Maximum set bit sum in array without considering adjacent elements
Last Updated :
11 Jul, 2025
Given an array of integers arr[]. The task is to find the maximum sum of set bits(of the array elements) without adding the set bits of adjacent elements of the array.
Examples:
Input : arr[] = {1, 2, 4, 5, 6, 7, 20, 25}
Output : 9
Input : arr[] = {5, 7, 9, 5, 13, 7, 20, 25}
Output : 11
Approach:
- First of all, find the total number of set bits for every element of the array and store them in a different array or the same array(to avoid using extra space).
- Now, the problem is reduced to find the maximum sum in the array such that no two elements are adjacent.
- Loop for all elements in arr[] and maintain two sums incl and excl where incl = Max sum including the previous element and excl = Max sum excluding the previous element.
- Max sum excluding the current element will be max(incl, excl) and max sum including the current element will be excl + current element (Note that only excl is considered because elements cannot be adjacent).
- At the end of the loop return max of incl and excl.
Below is the implementation of the above approach:
C++
// C++ program to maximum set bit sum in array
// without considering adjacent elements
#include<bits/stdc++.h>
using namespace std;
// Function to count total number
// of set bits in an integer
int bit(int n)
{
int count = 0;
while(n)
{
count++;
n = n & (n - 1);
}
return count;
}
// Maximum sum of set bits
int maxSumOfBits(int arr[], int n)
{
// Calculate total number of
// set bits for every element
// of the array
for(int i = 0; i < n; i++)
{
// find total set bits for
// each number and store
// back into the array
arr[i] = bit(arr[i]);
}
int incl = arr[0];
int excl = 0;
int excl_new;
for (int i = 1; i < n; i++)
{
// current max excluding i
excl_new = (incl > excl) ?
incl : excl;
// current max including i
incl = excl + arr[i];
excl = excl_new;
}
// return max of incl and excl
return ((incl > excl) ?
incl : excl);
}
// Driver code
int main()
{
int arr[] = {1, 2, 4, 5,
6, 7, 20, 25};
int n = sizeof(arr) / sizeof(arr[0]);
cout << maxSumOfBits(arr, n);
return 0;
}
// Java program to maximum set bit sum in array
// without considering adjacent elements
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{
// Function to count total number
// of set bits in an integer
static int bit(int n)
{
int count = 0;
while(n > 0)
{
count++;
n = n & (n - 1);
}
return count;
}
// Maximum sum of set bits
static int maxSumOfBits(int arr[], int n)
{
// Calculate total number of set bits
// for every element of the array
for(int i = 0; i < n; i++)
{
// find total set bits for
// each number and store
// back into the array
arr[i] = bit(arr[i]);
}
int incl = arr[0];
int excl = 0;
int excl_new;
for (int i = 1; i < n; i++)
{
// current max excluding i
excl_new = (incl > excl) ?
incl : excl;
// current max including i
incl = excl + arr[i];
excl = excl_new;
}
// return max of incl and excl
return ((incl > excl) ?
incl : excl);
}
// Driver code
public static void main(String args[])
{
int arr[] = {1, 2, 4, 5,
6, 7, 20, 25};
int n = arr.length;
System.out.print(maxSumOfBits(arr, n));
}
}
// This code is contributed
// by Subhadeep
# Python3 program to maximum set bit sum in
# array without considering adjacent elements
# Function to count total number
# of set bits in an integer
def bit(n):
count = 0
while(n):
count += 1
n = n & (n - 1)
return count
# Maximum sum of set bits
def maxSumOfBits(arr, n):
# Calculate total number of set bits
# for every element of the array
for i in range( n):
# find total set bits for each
# number and store back into the array
arr[i] = bit(arr[i])
incl = arr[0]
excl = 0
for i in range(1, n) :
# current max excluding i
if incl > excl:
excl_new = incl
else:
excl_new = excl
# current max including i
incl = excl + arr[i];
excl = excl_new
# return max of incl and excl
if incl > excl:
return incl
else :
return excl
# Driver code
if __name__ == "__main__":
arr = [1, 2, 4, 5,
6, 7, 20, 25]
n = len(arr)
print (maxSumOfBits(arr, n))
# This code is contributed by ita_c
// C# program to maximum set bit sum in array
// without considering adjacent elements
using System;
class GFG
{
// Function to count total number
// of set bits in an integer
static int bit(int n)
{
int count = 0;
while(n > 0)
{
count++;
n = n & (n - 1);
}
return count;
}
// Maximum sum of set bits
static int maxSumOfBits(int []arr, int n)
{
// Calculate total number of set bits
// for every element of the array
for(int i = 0; i < n; i++)
{
// find total set bits for
// each number and store
// back into the array
arr[i] = bit(arr[i]);
}
int incl = arr[0];
int excl = 0;
int excl_new;
for (int i = 1; i < n; i++)
{
// current max excluding i
excl_new = (incl > excl) ?
incl : excl;
// current max including i
incl = excl + arr[i];
excl = excl_new;
}
// return max of incl and excl
return ((incl > excl) ?
incl : excl);
}
// Driver code
public static void Main()
{
int []arr = {1, 2, 4, 5,
6, 7, 20, 25};
int n = arr.Length;
Console.WriteLine(maxSumOfBits(arr, n));
}
}
// This code is contributed
// by chandan_jnu.
<?php
// PHP program to maximum set bit sum in array
// without considering adjacent elements
// Function to count total number
// of set bits in an integer
function bit($n)
{
$count = 0;
while($n)
{
$count++;
$n = $n & ($n - 1);
}
return $count;
}
// Maximum sum of set bits
function maxSumOfBits($arr, $n)
{
// Calculate total number of
// set bits for every element
// of the array
for( $i = 0; $i < $n; $i++)
{
// find total set bits for
// each number and store
// back into the array
$arr[$i] = bit($arr[$i]);
}
$incl = $arr[0];
$excl = 0;
$excl_new;
for ($i = 1; $i < $n; $i++)
{
// current max excluding i
$excl_new = ($incl > $excl) ?
$incl : $excl;
// current max including i
$incl = $excl + $arr[$i];
$excl = $excl_new;
}
// return max of incl and excl
return (($incl > $excl) ?
$incl : $excl);
}
// Driver code
$arr = array(1, 2, 4, 5,
6, 7, 20, 25);
$n = sizeof($arr) / sizeof($arr[0]);
echo maxSumOfBits($arr, $n);
#This Code is Contributed by ajit
?>
<script>
// Javascript program to maximum
// set bit sum in array
// without considering adjacent elements
// Function to count total number
// of set bits in an integer
function bit(n)
{
let count = 0;
while(n > 0)
{
count++;
n = n & (n - 1);
}
return count;
}
// Maximum sum of set bits
function maxSumOfBits(arr, n)
{
// Calculate total number of set bits
// for every element of the array
for(let i = 0; i < n; i++)
{
// find total set bits for
// each number and store
// back into the array
arr[i] = bit(arr[i]);
}
let incl = arr[0];
let excl = 0;
let excl_new;
for (let i = 1; i < n; i++)
{
// current max excluding i
excl_new = (incl > excl) ? incl : excl;
// current max including i
incl = excl + arr[i];
excl = excl_new;
}
// return max of incl and excl
return ((incl > excl) ? incl : excl);
}
let arr = [1, 2, 4, 5, 6, 7, 20, 25];
let n = arr.length;
document.write(maxSumOfBits(arr, n));
</script>
complexity Analysis:
- Time Complexity: O(Nlogn)
- Auxiliary Space: O(1)
Note: Above code can be optimised to O(N) using __builtin_popcount function to count set bits in O(1) time.
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