How to arrange an array in decreasing order of frequency of each number?

Input : {5, 13, 6, 5, 13, 7, 8, 6, 5}

Output : {5, 5, 5, 13, 13, 6, 6, 7, 8}

The question is to arrange the nu开发者_高级运维mbers in the array in decreasing order of their frequency, preserving the order of their occurrence.

If there is a tie, like in this example between 13 and 6, then the number occurring first in the input array would come first in the output array.

---

I guess I'd do it like this:

Use a key-value data structure where the number itself is the key and the occurrence count and first occurrence index are the value.

Now traverse all numbers. If the number is not yet known (in the data structure), add it, remembering the current index as well as 1 as count. Otherwise, increment the count.

Now sort the data structure contents by the occurrence count (decreasing) and the occurrence index (increasing), and output the result (repeating the number using the occurrence count).

Processing space used <= N, time used depending on data structure and dictionary would probably be about O(N log N)

---

I can think of two solutions:

Make a first pass to count frequencies and store the first index and then sort according to that

This is simple to implement, uses linear (O(N)) additional memory in the worst case and takes O(N log N) time.

Do everything in one pass with some sort of priority queue, where the "priority" is the frequency count (changing as you pass through the input), and the index of the first occurrence to break ties

This does everything in a single pass, but I can't think of any advantage over the other solution. It still requires O(N) additional memory, and will also take O(N log N) time. Also, it requires a more complex data structure.

---

In Python2.7 or Python3.1

>>> from collections import Counter
>>> L=[5, 13, 6, 5, 13, 7, 8, 6, 5]
>>> c=Counter(L)
>>> def keyfunc(x):
...     return (-c.get(x),L.index(x))
... 
>>> sorted(L,key=keyfunc)
[5, 5, 5, 13, 13, 6, 6, 7, 8]

In Python2.6

>>> from collections import defaultdict
>>> L=[5, 13, 6, 5, 13, 7, 8, 6, 5]
>>> c=defaultdict(int)
>>> for x in L:
...     c[x]+=1
... 
>>> def keyfunc(x):
...     return (-c.get(x),L.index(x))
... 
>>> sorted(L,key=keyfunc)
[5, 5, 5, 13, 13, 6, 6, 7, 8]

Here is a version that doesn't use any library functions (weird constraint)

>>> L=[5, 13, 6, 5, 13, 7, 8, 6, 5]
>>> c={}
>>> for x in L:
...     c[x]=c.setdefault(x,0)+1
... 
>>> def keyfunc(x):
...     return (-c.get(x),L.index(x))
... 
>>> sorted(L,key=keyfunc)
[5, 5, 5, 13, 13, 6, 6, 7, 8]

In each case, keyfunc is used to control the ordering of the sort

keyfunc(5) returns (-3,0)
keyfunc(6) returns (-2,2)
keyfunc(7) returns (-1,5)
keyfunc(8) returns (-1,6)
keyfunc(13) returns (-2,1)

The list items are sorted according to the return value of keyfunc

---

You can probably do it in one pass with a bubble sort type algorithm where you keep a note of the count of the previous value and swap bunches of numbers when you find more of another number.

But - as a first step you should do a 2pass solution, with a std::map/pair to store number or if you are told the range of values in advance use a simple array.

---

C#. It takes the most obvious approach: sort by frequency and ordinal.

Output: 5 5 5 13 13 6 6 7 8. Space: O(n). Time: O(n log n).

class Program
{
    class FreqAndOrdinal
    {
        public int Frequency;
        public int Ordinal;
        public FreqAndOrdinal(int freq, int ord)
        {
            this.Frequency = freq;
            this.Ordinal = ord;
        }
    }

    static int Compare(FreqAndOrdinal x, FreqAndOrdinal y)
    {
        int result = y.Frequency.CompareTo(x.Frequency);
        return result == 0 ? x.Ordinal.CompareTo(y.Ordinal) : result;
    }

    static void Main(string[] args)
    {
        int[] nums = new int[] { 5, 13, 6, 5, 13, 7, 8, 6, 5 };
        var freqLookup = new Dictionary<int, FreqAndOrdinal>(nums.Length);
        for (int i = 0; i < nums.Length; i++)
        {
            FreqAndOrdinal tmp;
            if (freqLookup.TryGetValue(nums[i], out tmp))
                ++tmp.Frequency;
            else
                freqLookup[nums[i]] = new FreqAndOrdinal(1, i);
        }

        Array.Sort(nums, (x,y) => Compare(freqLookup[x], freqLookup[y]));

        for (int i = 0; i < nums.Length; i++)
        {
            Console.Write(" {0}", nums[i]);
        }
        Console.ReadKey();
    }
}

---

Sort the array, and in your sort function for x, y: sort by count(x) vs. count(y). If they are the same, sort by index(x) vs. index(y)

In python:

input = [5, 13, 6, 5, 13, 7, 8, 6, 5]
orig = list(input)

def cmp(x, y):
    if (orig.count(y) - orig.count(x) != 0):
        return orig.count(y) - orig.count(x)
    return orig.index(x) - orig.index(y)   

input.sort(cmp) 
print input

To make it more efficient, precalculate counts and indexes before sorting the array.

---

private static void sortByFrequency(int[] a)
{
    Map<Integer, Element> map = new HashMap<Integer, Element>();
    for(int i=0; i<a.length; i++)
    {
        if(map.get(a[i]) == null)
        {
            map.put(a[i], new Element(i));
        }
        else
        {
            Element e = map.get(a[i]);
            e.frequency++;
        }
    }

    Set<Integer> set = map.keySet();
    TreeSet<Element> treeSet = new TreeSet<Element>();
    for(int i : set)
    {
        treeSet.add(map.get(i));
    }

    for(Element e : treeSet)
    {
        for(int i=0; i<e.frequency;i++)
        {
            System.out.println(a[e.index]);
        }
    }
}

private static class Element implements Comparable<Element>
{
    private final int index;
    private int frequency;

    Element(int index)
    {
        this.index = index;
        this.frequency = 1;
    }

    @Override
    public int compareTo(Element o)
    {
        int k = o.frequency - this.frequency;
        if(k != 0) return k;
        else
        {
            return this.index - o.index;
        }
    }
}

public static void main(String[] args)
{
    int[] a = {5, 13, 6, 5, 13, 7, 8, 6, 5};
    sortByFrequency(a);
}