Given an array “arr[0..n-1]” of integers, calculate sum of “arr[i] & arr[j]” for all the pairs in the given where i < j. Here & is bitwise AND operator. Expected time complexity is O(n).
Examples :
Input: arr[] = {5, 10, 15}
Output: 15
Required Value = (5 & 10) + (5 & 15) + (10 & 15)
= 0 + 5 + 10
= 15
Input: arr[] = {1, 2, 3, 4}
Output: 3
Required Value = (1 & 2) + (1 & 3) + (1 & 4) +
(2 & 3) + (2 & 4) + (3 & 4)
= 0 + 1 + 0 + 2 + 0 + 0
= 3
A Brute Force approach is to run two loops and time complexity is O(n2).
C++
// A Simple C++ program to compute sum of bitwise AND// of all pairs#include <bits/stdc++.h>using namespace std;
// Returns value of "arr[0] & arr[1] + arr[0] & arr[2] +// ... arr[i] & arr[j] + ..... arr[n-2] & arr[n-1]"int pairAndSum(int arr[], int n)
{ int ans = 0; // Initialize result
// Consider all pairs (arr[i], arr[j) such that
// i < j
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
ans += arr[i] & arr[j];
return ans;
}// Driver program to test above functionint main()
{ int arr[] = { 5, 10, 15 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << pairAndSum(arr, n) << endl;
return 0;
}// This code is contributed by Aditya Kumar (adityakumar129) |
C
// A Simple C++ program to compute sum of bitwise AND// of all pairs#include <stdio.h>// Returns value of "arr[0] & arr[1] + arr[0] & arr[2] +// ... arr[i] & arr[j] + ..... arr[n-2] & arr[n-1]"int pairAndSum(int arr[], int n)
{ int ans = 0; // Initialize result
// Consider all pairs (arr[i], arr[j) such that
// i < j
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
ans += arr[i] & arr[j];
return ans;
}// Driver program to test above functionint main()
{ int arr[] = { 5, 10, 15 };
int n = sizeof(arr) / sizeof(arr[0]);
printf("%d\n",pairAndSum(arr, n));
return 0;
}// This code is contributed by Aditya Kumar (adityakumar129) |
Java
// A Simple Java program to compute// sum of bitwise AND of all pairsimport java.io.*;
class GFG {
// Returns value of "arr[0] & arr[1] +
// arr[0] & arr[2] + ... arr[i] & arr[j] +
// ..... arr[n-2] & arr[n-1]"
static int pairAndSum(int arr[], int n)
{
int ans = 0; // Initialize result
// Consider all pairs (arr[i], arr[j)
// such that i < j
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
ans += arr[i] & arr[j];
return ans;
}
// Driver program to test above function
public static void main(String args[])
{
int arr[] = { 5, 10, 15 };
int n = arr.length;
System.out.println(pairAndSum(arr, n));
}
}// This code is contributed by Aditya Kumar (adityakumar129) |
Python3
# A Simple Python 3 program to compute# sum of bitwise AND of all pairs# Returns value of "arr[0] & arr[1] +# arr[0] & arr[2] + ... arr[i] & arr[j] +# ..... arr[n-2] & arr[n-1]"def pairAndSum(arr, n) :
ans = 0 # Initialize result
# Consider all pairs (arr[i], arr[j)
# such that i < j
for i in range(0,n) :
for j in range((i+1),n) :
ans = ans + arr[i] & arr[j]
return ans
# Driver program to test above functionarr = [5, 10, 15]
n = len(arr)
print(pairAndSum(arr, n))
# This code is contributed by Nikita Tiwari. |
C#
// A Simple C# program to compute// sum of bitwise AND of all pairsusing System;
class GFG {
// Returns value of "arr[0] & arr[1] +
// arr[0] & arr[2] + ... arr[i] & arr[j] +
// ..... arr[n-2] & arr[n-1]"
static int pairAndSum(int []arr, int n)
{
int ans = 0; // Initialize result
// Consider all pairs (arr[i], arr[j)
// such that i < j
for (int i = 0; i < n; i++)
for (int j = i+1; j < n; j++)
ans += arr[i] & arr[j];
return ans;
}
// Driver program to test above function
public static void Main()
{
int []arr = {5, 10, 15};
int n = arr.Length;
Console.Write(pairAndSum(arr, n) );
}
} // This code is contributed by nitin mittal. |
PHP
<?php// A Simple PHP program to // compute sum of bitwise // AND of all pairs// Returns value of "arr[0] &// arr[1] + arr[0] & arr[2] + // ... arr[i] & arr[j] + .....// arr[n-2] & arr[n-1]"function pairAndSum($arr, $n)
{ // Initialize result
$ans = 0;
// Consider all pairs (arr[i],
// arr[j) such that i < j
for ($i = 0; $i < $n; $i++)
for ( $j = $i + 1; $j < $n; $j++)
$ans += $arr[$i] & $arr[$j];
return $ans;
}// Driver Code$arr = array(5, 10, 15);
$n = sizeof($arr) ;
echo pairAndSum($arr, $n), "\n";
// This code is contributed by m_kit?> |
Javascript
<script> // A Simple Javascript program to compute
// sum of bitwise AND of all pairs
// Returns value of "arr[0] & arr[1] +
// arr[0] & arr[2] + ... arr[i] & arr[j] +
// ..... arr[n-2] & arr[n-1]"
function pairAndSum(arr, n)
{
let ans = 0; // Initialize result
// Consider all pairs (arr[i], arr[j)
// such that i < j
for (let i = 0; i < n; i++)
for (let j = i+1; j < n; j++)
ans += arr[i] & arr[j];
return ans;
}
let arr = [5, 10, 15];
let n = arr.length;
document.write(pairAndSum(arr, n));
</script> |
Output :
15
Time Complexity: O(n2)
Auxiliary Space: O(1)
An Efficient Solution can solve this problem in O(n) time. The assumption here is that integers are represented using 32 bits.
The idea is to count number of set bits at every i’th position (i>=0 && i<=31). Any i’th bit of the AND of two numbers is 1 if the corresponding bit in both the numbers is equal to 1.
Let k be the count of set bits at i’th position. Total number of pairs with i’th set bit would be kC2 = k*(k-1)/2 (Count k means there are k numbers which have i’th set bit). Every such pair adds 2i to total sum. Similarly, we work for all other places and add the sum to our final answer.
This idea is similar to this. Below is the implementation.
C++
// An efficient C++ program to compute sum of bitwise AND// of all pairs#include <bits/stdc++.h>using namespace std;
// Returns value of "arr[0] & arr[1] + arr[0] & arr[2] +// ... arr[i] & arr[j] + ..... arr[n-2] & arr[n-1]"int pairAndSum(int arr[], int n)
{ int ans = 0; // Initialize result
// Traverse over all bits
for (int i = 0; i < 32; i++) {
// Count number of elements with i'th bit set
int k = 0; // Initialize the count
for (int j = 0; j < n; j++)
if ((arr[j] & (1 << i)))
k++;
// There are k set bits, means k(k-1)/2 pairs.
// Every pair adds 2^i to the answer. Therefore,
// we add "2^i * [k*(k-1)/2]" to the answer.
ans += (1 << i) * (k * (k - 1) / 2);
}
return ans;
}// Driver program to test above functionint main()
{ int arr[] = { 5, 10, 15 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << pairAndSum(arr, n) << endl;
return 0;
} |
C
// An efficient C++ program to compute sum of bitwise AND// of all pairs#include <stdio.h>// Returns value of "arr[0] & arr[1] + arr[0] & arr[2] +// ... arr[i] & arr[j] + ..... arr[n-2] & arr[n-1]"int pairAndSum(int arr[], int n)
{ int ans = 0; // Initialize result
// Traverse over all bits
for (int i = 0; i < 32; i++) {
// Count number of elements with i'th bit set
int k = 0; // Initialize the count
for (int j = 0; j < n; j++)
if ((arr[j] & (1 << i)))
k++;
// There are k set bits, means k(k-1)/2 pairs.
// Every pair adds 2^i to the answer. Therefore,
// we add "2^i * [k*(k-1)/2]" to the answer.
ans += (1 << i) * (k * (k - 1) / 2);
}
return ans;
}// Driver program to test above functionint main()
{ int arr[] = { 5, 10, 15 };
int n = sizeof(arr) / sizeof(arr[0]);
printf("%d\n",pairAndSum(arr, n));
return 0;
}// This code is contributed by Aditya Kumar (adityakumar129) |
Java
// An efficient Java program to compute// sum of bitwise AND of all pairsimport java.io.*;
class GFG {
// Returns value of "arr[0] & arr[1] +
// arr[0] & arr[2] + ... arr[i] & arr[j] +
// ..... arr[n-2] & arr[n-1]"
static int pairAndSum(int arr[], int n)
{
int ans = 0; // Initialize result
// Traverse over all bits
for (int i = 0; i < 32; i++) {
// Count number of elements with i'th bit set
// Initialize the count
int k = 0;
for (int j = 0; j < n; j++) {
if ((arr[j] & (1 << i)) != 0)
k++;
}
// There are k set bits, means k(k-1)/2 pairs.
// Every pair adds 2^i to the answer. Therefore,
// we add "2^i * [k*(k-1)/2]" to the answer.
ans += (1 << i) * (k * (k - 1) / 2);
}
return ans;
}
// Driver program to test above function
public static void main(String args[])
{
int arr[] = { 5, 10, 15 };
int n = arr.length;
System.out.println(pairAndSum(arr, n));
}
}// An efficient C++ program to compute sum of bitwise AND// of all pairs#include <stdio.h>// Returns value of "arr[0] & arr[1] + arr[0] & arr[2] +// ... arr[i] & arr[j] + ..... arr[n-2] & arr[n-1]"int pairAndSum(int arr[], int n)
{ int ans = 0; // Initialize result
// Traverse over all bits
for (int i = 0; i < 32; i++) {
// Count number of elements with i'th bit set
int k = 0; // Initialize the count
for (int j = 0; j < n; j++)
if ((arr[j] & (1 << i)))
k++;
// There are k set bits, means k(k-1)/2 pairs.
// Every pair adds 2^i to the answer. Therefore,
// we add "2^i * [k*(k-1)/2]" to the answer.
ans += (1 << i) * (k * (k - 1) / 2);
}
return ans;
}// Driver program to test above functionint main()
{ int arr[] = { 5, 10, 15 };
int n = sizeof(arr) / sizeof(arr[0]);
printf("%d\n",pairAndSum(arr, n));
return 0;
}// This code is contributed by Aditya Kumar (adityakumar129) |
Python3
# An efficient Python 3 program to# compute sum of bitwise AND of all pairs# Returns value of "arr[0] & arr[1] +# arr[0] & arr[2] + ... arr[i] & arr[j] +# ..... arr[n-2] & arr[n-1]"def pairAndSum(arr, n) :
ans = 0 # Initialize result
# Traverse over all bits
for i in range(0,32) :
# Count number of elements with i'th bit set
# Initialize the count
k = 0
for j in range(0,n) :
if ( (arr[j] & (1 << i)) ) :
k = k + 1
# There are k set bits, means k(k-1)/2 pairs.
# Every pair adds 2^i to the answer. Therefore,
# we add "2^i * [k*(k-1)/2]" to the answer.
ans = ans + (1 << i) * (k * (k - 1) // 2)
return ans
# Driver program to test above functionarr = [5, 10, 15]
n = len(arr)
print(pairAndSum(arr, n))
# This code is contributed by Nikita Tiwari. |
C#
// An efficient C# program to compute// sum of bitwise AND of all pairsusing System;
class GFG {
// Returns value of "arr[0] & arr[1] +
// arr[0] & arr[2] + ... arr[i] & arr[j] +
// ..... arr[n-2] & arr[n-1]"
static int pairAndSum(int []arr, int n)
{
int ans = 0; // Initialize result
// Traverse over all bits
for (int i = 0; i < 32; i++)
{
// Count number of elements with
// i'th bit set Initialize the count
int k = 0;
for (int j = 0; j < n; j++)
{
if ((arr[j] & (1 << i))!=0)
k++;
}
// There are k set bits, means
// k(k-1)/2 pairs. Every pair
// adds 2^i to the answer.
// Therefore, we add "2^i *
// [k*(k-1)/2]" to the answer.
ans += (1 << i) * (k * (k - 1)/2);
}
return ans;
}
// Driver program to test above function
public static void Main()
{
int []arr = new int[]{5, 10, 15};
int n = arr.Length;
Console.Write(pairAndSum(arr, n));
}
}/* This code is contributed by smitha*/ |
PHP
<?php// An efficient PHP program to // compute sum of bitwise AND// of all pairs// Returns value of "arr[0] & // arr[1] + arr[0] & arr[2] + // ... arr[i] & arr[j] + ..... // arr[n-2] & arr[n-1]"function pairAndSum($arr, $n)
{ // Initialize result
$ans = 0;
// Traverse over all bits
for ($i = 0; $i < 32; $i++)
{
// Count number of elements
// with i'th bit set
// Initialize the count
$k = 0;
for ($j = 0; $j < $n; $j++)
if (($arr[$j] & (1 << $i)) )
$k++;
// There are k set bits,
// means k(k-1)/2 pairs.
// Every pair adds 2^i to
// the answer. Therefore,
// we add "2^i * [k*(k-1)/2]"
// to the answer.
$ans += (1 << $i) * ($k * ($k - 1) / 2);
}
return $ans;
} // Driver Code
$arr = array(5, 10, 15);
$n = sizeof($arr);
echo pairAndSum($arr, $n) ;
// This code is contributed by nitin mittal.?> |
Javascript
<script> // An efficient Javascript program to compute
// sum of bitwise AND of all pairs
// Returns value of "arr[0] & arr[1] +
// arr[0] & arr[2] + ... arr[i] & arr[j] +
// ..... arr[n-2] & arr[n-1]"
function pairAndSum(arr, n)
{
let ans = 0; // Initialize result
// Traverse over all bits
for (let i = 0; i < 32; i++)
{
// Count number of elements with
// i'th bit set Initialize the count
let k = 0;
for (let j = 0; j < n; j++)
{
if ((arr[j] & (1 << i))!=0)
k++;
}
// There are k set bits, means
// k(k-1)/2 pairs. Every pair
// adds 2^i to the answer.
// Therefore, we add "2^i *
// [k*(k-1)/2]" to the answer.
ans += (1 << i) * (k * (k - 1)/2);
}
return ans;
}
let arr = [5, 10, 15];
let n = arr.length;
document.write(pairAndSum(arr, n));
// This code is contributed by rameshtravel07.
</script> |
Output:
15
Time Complexity: O(n)
Auxiliary Space: O(1)
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