Given an array of of size n and a number k, finds all elements that appear more than n/k times

Given an array of size n, find all elements in array that appear more than n/k times. For example, if the input arrays is {3, 1, 2, 2, 1, 2, 3, 3} and k is 4, then the output should be [2, 3]. Note that size of array is 8 (or n = 8), so we need to find all elements that appear more than 2 (or 8/4) times. There are two elements that appear more than two times, 2 and 3.
A simple method 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. Time Complexity of this method would be O(n²).
A better solution is to use sorting. First, sort all elements using a O(nLogn) algorithm. Once the array is sorted, we can find all required elements in a linear scan of array. So overall time complexity of this method is O(nLogn) + O(n) which is O(nLogn).
Following is an interesting O(nk) solution:
We can solve the above problem in O(nk) time using O(k-1) extra space. Note that there can never be more than k-1 elements in output (Why?). There are mainly three steps in this algorithm.
1) 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). Following is structure of temporary array elements.

struct eleCount {
    int element;
    int count;
}; 
struct eleCount temp[]; 

This step takes O(k) time.
2) 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. This step takes O(nk) time.
3) Iterate through final (k-1) potential candidates (stored in temp[]). or every element, check if it actually has count more than n/k. This step takes O(nk) time.
The main step is step 2, how to maintain (k-1) potential candidates at every point?  
We increment the count of an element if it is already present in array, else if there is an empty position we nsert it there and set the count to 1,else we decrement count of all elements by 1.

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

i = 0
         3 _ _
temp[] has one element, 3 with count 1

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

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

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

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

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

Now the question arises, what to do when temp[] is full and we see a new element – we remove the bottom row from stacks of elements, i.e., we decrease count of every element by 1 in temp[]. We ignore the current element.

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

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

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

Finally, we have at most k-1 numbers in temp[]. The elements in temp are {3, 1, 2}. Note that the counts in temp[] are useless now, the counts were needed only in step 2. Now we need to check whether the actual counts of elements in temp[] are more than n/k (9/4) or not. The elements 3 and 2 have counts more than 9/4. So we print 3 and 2.
Note that the algorithm doesn’t miss any output element. There can be two possibilities, many occurrences are together or spread across the array. If occurrences are together, then count will be high and won’t become 0. If occurrences are spread, then the element would come again in temp[]. Following is C++ implementation of above algorithm.

// A C++ program to print elements with count more than n/k #include<iostream> using namespace std; // A structure to store an element and its current count struct eleCount { int e; // Element int c; // Count }; // Prints elements with more than n/k occurrences in arr[] of // size n. If there are no such elements, then it prints nothing. void moreThanNdK(int arr[], int n, int k) { // k must be greater than 1 to get some output if (k < 2) return; /* Step 1: Create a temporary array (contains element and count) of size k-1. Initialize count of all elements as 0 */ struct eleCount temp[k-1]; for (int i=0; i<k-1; i++) temp[i].c = 0; /* Step 2: Process all elements of input array */ for (int i = 0; i < n; i++) { int j; /* If arr[i] is already present in the element count array, then increment its count */ for (j=0; j<k-1; j++) { if (temp[j].e == arr[i]) { temp[j].c += 1; break; } } /* If arr[i] is not present in temp[] */ if (j == k-1) { int l; /* If there is position available in temp[], then place arr[i] in the first available position and set count as 1*/ for (l=0; l<k-1; l++) { if (temp[l].c == 0) { temp[l].e = arr[i]; temp[l].c = 1; break; } } /* If all the position in the temp[] are filled, then decrease count of every element by 1 */ if (l == k-1) for (l=0; l<k; l++) temp[l].c -= 1; } } /*Step 3: Check actual counts of potential candidates in temp[]*/ for (int i=0; i<k-1; i++) { // Calculate actual count of elements int ac = 0; // actual count for (int j=0; j<n; j++) if (arr[j] == temp[i].e) ac++; // If actual count is more than n/k, then print it if (ac > n/k) cout << "Number:" << temp[i].e << " Count:" << ac << endl; } } /* Driver program to test above function */ int main() { cout << "First Test\n"; int arr1[] = {4, 5, 6, 7, 8, 4, 4}; int size = sizeof(arr1)/sizeof(arr1[0]); int k = 3; moreThanNdK(arr1, size, k); cout << "\nSecond Test\n"; int arr2[] = {4, 2, 2, 7}; size = sizeof(arr2)/sizeof(arr2[0]); k = 3; moreThanNdK(arr2, size, k); cout << "\nThird Test\n"; int arr3[] = {2, 7, 2}; size = sizeof(arr3)/sizeof(arr3[0]); k = 2; moreThanNdK(arr3, size, k); cout << "\nFourth Test\n"; int arr4[] = {2, 3, 3, 2}; size = sizeof(arr4)/sizeof(arr4[0]); k = 3; moreThanNdK(arr4, size, k); return 0; }

Output:

First Test
Number:4 Count:3

Second Test
Number:2 Count:2

Third Test
Number:2 Count:2

Fourth Test
Number:2 Count:2
Number:3 Count:2

Time Complexity: O(nk)
Auxiliary Space: O(k)
Generally asked variations of this problem are, find all elements that appear n/3 times or n/4 times in O(n) time complexity.