# Sort an array according to count of set bits | Set 2

Given an array **arr[]** of positive integers, the task is to sort the array in decreasing order of count of set bits in binary representations of array elements.

For integers having same number of set bits in their binary representation, sort according to their position in the original array i.e., a stable sort. For example, if input array is {3, 5}, then output array should also be {3, 5}. Note that both 3 and 5 have same number set bits.

**Examples:**

Input:arr[] = {5, 2, 3, 9, 4, 6, 7, 15, 32}

Output:15 7 5 3 9 6 2 4 32

The integers in their binary representation are:

15 – 1111

7 – 0111

5 – 0101

3 – 0011

9 – 1001

6 – 0110

2 – 0010

4 – 0100

32 – 10000

Hence, the non-increasing sorted order is:

{15, 7, 5, 3, 9, 6, 2, 4, 32}

Input:arr[] = {1, 2, 3, 4, 5, 6};

Output:3 5 6 1 2 4

**Approach:** We have already discussed the method of sorting based on set bit count in the previous section with various methods. This post contains implementation using maps.

As we know that a map/multimap stores data in sorted manner. So if we store (32 – countsetbits(arr[i])) for an arr[i] in map, then the output will come out in decreasing order of set bit count which is the desired output.

Below is the implementation of the above approach:

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// function to sort the array according ` `// to the number of set bits in elements ` `void` `sortArr(` `int` `arr[], ` `int` `n) ` `{ ` ` ` `multimap<` `int` `, ` `int` `> map; ` ` ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` `int` `count = 0; ` ` ` `int` `k = arr[i]; ` ` ` ` ` `// Counting no of setBits in arr[i] ` ` ` `while` `(k) { ` ` ` `k = k & k - 1; ` ` ` `count++; ` ` ` `} ` ` ` ` ` `// The count is subtracted from 32 ` ` ` `// because the result needs ` ` ` `// to be in descending order ` ` ` `map.insert(make_pair(32 - count, arr[i])); ` ` ` `} ` ` ` ` ` `// Printing the numbers in descending ` ` ` `// order of set bit count ` ` ` `for` `(` `auto` `it = map.begin(); it != map.end(); it++) { ` ` ` `cout << (*it).second << ` `" "` `; ` ` ` `} ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `arr[] = { 5, 2, 3, 9, 4, 6, 7, 15, 32 }; ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` ` ` `sortArr(arr, n); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

15 7 5 3 9 6 2 4 32

## Recommended Posts:

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- Count Inversions in an array | Set 1 (Using Merge Sort)
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- Count set bits in an integer
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- Count pairs with set bits sum equal to K
- Range query for count of set bits

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