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Count subsequences with same values of Bitwise AND, OR and XOR

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  • Difficulty Level : Medium
  • Last Updated : 28 Nov, 2022
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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 subsequence
int 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 function
int 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 subsequence
static 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 function
public 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 subsequence
static 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 function
public 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 subsequence
def 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 function
arr= [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 subsequence
function 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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