PROBLEM LINK:
Practice
Contest: Division 1
Contest: Division 2
Contest: Division 3
Author: Kumar Angesh Singh
Tester: Anay Karnik
Editorialist: Mohan Abhyas
DIFFICULTY:
Simple
PREREQUISITES:
None
PROBLEM:
You are given an undirected graph with N nodes (numbered 1 through N). For each valid i, the i-th node has a weight W_i. Also, for each pair of nodes i and j, there is an edge connecting these nodes if j - i \neq W_j - W_i.
Find the number of connected components in this graph.
Hint:
Hint
j - i \neq W_j - W_i can be rewritten as W_i - i \neq W_j - j.
EXPLANATION:
Let’s say W_i^1 = W_i - i. There is an edge between node i and node j if W_i^1 \neq W_j^1.
Case 1 All W_i^1 are equal
All W_i^1 are equal => no edges exist => no of connected components = N
Case 2 All W_i^1 are not equal
All W_i^1 are not equal => there exist i and j such that W_i^1 \neq W_j^1 => i and j are connected by edge. Consider any node k, it will be connected to atleast one of i, j(W_k^1 = W_i^1 => W_k^1 \neq W_j^1).
i, j are connected and every other node is connected to one of them => it is a connected graph => no of connected components = 1
TIME COMPLEXITY:
\mathcal{O}(N) or \mathcal{O}(Nlog(N)) per testcase as per the implementation.
SOLUTIONS:
Setter's Solution
#include <bits/stdc++.h>
using namespace std;
signed main(){
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int T=1;
cin>>T;
while(T--){
int n;
cin>>n;
int w[n];
for(int i=0;i<n;i++){
cin>>w[i];
}
bool flag=1;
for(int i=1;i<n;i++){
if(w[0]!=w[i]-i){
flag=0;
break;
}
}
if(flag==0){
cout<<"1\n";
}
else{
cout<<n<<'\n';
}
}
return 0;
}
Tester's Solution
#include <iostream>
#include <set>
#define int long long
signed main() {
std::ios::sync_with_stdio(false);
std::cin.tie(0);
int t;
std::cin >> t;
while(t--) {
int n;
std::cin >> n;
std::set<int> set;
for(int i = 0; i < n; i++) {
int x;
std::cin >> x;
set.insert(x-i);
}
if(set.size() == 1)
std::cout << n << std::endl;
else
std::cout << 1 << std::endl;
}
return 0;
}