Cuente pares en una array de modo que ambos elementos tengan bits de conjunto iguales

Dada una array arr con elementos únicos, la tarea es contar el número total de pares de elementos que tienen el mismo conjunto de bits.
Ejemplos: 
 

Entrada: arr[] = {2, 5, 8, 1, 3} 
Salida: 4 
Los recuentos de bits establecidos para {2, 5, 8, 1, 3} son {1, 2, 1, 1, 2} 
Todos los pares con el mismo conjunto de bits son {2, 8}, {2, 1}, {5, 3}, {8, 1}
Entrada: arr[] = {1, 11, 7, 3} 
Salida: 1 
Solo par posible es {11, 7} 
 

Acercarse: 
 

• Recorra la array de izquierda a derecha y cuente el número total de bits establecidos de cada entero.
• Use un mapa para almacenar el número de elementos con el mismo recuento de bits establecidos con bits establecidos como clave y recuento como valor.
• Luego iterar a través de los elementos del mapa y calcular cuántos pares de dos elementos se pueden formar a partir de n elementos (para cada elemento del mapa), es decir , (n * (n-1)) / 2 .
• El resultado final será la suma de la salida del paso anterior para cada elemento del mapa.

A continuación se muestra la implementación del enfoque anterior: 
 

// 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):
     
    # map 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>

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