# Sum of array elements whose count of set bits are unique

• Last Updated : 14 Sep, 2021

Given an array arr[] consisting of N positive integers, the task is to find the sum of all array elements having a distinct count of set bits in the array.

Examples:

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Input: arr[] = {8, 3, 7, 5, 3}
Output: 15
Explanation:
The count of set bits in each array of elements is:

1. arr = 8 = (1000)2, has 1 set bits.
2. arr = 3 = (11)2, has 2 set bits.
3. arr = 7 = (111)2, has 3 set bits.
4. arr = 5 = (101)2, has 2 set bits.
5. arr = 3 = (11)2, has 2 set bits.

Therefore, the number of array elements whose count of set bits are unique are 8 and 7. Therefore, the required sum = 8 + 7 = 15.

Input: arr[] = {4, 5, 3, 5, 3, 2}
Output:

Approach: The idea is to store the element with the corresponding count of set bits on a map, then find the sum of elements having a unique count of set bits. Follow the steps below to solve the problem:

• Initialize a variable, say sum to store the resultant sum of elements, and a Map, say M that stores the elements having a particular count of set bits.
• Traverse the array arr[] and store the element arr[i] according to the set bit count in a Map M.
• Now, traverse the map and if the frequency of any set bit count is 1, then add the corresponding value associated with it in the variable sum.
• After completing the above steps, print the value of the sum as the result.

Below is the implementation of the above approach:

## C++

 `// C++ program for the approach``#include ``#include ``using` `namespace` `std;` `// Function to count the number``// of set bits in an integer N``int` `setBitCount(``int` `n)``{``    ` `    ``// Stores the count of set bits``    ``int` `ans = 0;``    ` `    ``// Iterate until N is non-zero``    ``while` `(n)``    ``{``        ``ans += n & 1;``        ``n >>= 1;``    ``}``    ` `    ``// Stores the resultant count``    ``return` `ans;``}` `// Function to calculate sum of all array``// elements whose count of set bits are unique``int` `getSum(``int` `*arr, ``int` `n)``{``    ` `    ``// Stores frequency of all possible``    ``// count of set bits``    ``map<``int``, ``int``> mp;``    ` `    ``// Stores the sum of array elements``    ``int` `ans = 0;``    ` `    ``// Traverse the array``    ``for``(``int` `i = 0; i < n; i++)``    ``{``        ` `        ``// Count the number of set bits``        ``int` `key = setBitCount(arr[i]);``        ``mp[key] += 1;``    ``}``    ` `    ``// Traverse the array``    ``// And Update the value of ans``    ``for``(``int` `i = 0; i < n; i++)``    ``{``        ``int` `key = setBitCount(arr[i]);``        ` `        ``// If frequency is 1``        ``if` `(mp[key] == 1)``            ``ans += arr[i];``    ``}``    ``cout << ans;``}` `// Driver Code``int` `main()``{``    ``int` `arr = {8, 3, 7, 5, 3};``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ` `    ``getSum(arr, n);``    ` `    ``return` `0;``}` `// This code is contributed by rohitsingh07052`

## Java

 `// Java program for the approach``import` `java.util.*;``class` `GFG``{` `  ``// Function to count the number``  ``// of set bits in an integer N``  ``static` `int` `setBitCount(``int` `n)``  ``{` `    ``// Stores the count of set bits``    ``int` `ans = ``0``;` `    ``// Iterate until N is non-zero``    ``while` `(n != ``0``)``    ``{``      ``ans += n & ``1``;``      ``n >>= ``1``;``    ``}` `    ``// Stores the resultant count``    ``return` `ans;``  ``}` `  ``// Function to calculate sum of all array``  ``// elements whose count of set bits are unique``  ``static` `void` `getSum(``int` `[]arr, ``int` `n)``  ``{` `    ``// Stores frequency of all possible``    ``// count of set bits``    ``Map mp = ``new` `HashMap();` `    ``// Stores the sum of array elements``    ``int` `ans = ``0``;` `    ``// Traverse the array``    ``for``(``int` `i = ``0``; i < n; i++)``    ``{` `      ``// Count the number of set bits``      ``int` `key = setBitCount(arr[i]);``      ``if``(mp.containsKey(key))``        ``mp.put(key,mp.get(key) + ``1``);``      ``else``        ``mp.put(key, ``1``);` `    ``}` `    ``// Traverse the array``    ``// And Update the value of ans``    ``for``(``int` `i = ``0``; i < n; i++)``    ``{``      ``int` `key = setBitCount(arr[i]);` `      ``// If frequency is 1``      ``if` `(mp.containsKey(key) && mp.get(key) == ``1``)``        ``ans += arr[i];``    ``}``    ``System.out.print(ans);``  ``}` `  ``// Driver Code``  ``public` `static` `void` `main(String args[])``  ``{``    ``int` `[]arr = {``8``, ``3``, ``7``, ``5``, ``3``};``    ``int` `n = arr.length;` `    ``getSum(arr, n);``  ``}``}` `// This code is contributed by SURENDRA_GANGWAR.`

## Python3

 `# Python3 program for the approach` `# Function to count the number``# of set bits in an integer N``def` `setBitCount(n):``  ` `    ``# Stores the count of set bits``    ``ans ``=` `0``    ` `    ``# Iterate until N is non-zero``    ``while` `n:` `        ``ans ``+``=` `n & ``1``        ``n >>``=` `1``        ` `    ``# Stores the resultant count``    ``return` `ans`  `# Function to calculate sum of all array``# elements whose count of set bits are unique``def` `getSum(arr):``  ` `    ``# Stores frequency of all possible``    ``# count of set bits``    ``mp ``=` `{}``    ` `    ``# Stores the sum of array elements``    ``ans ``=` `0``    ` `    ``# Traverse the array``    ``for` `i ``in` `arr:``      ` `        ``# Count the number of set bits``        ``key ``=` `setBitCount(i)``        ``mp[key] ``=` `[``0``, i]` `    ``# Traverse the array``    ``for` `i ``in` `arr:``        ``key ``=` `setBitCount(i)``        ``mp[key][``0``] ``+``=` `1``    ` `    ``# Update the value of ans``    ``for` `i ``in` `mp:``      ` `        ``# If frequency is 1``        ``if` `mp[i][``0``] ``=``=` `1``:``            ``ans ``+``=` `mp[i][``1``]` `    ``print``(ans)` `# Driver Code` `arr ``=` `[``8``, ``3``, ``7``, ``5``, ``3``]``getSum(arr)`

## C#

 `// C# program for the approach``using` `System;``using` `System.Collections.Generic;` `class` `solution{``  ` `// Function to count the number``// of set bits in an integer N``static` `int` `setBitCount(``int` `n)``{``    ` `    ``// Stores the count of set bits``    ``int` `ans = 0;``    ` `    ``// Iterate until N is non-zero``    ``while` `(n>0)``    ``{``        ``ans += n & 1;``        ``n >>= 1;``    ``}``    ` `    ``// Stores the resultant count``    ``return` `ans;``}` `// Function to calculate sum of all array``// elements whose count of set bits are unique``static` `void` `getSum(``int` `[]arr, ``int` `n)``{``    ` `    ``// Stores frequency of all possible``    ``// count of set bits``    ``Dictionary<``int``, ``int``> mp = ``new` `Dictionary<``int``,``int``>();``    ` `    ``// Stores the sum of array elements``    ``int` `ans = 0;``    ` `    ``// Traverse the array``    ``for``(``int` `i = 0; i < n; i++)``    ``{``        ` `        ``// Count the number of set bits``        ``int` `key = setBitCount(arr[i]);``        ``if``(mp.ContainsKey(key))``          ``mp[key] += 1;``        ``else``          ``mp[key] = 1;``    ``}``    ` `    ``// Traverse the array``    ``// And Update the value of ans``    ``for``(``int` `i = 0; i < n; i++)``    ``{``        ``int` `key = setBitCount(arr[i]);``        ` `        ``// If frequency is 1``        ``if` `(mp[key] == 1)``            ``ans += arr[i];``    ``}``   ``Console.Write(ans);``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `[]arr = {8, 3, 7, 5, 3};``    ``int` `n = arr.Length;``    ` `    ``getSum(arr, n);``}``}` `// This code is contributed by ipg2016107.`

## Javascript

 `        `
Output:
`15`

Time Complexity: O(N)
Auxiliary Space: O(1)

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