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Sort elements by frequency | Set 1

  • Difficulty Level : Medium
  • Last Updated : 01 Sep, 2021

Print the elements of an array in the decreasing frequency if 2 numbers have same frequency then print the one which came first. 

Examples:  

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Input:  arr[] = {2, 5, 2, 8, 5, 6, 8, 8}
Output: arr[] = {8, 8, 8, 2, 2, 5, 5, 6}

Input: arr[] = {2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8}
Output: arr[] = {8, 8, 8, 2, 2, 5, 5, 6, -1, 9999999}

METHOD 1 (Use Sorting) 

  • Use a sorting algorithm to sort the elements O(nlogn)
  • Scan the sorted array and construct a 2D array of element and count O(n).
  • Sort the 2D array according to count O(nlogn).

Example:  



  Input 2 5 2 8 5 6 8 8

  After sorting we get
  2 2 5 5 6 8 8 8

  Now construct the 2D array as
  2, 2
  5, 2
  6, 1
  8, 3

  Sort by count
  8, 3
  2, 2
  5, 2
  6, 1

How to maintain the order of elements if the frequency is the same?

The above approach doesn’t make sure order of elements if the frequency is the same. To handle this, we should use indexes in step 3, if two counts are same then we should first process(or print) the element with a lower index. In step 1, we should store the indexes instead of elements. 

  Input   2  5  2  8  5  6  8  8

  After sorting we get
  Element 2 2 5 5 6 8 8 8
  Index   0 2 1 4 5 3 6 7

  Now construct the 2D array as
  Index, Count
  0,      2
  1,      2
  5,      1
  3,      3

  Sort by count (consider indexes in case of tie)
  3, 3
  0, 2
  1, 2
  5, 1
  
  Print the elements using indexes in the above 2D array.

Below is the implementation of above approach –

CPP




// Sort elements by frequency. If two elements have same
// count, then put the elements that appears first
#include <bits/stdc++.h>
using namespace std;
 
// Used for sorting
struct ele {
    int count, index, val;
};
 
// Used for sorting by value
bool mycomp(struct ele a, struct ele b)
{
    return (a.val < b.val);
}
 
// Used for sorting by frequency. And if frequency is same,
// then by appearance
bool mycomp2(struct ele a, struct ele b)
{
    if (a.count != b.count)
        return (a.count < b.count);
    else
        return a.index > b.index;
}
 
void sortByFrequency(int arr[], int n)
{
    struct ele element[n];
    for (int i = 0; i < n; i++) {
 
        // Fill Indexes
        element[i].index = i;
 
        // Initialize counts as 0
        element[i].count = 0;
 
        // Fill values in structure
        // elements
        element[i].val = arr[i];
    }
 
    /* Sort the structure elements according to value,
       we used stable sort so relative order is maintained.
     */
    stable_sort(element, element + n, mycomp);
 
    /* initialize count of first element as 1 */
    element[0].count = 1;
 
    /* Count occurrences of remaining elements */
    for (int i = 1; i < n; i++) {
 
        if (element[i].val == element[i - 1].val) {
            element[i].count += element[i - 1].count + 1;
 
            /* Set count of previous element as -1, we are
               doing this because we'll again sort on the
               basis of counts (if counts are equal than on
               the basis of index)*/
            element[i - 1].count = -1;
 
            /* Retain the first index (Remember first index
               is always present in the first duplicate we
               used stable sort. */
            element[i].index = element[i - 1].index;
        }
 
        /* Else If previous element is not equal to current
          so set the count to 1 */
        else
            element[i].count = 1;
    }
 
    /* Now we have counts and first index for each element
       so now sort on the basis of count and in case of tie
       use index to sort.*/
    stable_sort(element, element + n, mycomp2);
    for (int i = n - 1, index = 0; i >= 0; i--)
        if (element[i].count != -1)
            for (int j = 0; j < element[i].count; j++)
                arr[index++] = element[i].val;
}
 
// Driver program
int main()
{
    int arr[] = { 2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    sortByFrequency(arr, n);
 
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
    return 0;
}

Output

8 8 8 2 2 5 5 6 -1 9999999

Thanks to Gaurav Ahirwar for providing above implementation.

METHOD 2 (Use Hashing and Sorting) 
Using a hashing mechanism, we can store the elements (also first index) and their counts in a hash. Finally, sort the hash elements according to their counts.
Below is the implementation of above approach – 

CPP




// CPP program for above approach
#include <bits/stdc++.h>
using namespace std;
 
// Compare function
bool fcompare(pair<int, pair<int, int> > p,
              pair<int, pair<int, int> > p1)
{
    if (p.second.second != p1.second.second)
        return (p.second.second > p1.second.second);
    else
        return (p.second.first < p1.second.first);
}
void sortByFrequency(int arr[], int n)
{
    unordered_map<int, pair<int, int> > hash; // hash map
    for (int i = 0; i < n; i++) {
        if (hash.find(arr[i]) != hash.end())
            hash[arr[i]].second++;
        else
            hash[arr[i]] = make_pair(i, 1);
    } // store the count of all the elements in the hashmap
 
    // Iterator to Traverse the Hashmap
    auto it = hash.begin();
 
    // Vector to store the Final Sortted order
    vector<pair<int, pair<int, int> > > b;
    for (it; it != hash.end(); ++it)
        b.push_back(make_pair(it->first, it->second));
 
    sort(b.begin(), b.end(), fcompare);
 
    // Printing the Sorted sequence
    for (int i = 0; i < b.size(); i++) {
        int count = b[i].second.second;
        while (count--)
            cout << b[i].first << " ";
    }
}
 
// Driver Function
int main()
{
    int arr[] = { 2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    sortByFrequency(arr, n);
 
    return 0;
}

Python3




# Python program for above approach
 
from collections import defaultdict
 
# Sort by Frequency
 
 
def sortByFreq(arr, n):
    # arr -> Array to be sorted
    # n   -> Length of Array
 
    # d is a hashmap(referred as dictionary in python)
    d = defaultdict(lambda: 0)
    for i in range(n):
        d[arr[i]] += 1
 
    # Sorting the array 'arr' where key
    # is the function based on which
    # the array is sorted
    # While sorting we want to give
    # first priority to Frequency
    # Then to value of item
    arr.sort(key=lambda x: (-d[x], x))
 
    return (arr)
 
 
# Driver Function
if __name__ == "__main__":
    arr = [2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8]
    n = len(arr)
    solution = sortByFreq(arr, n)
    print(*solution)

Java




/*package whatever //do not write package name here */
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.List;
class GFG {
 
    static Integer[] arr = { 2, 5, 2, 8, 5, 6, 8, 8 };
 
    // Driver Code
    public static void main(String[] args)
    {
        List<Integer> list = Arrays.asList(arr);
        sortBasedOnFrequencyAndValue(list);
    }
    // Compare Function
    public static void
    sortBasedOnFrequencyAndValue(List<Integer> list)
    {
        int n = arr.length;
        final HashMap<Integer, Integer> mapCount
            = new HashMap<Integer, Integer>();
        final HashMap<Integer, Integer> mapIndex
            = new HashMap<Integer, Integer>();
        for (int i = 0; i < n; i++) {
            if (mapCount.containsKey(arr[i])) {
                mapCount.put(arr[i],
                             mapCount.get(arr[i]) + 1);
            }
            else {
                mapCount.put(arr[i],1); // Map to capture Count of elements
                mapIndex.put(arr[i],i); // Map to capture 1st occurrence of elements
            }
        }
 
        Collections.sort(list, new Comparator<Integer>() {
            public int compare(Integer n1, Integer n2)
            {
                int freq1 = mapCount.get(n1);
                int freq2 = mapCount.get(n2);
                if (freq1 != freq2) {
                    return freq2 - freq1;
                }
                else {
                    return mapIndex.get(n1)
                        - mapIndex.get(
                            n2); // Elements with Lesser
                                 // Index gets Higher
                                 // Priority
                }
            }
        });
        System.out.println(list);
    }
}

Javascript




<script>
 
let arr=[2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8];
 
// Compare Function
function sortBasedOnFrequencyAndValue(list)
{
    let n = arr.length;
        let  mapCount
            = new Map();
        let  mapIndex
            = new Map();
        for (let i = 0; i < n; i++) {
            if (mapCount.has(arr[i])) {
                mapCount.set(arr[i],
                             mapCount.get(arr[i]) + 1);
            }
            else {
                mapCount.set(arr[i],1); // Map to capture Count of elements
                mapIndex.set(arr[i],i); // Map to capture 1st occurrence of elements
            }
        }
  
        list.sort(function(n1,n2){
             
                let freq1 = mapCount.get(n1);
                let freq2 = mapCount.get(n2);
                if (freq1 != freq2) {
                    return freq2 - freq1;
                }
                else {
                    return mapIndex.get(n1)
                        - mapIndex.get(
                            n2); // Elements with Lesser
                                 // Index gets Higher
                                 // Priority
                }
             
        });
        document.write(list.join(" "));
}
 
// Driver Code
sortBasedOnFrequencyAndValue(arr);
 
// This code is contributed by patel2127
</script>

Output

8 8 8 2 2 5 5 6 -1 9999999

This can also be solved by Using two maps, one for array element as an index and after this second map whose keys are frequency and value are array elements.

METHOD 3(Use BST and Sorting) 

  • Insert elements in BST one by one and if an element is already present then increment the count of the node. Node of the Binary Search Tree (used in this approach) will be as follows.

C




struct tree {
    int element;
    int first_index /*To handle ties in counts*/
        int count;
} BST;</div>
  • Store the first indexes and corresponding counts of BST in a 2D array.
  • Sort the 2D array according to counts (and use indexes in case of tie).

Time Complexity: O(nlogn) if a Self Balancing Binary Search Tree is used. This is implemented in Set 2.
 https://youtu.be/NBXf9vCksuM
 Set 2: 
Sort elements by frequency | Set 2
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above 




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