node 2.

Example 2:

Input: edges = [1,2,-1], node1 = 0, node2 = 2
Output: 2
Explanation: The distance from node 0 to node 2 is 2, and the distance from node 2 to itself is 0.
The maximum of those two distances is 2. It can be proven that we cannot get a node with a smaller maximum distance than 2, so we return node 2.

Constraints:

• n == edges.length
• 2 <= n <= 105
• -1 <= edges[i] < n
• edges[i] != i
• 0 <= node1, node2 < n

Find Closest Node to Given Two Nodes LeetCode Solution in C++

void bfs(int src, vector<int> &dist,vector<int>& edge,int n){
    queue<int> q;
    q.push(src);
    dist[src]=0;
    while(q.size()>0){
       auto p= q.front(); q.pop();
       if(edge[p]!=-1 and dist[edge[p]]==INT_MAX){
           q.push(edge[p]);  
           dist[edge[p]]= dist[p]+1;
       }
    }
}

int closestMeetingNode(vector<int>& e, int n1, int n2) {
    int n= e.size();
    vector<int> A(n,INT_MAX), B(n,INT_MAX);
    bfs(n1,A,e,n);   
    bfs(n2,B,e,n);
    int res= INT_MAX, node=-1;
    for(int i=0;i<n;i++){
        if(A[i]==INT_MAX or B[i]==INT_MAX) continue;
        if(res>max(A[i],B[i])) node=i,res= max(A[i],B[i]);
    }
    return node;
}

Find Closest Node to Given Two Nodes LeetCode Solution in Java

void bfs(int src, int[] dist,int[] edge,int n){
    Queue<Integer> q= new LinkedList<>();
    q.add(src);
    dist[src]=0;
    while(q.size()>0){
        var p= q.remove();
        if(edge[p]!=-1 && dist[edge[p]]==Integer.INT_MAX){
            q.add(edge[p]);
            dist[edge[p]]= dist[p]+1;
        }
    }
}

public int closestMeetingNode(int[] e, int n1, int n2) {
    int n= e.length;
    int[] A= new int[n]; 
    int[] B= new int[n];
    Arrays.fill(A,Integer.MAX_VALUE);
    Arrays.fill(B,Integer.MAX_VALUE);
    
    //call bfs
    bfs(n1,A,e,n);
    bfs(n2,B,e,n);
    
    //get node with minimum value of distance
    int res= Integer.MAX_VALUE, node=-1;
    for(int i=0;i<n;i++){
        if(A[i]==Integer.MAX_VALUE || B[i]==Integer.MAX_VALUE) continue;
        if(res>Math.max(A[i],B[i])){
            node=i; 
            res= Math.max(A[i],B[i]);
        }
    }
    return node;
}

Find Closest Node to Given Two Nodes LeetCode Solution in Python

class Solution:
    def closestMeetingNode(self, edges: List[int], node1: int, node2: int) -> int:
    
        res = float("inf")
        
        def dfs(node, arr, visited, counter=0):
            
			#assigning how many moves it takes to reach node 
            arr[node] = counter
            visited.add(node)
            
            next_node = edges[node]
            if next_node!=-1 and next_node not in visited:
				#going through each neighbor if exists and updating the counter 
                dfs(edges[node], arr, visited, counter+1)
            
            return arr
        
		#find moves to reach nodes from node1
        n1 = [-1 for i in range(len(edges))]
        dfs(node1, n1, set())
        
		#find moves to reach nodes from node1
        n2 = [-1 for i in range(len(edges))]
        dfs(node2, n2, set())
                    
        answer = -1
		
        for i in range(len(edges)):
		
			#check if the end node is reachable from both starting nodes
            if n1[i]!=-1 and n2[i]!=-1:
                maximum_distance = max(n1[i], n2[i])
				
				#update the distance and the final answer if relevant
                if maximum_distance<res:
                    res = maximum_distance
                    answer = i
                
        return answer

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