**Winning Lottery Ticket**

### Problem Statement :

The SuperBowl Lottery is about to commence, and there are several lottery tickets being sold, and each ticket is identified with a ticket ID. In one of the many winning scenarios in the Superbowl lottery, a winning pair of tickets is: Concatenation of the two ticket IDs in the pair, in any order, contains each digit from to at least once. For example, if there are distinct tickets with ticket ID and , is a winning pair. NOTE: The ticket IDs can be concantenated in any order. Digits in the ticket ID can occur in any order. Your task is to find the number of winning pairs of distinct tickets, such that concatenation of their ticket IDs (in any order) makes for a winning scenario. Complete the function winningLotteryTicket which takes a string array of ticket IDs as input, and return the number of winning pairs. Input Format The first line contains denoting the total number of lottery tickets in the super bowl. Each of the next lines contains a string, where string on a line denotes the ticket id of the ticket. Output Format Print the number of pairs in a new line.

### Solution :

` ````
Solution in C++ :
In C++ :
#include "bits/stdc++.h"
using namespace std;
const int N = 1e6 + 6;
int n;
int cnt[1 << 10];
void readInp() {
ios_base :: sync_with_stdio(false);
cin.tie(NULL);
string x;
cin >> n;
for(int i = 1; i <= n; ++i) {
cin >> x;
int mask = 0;
for(int j = 0; j < x.size(); ++j) mask |= (1 << (x[j] - '0'));
++cnt[mask];
}
}
long long solve() {
long long ans = 0;
for(int m1 = 0; m1 <= 1023; ++m1)
for(int m2 = 0; m2 <= 1023; ++m2)
if((m1 | m2) == 1023)
ans += m1 == m2 ? 1LL * cnt[m1] * (cnt[m1] - 1) : 1LL * cnt[m1] * cnt[m2];
return ans / 2LL;
}
void out(long long x) {
cout << x << endl;
}
int main() {
readInp();
out(solve());
return 0;
}
```

` ````
Solution in Python :
In Python3 :
n = int(raw_input())
p = [raw_input().strip() for _ in xrange(n)]
fullMask = 2**10-1
cntMask = [0 for _ in xrange(fullMask+1)]
for s in p:
mask = 0
for c in s:
mask |= 1 << (ord(c) - ord('0'))
cntMask[mask] += 1
res = 0
for i in xrange(fullMask+1):
for j in xrange(i+1, fullMask+1):
if i | j == fullMask:
res += cntMask[i] * cntMask[j]
res += cntMask[fullMask] * (cntMask[fullMask]-1) / 2
print res
```

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