Leetcode 2359. Find Closest Node to Given Two Nodes | Weekly Contest 304. Problem 3 Medium

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31 июл 2022

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Комментарии : 18

@probabilitycodingisfunis1 2 года назад

int closestMeetingNode(vector& edges, int node1, int node2) { vectordist_node1(edges.size(),-1); vectordist_node2(edges.size(),-1); vectorvisited(edges.size(),false); dfs(node1,dist_node1,visited,edges,0); visited.assign(edges.size(),false); dfs(node2,dist_node2,visited,edges,0); int ans = edges.size(); int result = -1; for(int i=0;i

@CodeWithKanav Год назад

You are defining vector twice for visited array and twice for distance but we don't need two dist arrays this can be done with only one dist array. class Solution { private: void dfs(vector &edges, int node, int val, vector &dist, vector &vis) { vis[node] = true; dist[node] = max(dist[node], val); if(edges[node]!=-1 && !vis[edges[node]]) dfs(edges, edges[node], val+1, dist, vis); } public: int closestMeetingNode(vector& edges, int node1, int node2) { vector dist(edges.size(),0); vector vis(edges.size(), false), vis2(edges.size(), false); dfs(edges,node1,0,dist,vis); dfs(edges,node2,0,dist,vis2); int min_val = INT_MAX, ind = -1; for(int i=0; i

@shivampatidar2559 2 года назад

We can initialize the distance vector with INT_MAX so that we don't have to check whether the node is reachable or not.

@riddhinilawar7875 Год назад

Explanation is soo good..👍

@codingwithanonymous890 Год назад

Explained so well

@riyashussain2473 Год назад

Thank you so much !!

@vaibhavgupta973 Год назад

thank you !!

@Idukhan-jj9kc 2 года назад

Alisha it is applicable

@oladimejijames9554 2 года назад

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@ankitghosh9873 2 года назад

JUSTTT WOWWW Explanation!!!

@nvm5706 2 года назад

Mam your way explaination is absolutely awesome mam if you have enough time upload daily leetcode problem mam

@iampatelajeet 2 года назад

Really nice explanation Di, btw your ending was very surprising.

@reshmaparveen6679 2 года назад

Tq Alisha 🎊🎉🎉

@pharmacybyanuj2459 2 года назад

Please Make a video on median in a row wise sorted matrix problem.

@dhruvpandey9804 2 года назад

Amazing explanation mam

@atirrasheedhashmi 2 года назад

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@sgcreations5638 Год назад

int closestMeetingNode(vector& edges, int node1, int node2) { int n = edges.size() ; vectoradj(n) ; for( int i = 0 ; i < n ; i++ ) { if( edges[i] != -1 ) { adj[i].push_back(edges[i]) ; } } vectordis1( n , INT_MAX ) , dis2( n , INT_MAX ) ; dis1[node1] = 0 ; dis2[node2] = 0 ; queueq ; q.push(node1) ; vectorvis( n , false ) ; vis[node1] = true ; while(!q.empty()) { int node = q.front() ; q.pop() ; for( auto c : adj[node] ) { if(!vis[c]) { dis1[c] = dis1[node] + 1 ; vis[c] = true ; q.push(c) ; } } } queueq2 ; q2.push(node2) ; vectorvis2( n , false ) ; vis2[node2] = true ; while(!q2.empty()) { int node = q2.front() ; q2.pop() ; for( auto c : adj[node] ) { if(!vis2[c]) { dis2[c] = dis2[node] + 1 ; vis2[c] = true ; q2.push(c) ; } } } int ans = n ; int res = -1 ; for( int i = 0 ; i < n ; i++ ) { if( max( dis1[i] , dis2[i]) < ans ) { ans = max( dis1[i] , dis2[i]) ; res = i ; } } return res ; }