2.4 years ago

suryansh jaiswal • 370

Lucknow

for question 1 

void solve()
{
int n,x;
cin>>n>>x;
int ar[n];
f(i,n)
cin>>ar[i];Q
sort(ar,ar+n);
int ans=0;
for(int i=0;i<n;i++)
{
    if(2*ar[i]==x)
        ans+=i;
}
cout<<ans<<endl;
}
int32_t main()
{
    
 solve();        
}

2.5 years ago

Rishabh Kumar • 10

for test case -1 0 2 4 and get the value 2 your answer would not work 

2.5 years ago

Smeet Chavan • 10

According to me, this would be a solution :
 

#include<bits/stdc++.h>
using namespace std;

int main(){

  vector<int> arr = {1 , -1 , 2 , -2 , 3 , -3 , 4 , -4};
  int sum = 3;

  if(sum % 2){
    return 0;
  }

  int count = 0;
  int num = 0;

  for(int i = 0 ; i < n ; ++i){

    if(abs(arr[i]) == sum/2){
         count++;
    }
    else if(abs(arr[i]) < sum/2){
         num++;
    }
  }

  int ans = count*num + (count * (count - 1))/2 ;
  cout<<ans<<endl;

  return 0;
}

First of all, if the sum is odd, then we can never find an integer = sum/2 in the array. Hence, No pair exists.

Here, count*num represents the value(sum / 2) pairing up with all the numbers less than sum/2 which is "num".
And (count*(count - 1))/2 are the pairs of sum/2 with sum/2 i.e. if arr[i] == arr[j] == sum/2.

2.5 years ago

JoJo • 10

Question 1

class Solution {
    public void groupSizes(int n, int[] from, int[] to, int[] queries) {
        List<List<Integer>> adj = new ArrayList<>();
        for (int i = 0; i < n+1; i++) {
            adj.add(new ArrayList<>());
        }

        for (int i = 0; i < from.length; i++) {
            adj.get(from[i]).add(to[i]);
            adj.get(to[i]).add(from[i]);
        }
        
        int[] colors = new int[n+1];

        Map<Integer, Integer> colorToSize = new HashMap<>();
        int color = 1;

        for (int i = 1; i <= n; i++) {
            if (colors[i] != 0) continue;
            int size = dfs(i, adj, colors, color);
            colorToSize.put(color, size);
            color++;
        }

        for (int query : queries) {
            System.out.println(colorToSize.get(colors[query]));
        }
    }

    private int dfs(int i, List<List<Integer>> adj, int[] colors, int color) {
        colors[i] = color;
        int size = 1;

        for (int next : adj.get(i)) {
            if (colors[next] != 0) continue;
            size += dfs(next, adj, colors, color);
        }

        return size;
    }
}