Maximum set bit sum in array without considering adjacent elements

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:

  1. 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).
  2. Now, the problem is reduced to find the maximum sum in the array such that no two elements are adjacent.
  3. 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.
  4. 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).
  5. At the end of the loop return max of incl and excl.

Below is the implementation of the above approach:

C++

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// 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;

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Java

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// 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

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Python3

# 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#

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// 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.

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PHP

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<?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    
?>

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Output:

9

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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