# Sort elements by frequency | Set 4 (Efficient approach using hash)

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

Examples:

```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}
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

We have discussed different approaches in below posts :
Sort elements by frequency | Set 1
Sort elements by frequency | Set 2
Sorting Array Elements By Frequency | Set 3 (Using STL)

All of the above approaches work in O(n Log n) time where n is total number of elements. In this post, a new approach is discussed that works in O(n + m Log m) time where n is total number of elements and m is total number of distinct elements.

The idea is to use hashing.

1. We insert all elements and their counts into a hash. This step takes O(n) time where n is number of elements.
2. We copy contents of hash to an array (or vector) and sort them by counts. This step takes O(m Log m) time where m is total number of distinct elements.
3. For maintaining the order of elements if the frequency is same, we use another hash which has the key as elements of the array and value as the index. If the frequency is same for two elements then sort elements according to the index.

Below image is a dry run of the above approach: Below is the implementation of the above approach:

## C++

 `// Sort elements by frequency. If two elements have same ` `// count, then put the elements that appears first ` `#include ` `using` `namespace` `std; ` ` `  `// Map m2 keeps track of indexes of elements in array ` `unordered_map<``int``, ``int``> m2; ` ` `  `// Used for sorting by frequency. And if frequency is same, ` `// then by appearance ` `bool` `sortByVal(``const` `pair<``int``, ``int``>& a, ``const` `pair<``int``, ``int``>& b) ` `{ ` `    ``// If frequency is same then sort by index ` `    ``if` `(a.second == b.second)  ` `        ``return` `m2[a.first] < m2[b.first]; ` `     `  `    ``return` `a.second > b.second; ` `} ` ` `  `// function to sort elements by frequency ` `void` `sortByFreq(``int` `a[], ``int` `n) ` `{ ` `    ``unordered_map<``int``, ``int``> m; ` `    ``vector > v; ` ` `  `    ``for` `(``int` `i = 0; i < n; ++i) { ` ` `  `        ``// Map m is used to keep track of count  ` `        ``// of elements in array ` `        ``m[a[i]]++; ` ` `  `        ``// Update the value of map m2 only once ` `        ``if` `(m2[a[i]] == 0)  ` `            ``m2[a[i]] = i + 1;         ` `    ``} ` ` `  `    ``// Copy map to vector ` `    ``copy(m.begin(), m.end(), back_inserter(v)); ` ` `  `    ``// Sort the element of array by frequency ` `    ``sort(v.begin(), v.end(), sortByVal); ` ` `  `    ``for` `(``int` `i = 0; i < v.size(); ++i)  ` `        ``for` `(``int` `j = 0; j < v[i].second; ++j)  ` `            ``cout << v[i].first << ``" "``;  ` `} ` ` `  `// Driver program ` `int` `main() ` `{ ` `    ``int` `a[] = { 2, 5, 2, 6, -1, 9999999, 5, 8, 8, 8 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a); ` ` `  `    ``sortByFreq(a, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `import` `java.io.IOException; ` `import` `java.util.*; ` ` `  `class` `GFG { ` `    ``public` `static` `StringBuffer sortByFrequency(``int` `arr1[], ``int` `l1) { ` `        ``// Build a map of array elements to its count ` `        ``Map countMap = getCountMap(arr1, l1); ` `        ``StringBuffer result = ``new` `StringBuffer(); ` `         `  `        ``// Sort the map using a comparingByValue comparator ` `        ``// build the result by appending keys the count times to the result ` `         `  `        ``countMap.entrySet().stream() ` `                ``.sorted(Map.Entry. comparingByValue().reversed()) ` `                ``.forEach(e -> { ` `                    ``int` `key = e.getKey(); ` `                    ``int` `val = e.getValue(); ` `                    ``for` `(``int` `i = ``0``; i < val; i++) { ` `                        ``result.append(key + ``" "``); ` `                    ``} ` `                ``}); ` `         `  `        ``return` `result; ` `    ``} ` ` `  `    ``// helper function to return the element count map ` `    ``private` `static` `Map getCountMap(``int``[] arr, ``int` `l1) { ` `        ``Map countMap = ``new` `LinkedHashMap<>(); ` `        ``for` `(``int` `i = ``0``; i < l1; i++) { ` `            ``if` `(countMap.containsKey(arr[i])) { ` `                ``countMap.put(arr[i], countMap.get(arr[i]) + ``1``); ` `            ``} ``else` `{ ` `                ``countMap.put(arr[i], ``1``); ` `            ``} ` `        ``} ` `        ``return` `countMap; ` `    ``} ` `     `  `    ``// Driver program ` `    ``public` `static` `void` `main(String[] args) ``throws` `IOException { ` `        ``int` `a[] = { ``2``, ``5``, ``2``, ``6``, -``1``, ``9999999``, ``5``, ``8``, ``8``, ``8` `}; ` `        ``System.out.println(sortByFrequency(a, a.length)); ` `    ``} ` `} `

Output:

```8 8 8 2 2 5 5 6 -1 9999999
```

Time Complexity : O(n) + O(m Log m) where n is total number of elements and m is total number of distinct elements

This article is contributed by Ankur Singh and improved by Ankur Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.