Friday, September 16, 2016

Sort an array according to count of set bits - GeeksforGeeks


Sort an array according to count of set bits - GeeksforGeeks
Given an array of positive integers, 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.

This problem can be solved in O(n) time. The idea is similar to counting sort.
Note: There can be minimum 1 set-bit and only a maximum of 31set-bits in any integer.
Steps (assuming that an integer takes 32 bits):
  1. Create a vector “count” of size 32. Each cell of count i.e., count[i] is another vector that stores all the elements whose set-bit-count is i
  2. Traverse the array and do following for each element:
    1. Count the number set-bits of this element. Let it be ‘setbitcount’
    2. count[setbitcount].push_back(element)
  3. Traverse ‘count’ in reverse fashion(as we need to sort in non-increasing order) and modify the array.


Capture
// Function to sort according to bit count
// This function assumes that there are 32
// bits in an integer.
void sortBySetBitCount(int arr[],int n)
{
    vector<vector<int> > count(32);
    int setbitcount = 0;
    for (int i=0; i<n; i++)
    {
        setbitcount = countBits(arr[i]);
        count[setbitcount].push_back(arr[i]);
    }
 
    int j = 0;  // Used as an index in final sorted array
 
    // Traverse through all bit counts (Note that we
    // sort array in decreasing order)
    for (int i=31; i>=0; i--)
    {
        vector<int> v1 = count[i];
        for (int i=0; i<v1.size(); i++)
            arr[j++] = v1[i];
    }
}
Using custom comparator of std::sort to sort the array according to set-bit count
// a utility function that returns total set bits
// count in an integer
int countBits(int a)
{
    int count = 0;
    while (a)
    {
        if (a & 1 )
            count+= 1;
        a = a>>1;
    }
    return count;
}
 
// custom comparator of std::sort
int cmp(int a,int b)
{
    int count1 = countBits(a);
    int count2 = countBits(b);
 
    // this takes care of the stability of
    // sorting algorithm too
    if (count1 <= count2)
        return false;
    return true;
}
 
// Function to sort according to bit count using
// std::sort
void sortBySetBitCount(int arr[], int n)
{
    stable_sort(arr, arr+n, cmp);
}

  1. Create an auxiliary array and store the set-bit counts of all integers in the aux array
  2. Simultaneously sort both arrays according to the non-increasing order of auxiliary array. (Note that we need to use a stable sort algorithm)
// a utility function that returns total set bits
// count in an integer
int countBits(int a)
{
    int count = 0;
    while (a)
    {
        if (a & 1 )
            count+= 1;
        a = a>>1;
    }
    return count;
}
 
// Function to simultaneously sort both arrays
// using insertion sort
void insertionSort(int arr[],int aux[], int n)
{
    for (int i = 1; i < n; i++)
    {
        // use 2 keys because we need to sort both
        // arrays simultaneously
        int key1 = aux[i];
        int key2 = arr[i];
        int j = i-1;
 
        /* Move elements of arr[0..i-1] and aux[0..i-1],
           such that elements of aux[0..i-1] are
           greater than key1, to one position ahead
           of their current position */
        while (j >= 0 && aux[j] < key1)
        {
            aux[j+1] = aux[j];
            arr[j+1] = arr[j];
            j = j-1;
        }
        aux[j+1] = key1;
        arr[j+1] = key2;
    }
}
 
// Function to sort according to bit count using
// an auxiliary array
void sortBySetBitCount(int arr[],int n)
{
    // Create an array and store count of
    // set bits in it.
    int aux[n];
    for (int i=0; i<n; i++)
        aux[i] = countBits(arr[i]);
 
    // Sort arr[] according to values in aux[]
    insertionSort(arr, aux, n);
}
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