Count pairs with equal Bitwise AND and Bitwise OR value
Given an array, arr[] of size N, the task is to count the number of unordered pairs such that Bitwise AND and Bitwise OR of each pair is equal.
Examples:
Input: arr[] = {1, 2, 1}
Output: 1
Explanation:
Bitwise AND value and Bitwise OR value all possible pairs are:
Bitwise AND of the pair(arr[0], arr[1]) is (arr[0] & arr[1]) = (1 & 2) = 0
Bitwise AND of the pair(arr[0], arr[1]) is (arr[0] | arr[1]) = (1 | 2) = 3
Bitwise AND of the pair(arr[0], arr[2]) is (arr[0] & arr[2]) = (1 & 2) = 1
Bitwise AND of the pair(arr[0], arr[1]) is (arr[0] | arr[1]) = (1 | 2) = 1
Bitwise AND of the pair(arr[1], arr[2]) is (arr[1] & arr[2]) = (2 & 1) = 0
Bitwise AND of the pair(arr[0], arr[1]) is arr[0] | arr[1] = (2 | 1) = 3
Therefore, the required output is 1.
Input: arr[] = {1, 2, 3, 1, 2, 2}
Output: 4
Naive Approach: The simplest approach to solve the problem is to traverse the array and generate all possible pairs of the given array. For each pair, check if Bitwise And of the pair is equal to Bitwise OR of that pair or not. If found to be true, then increment the counter. Finally, print the value of the counter.
Efficient Approach: To optimize the above approach the idea is based on the following observations:
0 & 0 = 0 and 0 | 0 = 0
0 & 1 = 0 and 0 | 1 = 1
1 & 0 = 0 and 1 | 0 = 1
1 & 1 = 1 and 1 | 1 = 1
Therefore, If both the elements of a pair are equal, only then, bitwise AND(&) and Bitwise OR(|) of the pair becomes equal.
Follow the steps below to solve the problem:
- Initialize a variable, say cntPairs to store the count of pairs whose Bitwise AND(&) value and Bitwise OR(|) value is equal.
- Create a map, say mp to store the frequency of all distinct elements of the given array.
- Traverse the given array and store the frequency of all distinct elements of the given array in mp.
- Traverse map and check if frequency, say freq is greater than 1 then update cntPairs += (freq * (freq – 1)) / 2.
- Finally, print the value of cntPairs.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int countPairs( int arr[], int N)
{
int cntPairs = 0;
map< int , int > mp;
for ( int i = 0; i < N; i++) {
mp[arr[i]]++;
}
for ( auto freq: mp) {
cntPairs += (freq.second *
(freq.second - 1)) / 2;
}
return cntPairs;
}
int main()
{
int arr[] = { 1, 2, 3, 1, 2, 2 };
int N = sizeof (arr) / sizeof (arr[0]);
cout<<countPairs(arr, N);
}
|
Java
import java.io.*;
import java.util.*;
class GFG{
static int countPairs( int [] arr, int N)
{
int cntPairs = 0 ;
HashMap<Integer, Integer> mp = new HashMap<>();
for ( int i = 0 ; i < N; i++)
{
mp.put(arr[i],
mp.getOrDefault(arr[i], 0 ) + 1 );
}
for (Map.Entry<Integer, Integer> freq : mp.entrySet())
{
cntPairs += (freq.getValue() *
(freq.getValue() - 1 )) / 2 ;
}
return cntPairs;
}
public static void main(String[] args)
{
int [] arr = { 1 , 2 , 3 , 1 , 2 , 2 };
int N = arr.length;
System.out.println(countPairs(arr, N));
}
}
|
Python3
def countPairs(arr, N):
cntPairs = 0
mp = {}
for i in range ( 0 , N):
if arr[i] in mp:
mp[arr[i]] = mp[arr[i]] + 1
else :
mp[arr[i]] = 1
for freq in mp:
cntPairs + = int ((mp[freq] *
(mp[freq] - 1 )) / 2 )
return cntPairs
if __name__ = = "__main__" :
arr = [ 1 , 2 , 3 , 1 , 2 , 2 ]
N = len (arr)
print (countPairs(arr, N))
|
C#
using System;
using System.Collections.Generic;
class GFG{
static int countPairs( int [] arr, int N)
{
int cntPairs = 0;
Dictionary< int ,
int > mp = new Dictionary< int ,
int >();
for ( int i = 0; i < N; i++)
{
if (!mp.ContainsKey(arr[i]))
mp.Add(arr[i], 1);
else
mp[arr[i]] = mp[arr[i]] + 1;
}
foreach (KeyValuePair< int , int > freq in mp)
{
cntPairs += (freq.Value *
(freq.Value - 1)) / 2;
}
return cntPairs;
}
public static void Main()
{
int [] arr = { 1, 2, 3, 1, 2, 2 };
int N = arr.Length;
Console.WriteLine(countPairs(arr, N));
}
}
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Javascript
<script>
function countPairs(arr, N)
{
var cntPairs = 0;
var mp = new Map();
for ( var i = 0; i < N; i++) {
if (mp.has(arr[i]))
mp.set(arr[i], mp.get(arr[i])+1)
else
mp.set(arr[i], 1);
}
mp.forEach((value, key) => {
cntPairs += parseInt((value *
(value - 1)) / 2);
});
return cntPairs;
}
var arr = [1, 2, 3, 1, 2, 2 ];
var N = arr.length;
document.write( countPairs(arr, N));
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(N)
Last Updated :
20 May, 2021
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