HackerEarth Equal bitwise operations problem solution

In this HackerEarth Equal bitwise operations problem solution A sequence of integers is said to be beautiful if the bitwise AND, OR, and XOR of its elements are pairwise equal to each other. you are given a sequence A of size N. Find the number of non-empty subsequences of A that are beautiful. Print the answer modulo 10 to power 9 plus 7.

HackerEarth Equal bitwise operations problem solution.

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.HashMap;
import java.util.Map;
import java.util.StringTokenizer;

public class solution {

    static long modExp(long x, long y, long mod) {
        if(y == 0)
            return 1 % mod;
        long ret = modExp(x, y >> 1, mod);
        ret = ret * ret % mod;
        if((y & 1) == 1)
            ret = ret * (x % mod) % mod;
        return ret;
    }

    public static void main(String[] args) {
        FastReader s = new FastReader();
        PrintWriter w = new PrintWriter(System.out);

        long mod = (long)1e9 + 7;
        long[] fact = new long[(int)2e5 + 1];
        long[] ifact = new long[(int)2e5 + 1];
        fact[0] = 1;
        ifact[0] = 1;
        for(int i = 1; i <= (int)2e5; i++) {
            fact[i] = fact[i - 1] * i % mod;
            ifact[i] = modExp(fact[i], mod - 2, mod);
        }

        int n = s.nextInt();
        int[] a = new int[n];
        for(int i = 0; i < n; i++)
            a[i] = s.nextInt();
        HashMap<Integer, Integer> hm = new HashMap<>();
        for(int i = 0; i < n; i++) {
            if(!hm.containsKey(a[i]))
                hm.put(a[i], 0);
            hm.put(a[i], hm.get(a[i]) + 1);
        }
        long res = 0;
        for(Map.Entry<Integer, Integer> e : hm.entrySet()) {
            int k = e.getKey();
            int c = e.getValue();
            if(k == 0) {
                res = ((modExp(2, c, mod) - 1 + mod) % mod + res) % mod;
                continue;
            }
            for(int i = 1; i <= c; i+=2) {
                res = (fact[c] * ifact[i] % mod * ifact[c - i] % mod + res) % mod;
            }
        }
        w.print(res);

        w.close();
    }
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;
        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }
        double nextDouble() {
            return Double.parseDouble(next());
        }
        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}

second solution

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 14, mod = 1e9 + 7;
int n, po[maxn] = {1};
int main(){
    ios::sync_with_stdio(0), cin.tie(0);
    for(int i = 1; i < maxn; i++)
        po[i] = po[i - 1] * 2 % mod;
    cin >> n;
    map<int, int> cnt;
    for(int i = 0; i < n; i++){
        int x;
        cin >> x;
        cnt[x]++;
    }
    int ans = 0;
    for(auto [x, y] : cnt)
        if(x)
            (ans += po[y - 1]) %= mod;
        else
            (ans += po[y] - 1) %= mod;
    cout << ans << '\n';
}