Count pairs in an array such that both elements has equal set bits
Last Updated :
03 Jan, 2023
Given an array arr [] of size N with unique elements, the task is to count the total number of pairs of elements that have equal set bits count.
Examples:
Input: arr[] = {2, 5, 8, 1, 3}
Output: 4
Set bits counts for {2, 5, 8, 1, 3} are {1, 2, 1, 1, 2}
All pairs with same set bits count are {2, 8}, {2, 1}, {5, 3}, {8, 1}
Input: arr[] = {1, 11, 7, 3}
Output: 1
Only possible pair is {11, 7}
Approach:
- Traverse the array from left to right and count total number of set bits of each integer.
- Use a map to store the number of elements with same count of set bits with set bits as key, and count as value.
- Then iterator through map elements, and calculate how many two-element pairs can be formed from n elements (for each element of the map) i.e. (n * (n-1)) / 2.
- The final result will be the sum of output from the previous step for every element of the map.
Below is the implementation of the above approach:
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count all pairs
// with equal set bits count
int totalPairs(int arr[], int n)
{
// map to store count of elements
// with equal number of set bits
map<int, int> m;
for (int i = 0; i < n; i++) {
// inbuilt function that returns the
// count of set bits of the number
m[__builtin_popcount(arr[i])]++;
}
map<int, int>::iterator it;
int result = 0;
for (it = m.begin(); it != m.end(); it++) {
// there can be (n*(n-1)/2) unique two-
// element pairs to choose from n elements
result
+= (*it).second * ((*it).second - 1) / 2;
}
return result;
}
// Driver code
int main()
{
int arr[] = { 7, 5, 3, 9, 1, 2 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << totalPairs(arr, n);
return 0;
}
import java.util.*;
class GFG {
/* Function to get no of set
bits in binary representation
of passed binary no. */
static int countSetBits(int n)
{
int count = 0;
while (n > 0)
{
n &= (n - 1) ;
count++;
}
return count;
}
// Function to count all pairs
// with equal set bits count
static int totalPairs(int arr[], int n)
{
// map to store count of elements
// with equal number of set bits
HashMap<Integer, Integer> m = new HashMap<>();
for (int i = 0; i < n; i++) {
// function that returns the
// count of set bits of the number
int count = countSetBits(arr[i]);
if(m.containsKey(count))
m.put(count, m.get(count) + 1);
else
m.put(count, 1);
}
int result = 0;
for (Map.Entry<Integer, Integer> entry : m.entrySet()) {
int value = entry.getValue();
// there can be (n*(n-1)/2) unique two-
// element pairs to choose from n elements
result += ((value * (value -1)) / 2);
}
return result;
}
public static void main (String[] args) {
int arr[] = { 7, 5, 3, 9, 1, 2 };
int n = arr.length;
System.out.println(totalPairs(arr, n));
}
}
# Python3 implementation of the above approach
# Function to count all pairs
# with equal set bits count
def totalPairs(arr, n):
# dictionary to store count of elements
# with equal number of set bits
m = dict()
for i in range(n):
# inbuilt function that returns the
# count of set bits of the number
x = bin(arr[i]).count('1')
m[x] = m.get(x, 0) + 1;
result = 0
for it in m:
# there can be (n*(n-1)/2) unique two-
# element pairs to choose from n elements
result+= (m[it] * (m[it] - 1)) // 2
return result
# Driver code
arr = [7, 5, 3, 9, 1, 2]
n = len(arr)
print(totalPairs(arr, n))
# This code is contributed by mohit kumar
// C# program to rearrange a string so that all same
// characters become atleast d distance away
using System;
using System.Collections.Generic;
class GFG
{
/* Function to get no of set
bits in binary representation
of passed binary no. */
static int countSetBits(int n)
{
int count = 0;
while (n > 0)
{
n &= (n - 1) ;
count++;
}
return count;
}
// Function to count all pairs
// with equal set bits count
static int totalPairs(int []arr, int n)
{
// map to store count of elements
// with equal number of set bits
Dictionary<int,int> mp = new Dictionary<int,int>();
for (int i = 0 ; i < n; i++)
{
// function that returns the
// count of set bits of the number
int count = countSetBits(arr[i]);
if(mp.ContainsKey(count))
{
var val = mp[count];
mp.Remove(count);
mp.Add(count, val + 1);
}
else
{
mp.Add(count, 1);
}
}
int result = 0;
foreach(KeyValuePair<int, int> entry in mp){
int values = entry.Value;
// there can be (n*(n-1)/2) unique two-
// element pairs to choose from n elements
result += ((values * (values -1)) / 2);
}
return result;
}
// Driver code
public static void Main (String[] args)
{
int []arr = { 7, 5, 3, 9, 1, 2 };
int n = arr.Length;
Console.WriteLine(totalPairs(arr, n));
}
}
// This code is contributed by Princi Singh
<script>
// Javascript implementation of above approach
/* Function to get no of set
bits in binary representation
of passed binary no. */
function countSetBits(n)
{
var count = 0;
while (n > 0)
{
n &= (n - 1) ;
count++;
}
return count;
}
// Function to count all pairs
// with equal set bits count
function totalPairs(arr, n)
{
// map to store count of elements
// with equal number of set bits
var m = new Map();
for (var i = 0; i < n; i++) {
// inbuilt function that returns the
// count of set bits of the number
if(m.has(arr[i]))
{
m.set(countSetBits(arr[i]), m.get(countSetBits(arr[i]))+1);
}
else{
m.set(countSetBits(arr[i]), 1);
}
}
var result = 0;
m.forEach((value, key) => {
// there can be (n*(n-1)/2) unique two-
// element pairs to choose from n elements
result += parseInt(key * (key - 1) / 2);
});
return result;
}
// Driver code
var arr = [7, 5, 3, 9, 1, 2 ];
var n = arr.length;
document.write( totalPairs(arr, n));
</script>
Time Complexity: O(n log n), where n is the number of elements in given array
Auxiliary Space: O(n)
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