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

Sum of array elements whose count of set bits are unique

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:

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

arr[0] = 8 = (1000)2, has 1 set bits.

arr[1] = 3 = (11)2, has 2 set bits.

arr[2] = 7 = (111)2, has 3 set bits.

arr[3] = 5 = (101)2, has 2 set bits.

arr[4] = 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, required sum = 8 + 7 = 15.

Input: arr[] = {4, 5, 3, 5, 3, 2}Output: 0

Approach: The idea is to store the element with the corresponding count of set bit in a map then find the sum of elements having a unique count of set bit. 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 bit.

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:

Python3

def setBitCount(n):

ans = 0

while n:

ans += n & 1

n >>= 1

return ans

def getSum(arr):

mp = {}

ans = 0

for i in arr:

key = setBitCount(i)

mp[key] = [0, i]

for i in arr:

key = setBitCount(i)

mp[key][0] += 1

for i in mp:

if mp[i][0] == 1:

ans += mp[i][1]

print(ans)

arr = [8, 3, 7, 5, 3]

getSum(arr)

Output:

15

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

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