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:Â
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.Â
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;Â
// function for finding count of possible subsequenceint countSubseq(int arr[], int n){    int count = 0;Â
    // creating a map to count the frequency of each element    unordered_map<int, int> mp;Â
    // store frequency of each element    for (int i = 0; i < n; i++)        mp[arr[i]]++;Â
    // iterate through the map    for (auto i : mp) {Â
        // add all possible combination for key equal zero        if (i.first == 0)            count += pow(2, i.second) - 1;Â
        // add all (odd number of elements) possible         // combination for key other than zero        else            count += pow(2, i.second - 1);    }    return count;}Â
// driver functionint main(){Â Â Â Â int arr[] = { 2, 2, 2, 5, 6 };Â Â Â Â int n = sizeof(arr) / sizeof(arr[0]);Â Â Â Â cout << countSubseq(arr, n);Â Â Â Â return 0;} |
Java
import java .io.*; import java.util.*;Â
Â
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));}}Â
// This code is contributed by apurva raj |
C#
using System;using System.Collections.Generic;class GFG{Â
// function for finding count of possible subsequencestatic int countSubseq(int []arr, int n){Â Â Â Â int count = 0;Â
    // creating a map to count the frequency of each element     Dictionary<int, int> mp = new Dictionary<int,int>();Â
    // 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);             }         }Â
    // iterate through the map    foreach(KeyValuePair<int, int> entry in mp) {Â
        // add all possible combination for key equal zero        if (entry.Key == 0)            count += (int)(Math.Pow(2, entry.Value - 1));Â
        // add all (odd number of elements) possible         // combination for key other than zero        else            count += (int)(Math.Pow(2, entry.Value - 1));    }    return count;}Â
// Driver functionpublic static void Main(String []args)Â Â Â Â Â {Â Â Â Â int []arr = { 2, 2, 2, 5, 6 };Â Â Â Â int n = arr.Length;Â Â Â Â Console.WriteLine(countSubseq(arr, n));}}Â
// This code is contributed by shivanisinghss2110 |
Python3
# function for finding count of possible subsequencedef countSubseq(arr, n):Â Â Â Â count = 0Â
    # creating a map to count the frequency of each element    mp = {}Â
    # store frequency of each element    for x in arr:        if x in mp.keys():            mp[x]+=1        else:            mp[x]=1Â
    # iterate through the map    for i in mp.keys():Â
        # add all possible combination for key equal zero        if (i == 0):            count += pow(2, mp[i]) - 1Â
        # add all (odd number of elements) possible         # combination for key other than zero        else:            count += pow(2, mp[i] - 1)    return countÂ
# Driver functionarr= [2, 2, 2, 5, 6 ]n = len(arr)print(countSubseq(arr, n))Â
# This code is contributed by apurva raj |
Javascript
<script>// function for finding count of possible subsequencefunction countSubseq(arr, n){Â Â Â Â let count = 0;Â
    // creating a map to count the frequency of each element    let mp = new Map();Â
    // store frequency of each element    for (let i = 0; i < n; i++){        mp[arr[i]]++;Â
        if(mp.has(arr[i])){            mp.set(arr[i], mp.get(arr[i]) + 1)        }else{            mp.set(arr[i], 1)        }    }Â
    // iterate through the map    for (let i of mp) {Â
        // add all possible combination for key equal zero        if (i[0] == 0)            count += Math.pow(2, i[1]) - 1;Â
        // add all (odd number of elements) possible        // combination for key other than zero        else            count += Math.pow(2, i[1] - 1);    }    return count;}Â
// driver function    let arr = [ 2, 2, 2, 5, 6 ];    let n = arr.length;    document.write(countSubseq(arr, n));Â
// This code is contributed by _saurabh_jaiswal</script> |
Output:Â
6
Â
Time complexity: O(N)
Auxiliary Space: O(N)
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