Find if there is an element repeating itself n/k times

Question

You have an array size n and a constant k (whatever)

You can assume the the array is of int type (although it could be of any type)

Describe an algorithm that finds if there is an element(s) that repeats itself at least n/k times... if there is return one. Do so in linear time (O(n))

The catch: do this algorithm (or even pseudo-code) using constant memory and running over the array only twice

Answer 1

I'm not 100% sure, but it sounds like you want to solve the Britney Spears problem—finding an item that makes up a certain fraction of a sample using constant memory.

Here is a statement of the problem in English, with a sketch of the solution:

… from a 2002 article by Erik D. Demaine of MIT and Alejandro López-Ortiz and J. Ian Munro of the University of Waterloo in Canada. Demaine and his colleagues have extended the algorithm to cover a more-general problem: Given a stream of length n, identify a set of size m that includes all the elements occurring with a frequency greater than n /( m +1). (In the case of m =1, this reduces to the majority problem.) The extended algorithm requires m registers for the candidate elements as well as m counters. The basic scheme of operation is analogous to that of the majority algorithm. When a stream ele­ment matches one of the candidates, the corresponding counter is incremented; when there is no match to any candidate, all of the counters are decremented; if a counter is at 0, the associated candidate is replaced by a new element from the stream.

Answer 2

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).

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? The steps used in step 2 are like famous game: Tetris. We treat each number as a piece in Tetris, which falls down in our temporary array temp[]. Our task is to try to keep the same number stacked on the same column (count in temporary array is incremented).

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;
}