We are given an array arr of n element. We need to count number of non-empty subsequences such that these individual subsequences have same values of bitwise AND, OR and XOR. For example, we need to count a subsequence (x, y, z) if (x | y | z) is equal to (x & y & z) and (x ^ y ^ z). For a single element subsequence, we consider the element itself as result of XOR, AND and OR. Therefore all single-element subsequences are always counted as part of result.
Examples:
Input : a = [1, 3, 7]
Output : 3
Explanation:
There are 7 non empty subsequence .
subsequence OR AND XOR
{1} 1 1 1
{3} 3 3 3
{7} 7 7 7
{1, 3} 3 1 2
{1, 7} 7 1 6
{3, 7} 7 3 4
{1, 3, 7} 7 1 5
Out of 7, there are 3 subsequences {1}
{3} {7} which have same values of AND,
OR and XOR.
Input : a[] = [0, 0, 0]
Output : 7
Explanation: All 7 non empty subsequences
have same values of AND, OR and XOR.
Input : a[] = [2, 2, 2, 3, 4]
Output : 6
Explanation: subsequence {2}, {2}, {2},
{2, 2, 2}, {3}, {4} have same values of
AND, OR and XOR.
1) If there are n occurrences of zeroes in the given array, then will be 2n – 1 subsequences contributed by these zeroes.
2) If there are n occurrences of a non-zero element x, then there will be 2n-1 subsequences contributed by occurrences of this element. Please note that, in case of non-zero elements, only odd number of occurrences can cause same results for bitwise operators.
Find count of each element in the array then apply the above formulas.
C++
#include <bits/stdc++.h>using namespace std;// function for finding count of possible subsequenceint countSubseq(int arr[], int n){ int count = 0; // creating a map to count the frequency of each element unordered_map<int, int> mp; // store frequency of each element for (int i = 0; i < n; i++) mp[arr[i]]++; // iterate through the map for (auto i : mp) { // add all possible combination for key equal zero if (i.first == 0) count += pow(2, i.second) - 1; // add all (odd number of elements) possible // combination for key other than zero else count += pow(2, i.second - 1); } return count;}// driver functionint main(){ int arr[] = { 2, 2, 2, 5, 6 }; int n = sizeof(arr) / sizeof(arr[0]); cout << countSubseq(arr, n); return 0;} |
Java
import java .io.*; import java.util.*;class GFG { // function for finding count of possible subsequencestatic int countSubseq(int arr[], int n){ int count = 0; // creating a map to count the frequency of each element HashMap<Integer,Integer>mp=new HashMap<Integer,Integer>(); // store frequency of each element for (int i = 0; i < n; i++) if (mp.containsKey(arr[i])) mp.put(arr[i],mp.get(arr[i])+1); else mp.put(arr[i],1); // iterate through the map for (Map.Entry<Integer,Integer>entry:mp.entrySet()) { // add all possible combination for key equal zero if (entry.getKey() == 0) count += Math.pow(2, entry.getValue()) - 1; // add all (odd number of elements) possible // combination for key other than zero else count += Math.pow(2, entry.getValue()- 1); } return count;} // driver functionpublic static void main(String[] args){ int arr[] = { 2, 2, 2, 5, 6 }; int n=arr.length; System.out.println(countSubseq(arr, n));}}// This code is contributed by apurva raj |
C#
using System;using System.Collections.Generic;class GFG{// function for finding count of possible subsequencestatic int countSubseq(int []arr, int n){ int count = 0; // creating a map to count the frequency of each element Dictionary<int, int> mp = new Dictionary<int,int>(); // store frequency of each element for (int i = 0; i < n; i++) { if (mp.ContainsKey(arr[i])) { var val = mp[arr[i]]; mp.Remove(arr[i]); mp.Add(arr[i], val + 1); } else { mp.Add(arr[i], 1); } } // iterate through the map foreach(KeyValuePair<int, int> entry in mp) { // add all possible combination for key equal zero if (entry.Key == 0) count += (int)(Math.Pow(2, entry.Value - 1)); // add all (odd number of elements) possible // combination for key other than zero else count += (int)(Math.Pow(2, entry.Value - 1)); } return count;}// Driver functionpublic static void Main(String []args) { int []arr = { 2, 2, 2, 5, 6 }; int n = arr.Length; Console.WriteLine(countSubseq(arr, n));}}// This code is contributed by shivanisinghss2110 |
Python3
# function for finding count of possible subsequencedef countSubseq(arr, n): count = 0 # creating a map to count the frequency of each element mp = {} # store frequency of each element for x in arr: if x in mp.keys(): mp[x]+=1 else: mp[x]=1 # iterate through the map for i in mp.keys(): # add all possible combination for key equal zero if (i == 0): count += pow(2, mp[i]) - 1 # add all (odd number of elements) possible # combination for key other than zero else: count += pow(2, mp[i] - 1) return count# Driver functionarr= [2, 2, 2, 5, 6 ]n = len(arr)print(countSubseq(arr, n))# This code is contributed by apurva raj |
Javascript
<script>// function for finding count of possible subsequencefunction countSubseq(arr, n){ let count = 0; // creating a map to count the frequency of each element let mp = new Map(); // store frequency of each element for (let i = 0; i < n; i++){ mp[arr[i]]++; if(mp.has(arr[i])){ mp.set(arr[i], mp.get(arr[i]) + 1) }else{ mp.set(arr[i], 1) } } // iterate through the map for (let i of mp) { // add all possible combination for key equal zero if (i[0] == 0) count += Math.pow(2, i[1]) - 1; // add all (odd number of elements) possible // combination for key other than zero else count += Math.pow(2, i[1] - 1); } return count;}// driver function let arr = [ 2, 2, 2, 5, 6 ]; let n = arr.length; document.write(countSubseq(arr, n));// This code is contributed by _saurabh_jaiswal</script> |
Output:
6
Time complexity: O(N)
Auxiliary Space: O(N)
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