# Sorting array elements with set bits equal to K

Given an array of integers and a number . The task is to sort only those elements of the array whose total set bits are equal to K. Sorting must be done at their relative positions only without affecting any other elements.

**Examples**:

Input: arr[] = {32, 1, 9, 4, 64, 2}, K = 1Output: 1 2 9 4 32 64 All of the elements except 9 has exactly 1 bit set. So, all elements except 9 are sorted without affecting the position of 9 in the input array.Input: arr[] = {2, 15, 12, 1, 3, 9}, K = 2Output: 2 15 3 1 9 12

**Approach:**

- Initialise two empty vectors.
- Traverse the array, from left to right and check the set bits of each element.
- Here, C++ inbuilt function __builtin_popcount() to count setbits.
- In first vector, insert the index of all elements with set bits equal to K.
- In second vector, insert the elements with set bits equal to K.
- Sort the second vector.
- Now, we have the index of all elements with set bit equals to K in sorted order and also all of the elements with set bit as K in sorted order.
- So, insert the elements of the second vector into the array at the indices present in first vector one by one.

Below is the implementation of the above approach:

## C++

`// C++ program for sorting array elements ` `// with set bits equal to K ` ` ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to sort elements with ` `// set bits equal to k ` `void` `sortWithSetbits(` `int` `arr[], ` `int` `n, ` `int` `k) ` `{ ` ` ` `// initialise two vectors ` ` ` `vector<` `int` `> v1, v2; ` ` ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` `if` `(__builtin_popcount(arr[i]) == k) { ` ` ` ` ` `// first vector contains indices of ` ` ` `// required element ` ` ` `v1.push_back(i); ` ` ` ` ` `// second vector contains ` ` ` `// required elements ` ` ` `v2.push_back(arr[i]); ` ` ` `} ` ` ` `} ` ` ` ` ` `// sorting the elements in second vector ` ` ` `sort(v2.begin(), v2.end()); ` ` ` ` ` `// replacing the elements with k set bits ` ` ` `// with the sorted elements ` ` ` `for` `(` `int` `i = 0; i < v1.size(); i++) ` ` ` `arr[v1[i]] = v2[i]; ` ` ` ` ` `// printing the new sorted array elements ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `cout << arr[i] << ` `" "` `; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `arr[] = { 14, 255, 1, 7, 13 }; ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` `int` `k = 3; ` ` ` ` ` `sortWithSetbits(arr, n, k); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## Python3

# Python 3 program for sorting array

# elements with set bits equal to K

# Function to sort elements with

# set bits equal to k

def sortWithSetbits(arr, n, k):

# initialise two vectors

v1 = []

v2 = []

for i in range(0, n, 1):

if (bin(arr[i]).count(‘1’) == k):

# first vector contains indices

# of required element

v1.append(i)

# second vector contains

# required elements

v2.append(arr[i])

# sorting the elements in second vector

v2.sort(reverse = False)

# replacing the elements with k set

# bits with the sorted elements

for i in range(0, len(v1), 1):

arr[v1[i]] = v2[i]

# printing the new sorted array elements

for i in range(0, n, 1):

print(arr[i], end = ” “)

# Driver code

if __name__ == ‘__main__’:

arr = [14, 255, 1, 7, 13]

n = len(arr)

k = 3

sortWithSetbits(arr, n, k)

# This code is contributed by

# Surendra_Gangwar

**Output:**

7 255 1 13 14

## Recommended Posts:

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- Sorting without comparison of elements
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- Sorting array using Stacks
- Row wise sorting in 2D array

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