Given an array of **n** numbers and a positive integer **k**. The problem is to find **k** numbers with most occurrences, i.e., the top **k** numbers having the maximum frequency. If two numbers have the same frequency then the larger number should be given preference. The numbers should be displayed in decreasing order of their frequencies. It is assumed that the array consists of **k** numbers with most occurrences.

**Examples:**

Input:

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

k = 2Output:4 1Explanation:

Frequency of4= 2

Frequency of1= 2

These two have the maximum frequency and4is larger than1.

Input :

arr[] = {7, 10, 11, 5, 2, 5, 5, 7, 11, 8, 9},

k = 4Output:5 11 7 10Explanation:

Frequency of5= 3

Frequency of11= 2

Frequency of7= 2

Frequency of10= 1

These four have the maximum frequency and5is largest among rest.

*Asked in Amazon Interview*

__Method 1:__

**Approach:**The thought process should begin from creating a HashMap to store element-frequency pair in the HashMap. HashMap is used to perform insertion and updation in constant time. Then sort the element-frequency pair in decreasing order of frequency. This gives the information about each element and the number of times they are present in the array. To get k elements of the array, print the first k elements of the sorted array.**Hashmap:**HashMap is a part of Java’s collection since Java 1.2. It provides the basic implementation of the Map interface of Java. It stores the data in (Key, Value) pairs. To access a value one must know its key. HashMap is known as HashMap because it uses a technique called Hashing. Hashing is a technique of converting a large String to small String that represents the same String. A shorter value helps in indexing and faster searches. HashSet also uses HashMap internally. It internally uses a link list to store key-value pairs already explained in HashSet in detail and further articles.*More on HashMap >>***Algorithm:**- Create a Hashmap
*hm*, to store key-value pair, i.e. element-frequency pair. - Traverse the array from start to end.
- For every element in the array update
*hm[array[i]]++* - Store the element-frequency pair in a vector and sort the vector in decreasing order of frequency.
- Print the first k elements of sorted array.

- Create a Hashmap

Below is the Implementation of the above algorithm:

## C++

`// C++ implementation to find k numbers with most` `// occurrences in the given array` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// comparison function to sort the 'freq_arr[]'` `bool` `compare(pair<` `int` `, ` `int` `> p1, pair<` `int` `, ` `int` `> p2)` `{` ` ` `// if frequencies of two elements are same` ` ` `// then the larger number should come first` ` ` `if` `(p1.second == p2.second)` ` ` `return` `p1.first > p2.first;` ` ` `// sort on the basis of decreasing order` ` ` `// of frequencies` ` ` `return` `p1.second > p2.second;` `}` `// funnction to print the k numbers with most occurrences` `void` `print_N_mostFrequentNumber(` `int` `arr[], ` `int` `n, ` `int` `k)` `{` ` ` `// unordered_map 'um' implemented as frequency hash table` ` ` `unordered_map<` `int` `, ` `int` `> um;` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `um[arr[i]]++;` ` ` `// store the elements of 'um' in the vector 'freq_arr'` ` ` `vector<pair<` `int` `, ` `int` `> > freq_arr(um.begin(), um.end());` ` ` `// sort the vector 'freq_arr' on the basis of the` ` ` `// 'compare' function` ` ` `sort(freq_arr.begin(), freq_arr.end(), compare);` ` ` `// display the top k numbers` ` ` `cout << k << ` `" numbers with most occurrences are:\n"` `;` ` ` `for` `(` `int` `i = 0; i < k; i++)` ` ` `cout << freq_arr[i].first << ` `" "` `;` `}` `// Driver program to test above` `int` `main()` `{` ` ` `int` `arr[] = { 3, 1, 4, 4, 5, 2, 6, 1 };` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `int` `k = 2;` ` ` `print_N_mostFrequentNumber(arr, n, k);` ` ` `return` `0;` `}` |

## Java

`// Java implementation to find` `// k elements with max occurence.` `import` `java.util.*;` `public` `class` `KFrequentNumbers` `{` ` ` `static` `void` `print_N_mostFrequentNumber(` `int` `[] arr,` ` ` `int` `n,` ` ` `int` `k)` ` ` `{` ` ` `Map<Integer, Integer> mp` ` ` `= ` `new` `HashMap<Integer, Integer>();` ` ` `// Put count of all the` ` ` `// distinct elements in Map` ` ` `// with element as the key &` ` ` `// count as the value.` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) {` ` ` `// Get the count for the` ` ` `// element if already present in the` ` ` `// Map or get the default value which is 0.` ` ` `mp.put(arr[i],` ` ` `mp.getOrDefault(arr[i], ` `0` `) + ` `1` `);` ` ` `}` ` ` `// Create a list from elements of HashMap` ` ` `List<Map.Entry<Integer, Integer> > list` ` ` `= ` `new` `ArrayList<Map.Entry<Integer,` ` ` `Integer> >(` ` ` `mp.entrySet());` ` ` `// Sort the list` ` ` `Collections.sort(` ` ` `list,` ` ` `new` `Comparator<Map.Entry<Integer,` ` ` `Integer> >()` ` ` `{` ` ` `public` `int` `compare(` ` ` `Map.Entry<Integer, Integer> o1,` ` ` `Map.Entry<Integer, Integer> o2)` ` ` `{` ` ` `if` `(o1.getValue() == o2.getValue())` ` ` `return` `o2.getKey() - o1.getKey();` ` ` `else` ` ` `return` `o2.getValue()` ` ` `- o1.getValue();` ` ` `}` ` ` `});` ` ` `for` `(` `int` `i = ` `0` `; i < k; i++)` ` ` `System.out.println(list.get(i).getKey());` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `arr[] = { ` `3` `, ` `1` `, ` `4` `, ` `4` `, ` `5` `, ` `2` `, ` `6` `, ` `1` `};` ` ` `int` `n = arr.length;` ` ` `int` `k = ` `2` `;` ` ` ` ` `// Function call` ` ` `print_N_mostFrequentNumber(arr, n, k);` ` ` `}` `}` |

## Python3

`# Python3 implementation to find k numbers` `# with most occurrences in the given array` `# funnction to print the k numbers with` `# most occurrences` `def` `pr_N_mostFrequentNumber(arr, n, k):` ` ` `um ` `=` `{}` ` ` `for` `i ` `in` `range` `(n):` ` ` `if` `arr[i] ` `in` `um:` ` ` `um[arr[i]] ` `+` `=` `1` ` ` `else` `:` ` ` `um[arr[i]] ` `=` `1` ` ` `a ` `=` `[` `0` `] ` `*` `(` `len` `(um))` ` ` `j ` `=` `0` ` ` `for` `i ` `in` `um:` ` ` `a[j] ` `=` `[i, um[i]]` ` ` `j ` `+` `=` `1` ` ` `a ` `=` `sorted` `(a, key` `=` `lambda` `x: x[` `0` `],` ` ` `reverse` `=` `True` `)` ` ` `a ` `=` `sorted` `(a, key` `=` `lambda` `x: x[` `1` `],` ` ` `reverse` `=` `True` `)` ` ` `# display the top k numbers` ` ` `print` `(k, ` `"numbers with most occurrences are:"` `)` ` ` `for` `i ` `in` `range` `(k):` ` ` `print` `(a[i][` `0` `], end` `=` `" "` `)` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `arr ` `=` `[` `3` `, ` `1` `, ` `4` `, ` `4` `, ` `5` `, ` `2` `, ` `6` `, ` `1` `]` ` ` `n ` `=` `8` ` ` `k ` `=` `2` ` ` `pr_N_mostFrequentNumber(arr, n, k)` `# This code is contributed by` `# Shubham Singh(SHUBHAMSINGH10)` |

**Output**

2 numbers with most occurrences are: 4 1

**Complexity Analysis:**

**Time Complexity:**O(d log d), where**d**is the count of distinct elements in the array. To sort the array O(d log d) time is needed.**Auxiliary Space:**O(d), where**d**is the count of distinct elements in the array. To store the elements in HashMap O(d) space complexity is needed.

__Method 2:__

**Approach:**Create a HashMap to store element-frequency pair in the HashMap. HashMap is used to perform insertion and updation in constant time. Then use a priority queue to store the element-frequency pair (Max-Heap). This gives the element which has maximum frequency at the root of the Priority Queue. Remove the top or root of Priority Queue K times and print the element. To insert and delete the top of the priority queue*O(log n)*time is required.**Priority Queue:**Priority queues are a type of container adapters, specifically designed such that the first element of the queue is the greatest of all elements in the queue and elements are in non increasing order(hence we can see that each element of the queue has a priority{fixed order}).*More Info about Priority Queue:*C++ STL priority_queue**Algorithm :**- Create a Hashmap
*hm*, to store key-value pair, i.e. element-frequency pair. - Traverse the array from start to end.
- For every element in the array update
*hm[array[i]]++* - Store the element-frequency pair in a Priority Queue and create the Priority Queue, this takes O(n) time.
- Run a loop k times, and in each iteration remove the top of the priority queue and print the element.

- Create a Hashmap

Below is the Implementation of the above algorithm:

## C++

`// C++ implementation to find k numbers with most` `// occurrences in the given array` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// comparison function defined for the priority queue` `struct` `compare {` ` ` `bool` `operator()(pair<` `int` `, ` `int` `> p1, pair<` `int` `, ` `int` `> p2)` ` ` `{` ` ` `// if frequencies of two elements are same` ` ` `// then the larger number should come first` ` ` `if` `(p1.second == p2.second)` ` ` `return` `p1.first < p2.first;` ` ` `// insert elements in the priority queue on the basis of` ` ` `// decreasing order of frequencies` ` ` `return` `p1.second < p2.second;` ` ` `}` `};` `// funnction to print the k numbers with most occurrences` `void` `print_N_mostFrequentNumber(` `int` `arr[], ` `int` `n, ` `int` `k)` `{` ` ` `// unordered_map 'um' implemented as frequency hash table` ` ` `unordered_map<` `int` `, ` `int` `> um;` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `um[arr[i]]++;` ` ` ` ` `// priority queue 'pq' implemented as max heap on the basis` ` ` `// of the comparison operator 'compare'` ` ` `// element with the highest frequency is the root of 'pq'` ` ` `// in case of conflicts, larger element is the root` ` ` `priority_queue<pair<` `int` `, ` `int` `>, vector<pair<` `int` `, ` `int` `> >,` ` ` `compare>` ` ` `pq(um.begin(), um.end());` ` ` `// display the top k numbers` ` ` `cout << k << ` `" numbers with most occurrences are:\n"` `;` ` ` `for` `(` `int` `i = 1; i <= k; i++) {` ` ` `cout << pq.top().first << ` `" "` `;` ` ` `pq.pop();` ` ` `}` `}` `// Driver program to test above` `int` `main()` `{` ` ` `int` `arr[] = { 3, 1, 4, 4, 5, 2, 6, 1 };` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `int` `k = 2;` ` ` `print_N_mostFrequentNumber(arr, n, k);` ` ` `return` `0;` `}` |

## Java

`// Java implementation to find k` `// elements with max occurence.` `import` `java.util.*;` `public` `class` `KFrequentNumbers {` ` ` `static` `void` `print_N_mostFrequentNumber(` `int` `[] arr,` ` ` `int` `n,` ` ` `int` `k)` ` ` `{` ` ` `Map<Integer, Integer> mp` ` ` `= ` `new` `HashMap<Integer, Integer>();` ` ` `// Put count of all the` ` ` `// distinct elements in Map` ` ` `// with element as the key &` ` ` `// count as the value.` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) {` ` ` `// Get the count for the` ` ` `// element if already` ` ` `// present in the Map or` ` ` `// get the default value` ` ` `// which is 0.` ` ` `mp.put(arr[i],` ` ` `mp.getOrDefault(arr[i], ` `0` `) + ` `1` `);` ` ` `}` ` ` `// Create a Priority Queue` ` ` `// to sort based on the` ` ` `// count or on the key if the` ` ` `// count is same` ` ` `PriorityQueue<Map.Entry<Integer,` ` ` `Integer>> queue` ` ` `= ` `new` `PriorityQueue<>(` ` ` `(a, b)` ` ` `-> a.getValue().equals(b.getValue())` ` ` `? Integer.compare(b.getKey(),` ` ` `a.getKey())` ` ` `: Integer.compare(b.getValue(),` ` ` `a.getValue()));` ` ` `// Insert the data from the map` ` ` `// to the Priority Queue.` ` ` `for` `(Map.Entry<Integer, Integer> entry :` ` ` `mp.entrySet())` ` ` `queue.offer(entry);` ` ` `// Print the top k elements` ` ` `for` `(` `int` `i = ` `0` `; i < k; i++)` ` ` `{` ` ` `System.out.println(queue.poll().getKey());` ` ` `}` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `arr[] = { ` `3` `, ` `1` `, ` `4` `, ` `4` `, ` `5` `, ` `2` `, ` `6` `, ` `1` `};` ` ` `int` `n = arr.length;` ` ` `int` `k = ` `2` `;` ` ` ` ` `// Function call` ` ` `print_N_mostFrequentNumber(arr, n, k);` ` ` `}` `}` `// This code is contributed by Shubham Kumar Shah` |

## Python3

`# Python3 implementation to find k` `# numbers with most occurrences in` `# the given array` `import` `heapq` `# Function to print the k numbers with` `# most occurrences` `def` `print_N_mostFrequentNumber(arr, n, k):` ` ` ` ` `mp ` `=` `dict` `()` ` ` ` ` `# Put count of all the distinct elements` ` ` `# in dictionary with element as the` ` ` `# key & count as the value.` ` ` `for` `i ` `in` `range` `(` `0` `, n):` ` ` `if` `arr[i] ` `not` `in` `mp:` ` ` `mp[arr[i]] ` `=` `0` ` ` `else` `:` ` ` `mp[arr[i]] ` `+` `=` `1` ` ` ` ` `# Using heapq data structure` ` ` `heap ` `=` `[(value, key) ` `for` `key,` ` ` `value ` `in` `mp.items()]` ` ` ` ` `# Get the top k elements` ` ` `largest ` `=` `heapq.nlargest(k, heap)` ` ` `# Insert the data from the map to` ` ` `# the priority queue` ` ` `print` `(k, ` `" numbers with most "` ` ` `"occurrences are:"` `, sep ` `=` `"")` ` ` ` ` `# Print the top k elements` ` ` `for` `i ` `in` `range` `(k):` ` ` `print` `(largest[i][` `1` `], end ` `=` `" "` `)` `# Driver code` `if` `__name__` `=` `=` `"__main__"` `:` ` ` ` ` `arr ` `=` `[ ` `3` `, ` `1` `, ` `4` `, ` `4` `, ` `5` `, ` `2` `, ` `6` `, ` `1` `]` ` ` `n ` `=` `len` `(arr)` ` ` `k ` `=` `2` ` ` ` ` `print_N_mostFrequentNumber(arr, n, k)` ` ` `# This code is contributed by MuskanKalra1` |

**Output**

2 numbers with most occurrences are: 4 1

**Complexity Analysis:**

**Time Complexity:**O(k log d + d), where**d**is the count of distinct elements in the array.

To remove the top of priority queue O(log d) time is required, so if k elements are removed then O(k log d) time is required and to traverse the distinct elements O(d) time is required.**Auxiliary Space:**O(d), where**d**is the count of distinct elements in the array.

To store the elements in HashMap O(d) space complexity is needed.

**Find k most frequent in linear time****References:** https://www.careercup.com/question?id=5082885552865280

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