Can we solve this puzzle with the help of AI technology. دعاء لرجوع الزوج

https://zeroplusfour.com/similar-pair-hackerrank-solution/

Here's how I did in all languages Java 8 , C++ , C , Python 3, Python 2.

Thanks, that is a very good idea. In fact where we keep a record of deleted values. powerful revenge spells

this is my final code all test cases passed except two if anyone can help me in it.

#include <bits/stdc++.h>
using namespace std;
vector<int> adj[100001];
long long pairs=0;
int k;

void dfs(int x,vector<int>parent){
    parent.push_back(x);
    for(auto i:adj[x]){
        int j=0;
        while(j<parent.size()){
            if(abs(i-parent[j])<=k){
                pairs++;
            }
            j++;
        }
        if(!adj[i].empty()){
            dfs(i,parent);
        }
    }
}

int main()
{
    int n;
    cin>>n;
    cin>>k;
    long long sum=(n*(n+1))/2;
    for(int i=0;i<n-1;i++){
        int u,v;
        cin>>u>>v;
        sum-=v;
        adj[u].push_back(v);
    }
    int root=(int)sum;
    vector<int>parent;
    dfs(root,parent);
    cout<<pairs<<endl;
    return 0;
}

It's definitely worth the time to learn about Fenwick trees. These will allow you to add items, remove items, and count a range of items in O(log(n)) time. After that it's fairly straightforward.

This is the approach I took:
- Built the adjacency list from the directed edges.
- Found the root using a math trick: n(n+1)/2 - sum(edges[i][1]).
- Created a FT starting with an array of n nodes of 0.
- DFS through the tree:
-- Count the nodes in the tree between current-k and current+k (truncated to 1..n ).
-- If the current node isn't a leaf:
--- Set the current node.
--- Add the counts from the children.
--- Clear the current node.