Given an array of n numbers. Your task is to read numbers from the array and keep at-most K numbers at the top (According to their decreasing frequency) every time a new number is read. We basically need to print top k numbers sorted by frequency when input stream has included k distinct elements, else need to print all distinct elements sorted by frequency.

**Examples:**

Input :arr[] = {5, 2, 1, 3, 2}

k = 4

Output :5 2 5 1 2 5 1 2 3 5 2 1 3 5

Explanation:

- After reading 5, there is only one element 5 whose frequency is max till now.

so print 5.- After reading 2, we will have two elements 2 and 5 with the same frequency.

As 2, is smaller than 5 but their frequency is the same so we will print 2 5.- After reading 1, we will have 3 elements 1, 2 and 5 with the same frequency,

so print 1 2 5.- Similarly after reading 3, print 1 2 3 5
- After reading last element 2 since 2 has already occurred so we have now a

frequency of 2 as 2. So we keep 2 at the top and then rest of the element

with the same frequency in sorted order. So print, 2 1 3 5.

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

k = 4

Output :5 2 5 1 2 5 1 2 3 5 1 2 3 4

Explanation:

- After reading 5, there is only one element 5 whose frequency is max till now.

so print 5.- After reading 2, we will have two elements 2 and 5 with the same frequency.

As 2, is smaller than 5 but their frequency is the same so we will print 2 5.- After reading 1, we will have 3 elements 1, 2 and 5 with the same frequency,

so print 1 2 5.

Similarly after reading 3, print 1 2 3 5- After reading last element 4, All the elements have same frequency

So print, 1 2 3 4.

__ Approach:__ The idea is to store the top k elements with maximum frequency. To store them a vector or an array can be used. To keep the track of frequencies of elements create a HashMap to store element-frequency pair. Given a stream of numbers, when a new element appears in the stream update the frequency of that element in HashMap and put that element at the end of the list of K numbers (total k+1 elements) now compare adjacent elements of the list and swap if higher frequency element is stored next to it.

**Algorithm:**

- Create a Hashmap
*hm*, and an array of*k + 1*length. - Traverse the input array from start to end.
- Insert the element at k+1 th position of the array, update the frequency of that element in HashMap.
- Now, traverse the temp array from start to end – 1
- For very element, compare the frequency and swap if higher frequency element is stored next to it, if the frequency is same then swap is the next element is greater.
- print the top k element in each traversal of original array.

**Implementation:**

## C++

`// C++ program to find top k elements in a stream ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to print top k numbers ` `void` `kTop(` `int` `a[], ` `int` `n, ` `int` `k) ` `{ ` ` ` `// vector of size k+1 to store elements ` ` ` `vector<` `int` `> top(k + 1); ` ` ` ` ` `// array to keep track of frequency ` ` ` `unordered_map<` `int` `, ` `int` `> freq; ` ` ` ` ` `// iterate till the end of stream ` ` ` `for` `(` `int` `m = 0; m < n; m++) { ` ` ` `// increase the frequency ` ` ` `freq[a[m]]++; ` ` ` ` ` `// store that element in top vector ` ` ` `top[k] = a[m]; ` ` ` ` ` `// search in top vector for same element ` ` ` `auto` `it = find(top.begin(), top.end() - 1, a[m]); ` ` ` ` ` `// iterate from the position of element to zero ` ` ` `for` `(` `int` `i = distance(top.begin(), it) - 1; i >= 0; --i) { ` ` ` `// compare the frequency and swap if higher ` ` ` `// frequency element is stored next to it ` ` ` `if` `(freq[top[i]] < freq[top[i + 1]]) ` ` ` `swap(top[i], top[i + 1]); ` ` ` ` ` `// if frequency is same compare the elements ` ` ` `// and swap if next element is high ` ` ` `else` `if` `((freq[top[i]] == freq[top[i + 1]]) ` ` ` `&& (top[i] > top[i + 1])) ` ` ` `swap(top[i], top[i + 1]); ` ` ` `else` ` ` `break` `; ` ` ` `} ` ` ` ` ` `// print top k elements ` ` ` `for` `(` `int` `i = 0; i < k && top[i] != 0; ++i) ` ` ` `cout << top[i] << ` `' '` `; ` ` ` `} ` ` ` `cout << endl; ` `} ` ` ` `// Driver program to test above function ` `int` `main() ` `{ ` ` ` `int` `k = 4; ` ` ` `int` `arr[] = { 5, 2, 1, 3, 2 }; ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` `kTop(arr, n, k); ` ` ` `return` `0; ` `} ` |

## Java

`import` `java.io.*; ` `import` `java.util.*; ` `class` `GFG { ` ` ` ` ` `// function to search in top vector for element ` ` ` `static` `int` `find(` `int` `[] arr, ` `int` `ele) ` ` ` `{ ` ` ` `for` `(` `int` `i = ` `0` `; i < arr.length; i++) ` ` ` `if` `(arr[i] == ele) ` ` ` `return` `i; ` ` ` `return` `-` `1` `; ` ` ` `} ` ` ` ` ` `// Function to print top k numbers ` ` ` `static` `void` `kTop(` `int` `[] a, ` `int` `n, ` `int` `k) ` ` ` `{ ` ` ` `// vector of size k+1 to store elements ` ` ` `int` `[] top = ` `new` `int` `[k + ` `1` `]; ` ` ` ` ` `// array to keep track of frequency ` ` ` `HashMap<Integer, Integer> freq = ` `new` `HashMap<>(); ` ` ` `for` `(` `int` `i = ` `0` `; i < k + ` `1` `; i++) ` ` ` `freq.put(i, ` `0` `); ` ` ` ` ` `// iterate till the end of stream ` ` ` `for` `(` `int` `m = ` `0` `; m < n; m++) { ` ` ` `// increase the frequency ` ` ` `if` `(freq.containsKey(a[m])) ` ` ` `freq.put(a[m], freq.get(a[m]) + ` `1` `); ` ` ` `else` ` ` `freq.put(a[m], ` `1` `); ` ` ` ` ` `// store that element in top vector ` ` ` `top[k] = a[m]; ` ` ` ` ` `// search in top vector for same element ` ` ` `int` `i = find(top, a[m]); ` ` ` `i -= ` `1` `; ` ` ` ` ` `// iterate from the position of element to zero ` ` ` `while` `(i >= ` `0` `) { ` ` ` `// compare the frequency and swap if higher ` ` ` `// frequency element is stored next to it ` ` ` `if` `(freq.get(top[i]) < freq.get(top[i + ` `1` `])) { ` ` ` `int` `temp = top[i]; ` ` ` `top[i] = top[i + ` `1` `]; ` ` ` `top[i + ` `1` `] = temp; ` ` ` `} ` ` ` ` ` `// if frequency is same compare the elements ` ` ` `// and swap if next element is high ` ` ` `else` `if` `((freq.get(top[i]) == freq.get(top[i + ` `1` `])) && (top[i] > top[i + ` `1` `])) { ` ` ` `int` `temp = top[i]; ` ` ` `top[i] = top[i + ` `1` `]; ` ` ` `top[i + ` `1` `] = temp; ` ` ` `} ` ` ` ` ` `else` ` ` `break` `; ` ` ` `i -= ` `1` `; ` ` ` `} ` ` ` ` ` `// print top k elements ` ` ` `for` `(` `int` `j = ` `0` `; j < k && top[j] != ` `0` `; ++j) ` ` ` `System.out.print(top[j] + ` `" "` `); ` ` ` `} ` ` ` `System.out.println(); ` ` ` `} ` ` ` ` ` `// Driver program to test above function ` ` ` `public` `static` `void` `main(String args[]) ` ` ` `{ ` ` ` `int` `k = ` `4` `; ` ` ` `int` `[] arr = { ` `5` `, ` `2` `, ` `1` `, ` `3` `, ` `2` `}; ` ` ` `int` `n = arr.length; ` ` ` `kTop(arr, n, k); ` ` ` `} ` `} ` ` ` `// This code is contributed by rachana soma ` |

## Python

`# Python program to find top k elements in a stream ` ` ` `# Function to print top k numbers ` `def` `kTop(a, n, k): ` ` ` ` ` `# list of size k + 1 to store elements ` ` ` `top ` `=` `[` `0` `for` `i ` `in` `range` `(k ` `+` `1` `)] ` ` ` ` ` `# dictionary to keep track of frequency ` ` ` `freq ` `=` `{i:` `0` `for` `i ` `in` `range` `(k ` `+` `1` `)} ` ` ` ` ` `# iterate till the end of stream ` ` ` `for` `m ` `in` `range` `(n): ` ` ` ` ` `# increase the frequency ` ` ` `if` `a[m] ` `in` `freq.keys(): ` ` ` `freq[a[m]] ` `+` `=` `1` ` ` `else` `: ` ` ` `freq[a[m]] ` `=` `1` ` ` ` ` `# store that element in top vector ` ` ` `top[k] ` `=` `a[m] ` ` ` ` ` `i ` `=` `top.index(a[m]) ` ` ` `i ` `-` `=` `1` ` ` ` ` `while` `i >` `=` `0` `: ` ` ` ` ` `# compare the frequency and swap if higher ` ` ` `# frequency element is stored next to it ` ` ` `if` `(freq[top[i]] < freq[top[i ` `+` `1` `]]): ` ` ` `t ` `=` `top[i] ` ` ` `top[i] ` `=` `top[i ` `+` `1` `] ` ` ` `top[i ` `+` `1` `] ` `=` `t ` ` ` ` ` `# if frequency is same compare the elements ` ` ` `# and swap if next element is high ` ` ` `elif` `((freq[top[i]] ` `=` `=` `freq[top[i ` `+` `1` `]]) ` `and` `(top[i] > top[i ` `+` `1` `])): ` ` ` `t ` `=` `top[i] ` ` ` `top[i] ` `=` `top[i ` `+` `1` `] ` ` ` `top[i ` `+` `1` `] ` `=` `t ` ` ` `else` `: ` ` ` `break` ` ` `i ` `-` `=` `1` ` ` ` ` `# print top k elements ` ` ` `i ` `=` `0` ` ` `while` `i < k ` `and` `top[i] !` `=` `0` `: ` ` ` `print` `top[i], ` ` ` `i ` `+` `=` `1` ` ` `print` ` ` `# Driver program to test above function ` `k ` `=` `4` `arr ` `=` `[ ` `5` `, ` `2` `, ` `1` `, ` `3` `, ` `2` `] ` `n ` `=` `len` `(arr) ` `kTop(arr, n, k) ` ` ` `# This code is contributed by Sachin Bisht ` |

## C#

`// C# program to find top k elements in a stream ` `using` `System; ` `using` `System.Collections.Generic; ` ` ` `class` `GFG { ` ` ` `// function to search in top vector for element ` ` ` `static` `int` `find(` `int` `[] arr, ` `int` `ele) ` ` ` `{ ` ` ` `for` `(` `int` `i = 0; i < arr.Length; i++) ` ` ` `if` `(arr[i] == ele) ` ` ` `return` `i; ` ` ` `return` `-1; ` ` ` `} ` ` ` ` ` `// Function to print top k numbers ` ` ` `static` `void` `kTop(` `int` `[] a, ` `int` `n, ` `int` `k) ` ` ` `{ ` ` ` `// vector of size k+1 to store elements ` ` ` `int` `[] top = ` `new` `int` `[k + 1]; ` ` ` ` ` `// array to keep track of frequency ` ` ` `Dictionary<` `int` `, ` ` ` `int` `> ` ` ` `freq = ` `new` `Dictionary<` `int` `, ` ` ` `int` `>(); ` ` ` `for` `(` `int` `i = 0; i < k + 1; i++) ` ` ` `freq.Add(i, 0); ` ` ` ` ` `// iterate till the end of stream ` ` ` `for` `(` `int` `m = 0; m < n; m++) { ` ` ` `// increase the frequency ` ` ` `if` `(freq.ContainsKey(a[m])) ` ` ` `freq[a[m]]++; ` ` ` `else` ` ` `freq.Add(a[m], 1); ` ` ` ` ` `// store that element in top vector ` ` ` `top[k] = a[m]; ` ` ` ` ` `// search in top vector for same element ` ` ` `int` `i = find(top, a[m]); ` ` ` `i--; ` ` ` ` ` `// iterate from the position of element to zero ` ` ` `while` `(i >= 0) { ` ` ` `// compare the frequency and swap if higher ` ` ` `// frequency element is stored next to it ` ` ` `if` `(freq[top[i]] < freq[top[i + 1]]) { ` ` ` `int` `temp = top[i]; ` ` ` `top[i] = top[i + 1]; ` ` ` `top[i + 1] = temp; ` ` ` `} ` ` ` ` ` `// if frequency is same compare the elements ` ` ` `// and swap if next element is high ` ` ` `else` `if` `(freq[top[i]] == freq[top[i + 1]] && top[i] > top[i + 1]) { ` ` ` `int` `temp = top[i]; ` ` ` `top[i] = top[i + 1]; ` ` ` `top[i + 1] = temp; ` ` ` `} ` ` ` `else` ` ` `break` `; ` ` ` ` ` `i--; ` ` ` `} ` ` ` ` ` `// print top k elements ` ` ` `for` `(` `int` `j = 0; j < k && top[j] != 0; ++j) ` ` ` `Console.Write(top[j] + ` `" "` `); ` ` ` `} ` ` ` `Console.WriteLine(); ` ` ` `} ` ` ` ` ` `// Driver Code ` ` ` `public` `static` `void` `Main(String[] args) ` ` ` `{ ` ` ` `int` `k = 4; ` ` ` `int` `[] arr = { 5, 2, 1, 3, 2 }; ` ` ` `int` `n = arr.Length; ` ` ` `kTop(arr, n, k); ` ` ` `} ` `} ` ` ` `// This code is contributed by ` `// sanjeev2552 ` |

**Output:**

5 2 5 1 2 5 1 2 3 5 2 1 3 5

**Complexity Analysis:**

**Time Complexity:**O( n * k ).

In each traversal the temp array of size k is traversed, So the time Complexity is O( n * k ).**Space Complexity:**O(n).

To store the elements in HashMap O(n) space is required.

This article is contributed by **Niteesh Kumar**. 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.

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