Elements Appearing More Than n/k Times

Given an array arr[] of size n and an integer k, determine the number of elements that appear more than n/k times in the array.

Examples:

Input: arr[ ] = [3, 4, 2, 2, 1, 2, 3, 3], k = 4
Output: 2
Explanation: Here n/k is 8/4 = 2, therefore 2 appears 3 times in the array that is greater than 2 and 3 appears 3 times in the array that is greater than 2

Input: arr[ ] = [9, 10, 7, 9, 2, 9, 10], k = 3
Output: 1
Explanation: Here n/k is 7/3 = 2, therefore 9 appears 3 times in the array that is greater than 2.

[Expected Solution] Use Hashing - O(n) Time and O(n) Space

The idea is to pick all elements one by one. For every picked element, count its occurrences by traversing the array, if count becomes more than n/k, then print the element.

• First, make a frequency map of all the elements in the array
• Then traverse the map and check the frequency of every element
• If the frequency is greater than n/k then print the element.

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[Expected Solution for Small K] - Moore's Voting Algorithm - O(n x k) Time and O(k) Space

The idea is to apply Moore's Voting algorithm, as there can be at max k - 1 elements present in the array which appears more than n/k times so their will be k - 1 candidates. When we encounter an element which is one of our candidates then increment the count else decrement the count.

Illustration:

Consider k = 4, n = 9 
Given array: 3 1 2 2 2 1 4 3 3 

i = 0
temp[] has one element {3} with count 1

i = 1
temp[] has two elements {3, 1} with counts 1 and 1 respectively

i = 2
temp[] has three elements, {3, 1, 2} with counts as 1, 1 and 1 respectively.

i = 3
temp[] has three elements, {3, 1, 2} with counts as 1, 1 and 2 respectively.

i = 4
temp[] has three elements, {3, 1, 2} with counts as 1, 1 and 3 respectively.

i = 5
temp[] has three elements, {3, 1, 2 with counts as 1, 2 and 3 respectively.

i = 6
temp[] has two elements, {1, 2} with counts as 1 and 2 respectively.

i = 7
temp[] has three elements, {3, 1, 2} with counts as 1, 1 and 2 respectively.

i = 8 
temp[] has three elements, {3, 1, 2} with counts as 2, 1 and 2 respectively.

Follow the steps below to solve the problem:

• Create a temporary array of size (k - 1) to store elements and their counts (The output elements are going to be among these k-1 elements).
• Traverse through the input array and update temp[] (add/remove an element or increase/decrease count) for every traversed element. The array temp[] stores potential (k-1) candidates at every step.
• Iterate through final (k-1) potential candidates (stored in temp[]). or every element, check if it actually has counted of more than n/k.

Number:3 Count:3
Number:2 Count:3

Time Complexity: O(n * k), Checking for each element of the array(size N) in the candidate array of size K
Auxiliary Space: O(k) Space required to store the candidates.

Using Built-In Counter in Python

This approach is same the first approach but here in python there is a counter() that calculates the frequency array.

• Count the frequencies of every element using Counter() function.
• Traverse the frequency array and print all the elements which occur at more than n/k times.

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