Hashing right angled triangle

#1

Bhaiya , how can we make hashmap in this question?

#2

Hello @CODER_JATIN

U just have to store count of occurences of x and y so u can use two map of type map < int , int > or u can use array/ vector as hashmap where array indices can be treated as key and content as its value

#3

bhaiya ye samajh ni aaya?

#4

Tumhe is problem ka logic to pata hai na?

#5

haa bs ye smjh ni aaya hashmap kese use krna isme

#6

yaha pe mai ye bol raha tha ->

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


int main()
{
	int n;
	cin >> n;
      map<int, int> x_cnt, y_cnt;

	vector<int> x(n), y(n);

	for (int i = 0; i < n; ++i) {
		cin >> x[i] >> y[i];
		x_cnt[x[i]]++;
		y_cnt[y[i]]++;
	}

	int ans = 0;
	for (int i = 0; i < n; ++i) {
		ans += (x_cnt[x[i]] - 1) * (y_cnt[y[i]] - 1);
	}
	cout << ans;
	return 0;
}

agar ab bhi samjh nahi aaya to pucho

#7

BHAIYA, hashmap mai bydefault value 0 hoti hai key ki ?

#8

jese ham ye kar rhe hai x_cnt[x[i]]++; y_cnt[y[i]]++;

#9

ha agar value hamari int hai to bydefault 0 hota hai

#10

thik hai bhaiya ,
to Answer seedha yeh hi hoga na … (x_cnt[x[i]] - 1) * (y_cnt[y[i]] - 1);
loop ki kya jarurat

#11

jarurat hai , tumhe har corrdinate ke corresponding ye formula lagana hoga na.

isme (x[i],y[i]) coordinate hi to hai.

#12

hmm aa gya smjh bhaiya, thank you!

#13

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#14

bhaiya jese isme aapne do alag alag vector liye , to vector of pairr bhi le skte naa?

#15

ha . . . . . . . . . . . . .

#16

okay bhaiya . . . …

#17

#include

#include

#include

using namespace std;

int main()

{

int n;

vector<pair<int , int > > p;

cin>>n;

int x, y;

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

{

    cin>>x>>y;

    p.push_back({x,y});

}

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

{

    cout<<p[i].first<<",";

    cout<<p[i].second<<"\n";

}

return 0;

}

dekhna bhaiya ese galat kyo aari output

#18

bro code ko ide pe save karke send karo link

#19
#20

save nahi kiya hai tumne