Question

Given an array with N elements. We know that one of those elements repeats itself at least N/2 times.

We don’t know anything about the other elements . They may repeat or may be unique .

Is there a way to find out the element that repeats at least N/2 times in a single pass or may be O(N)?

No extra space is to be used .

Answer 1

st0le answered the question, but here’s a 5minute implementation:

#include <stdio.h>

#define SIZE 13

int boyerMoore(int arr[]) {
    int current_candidate = arr[0], counter = 0, i;
    for (i = 0; i < SIZE; ++i) {
        if (current_candidate == arr[i]) {
            ++counter;
            printf("candidate: %i, counter: %i\n",current_candidate,counter);
        } else if (counter == 0) {
            current_candidate = arr[i];
            ++counter;
            printf("candidate: %i, counter: %i\n",current_candidate,counter);
        } else {
            --counter;
            printf("candidate: %i, counter: %i\n",current_candidate,counter);
        }
    }
    return current_candidate;
}

int main() {
    int numbers[SIZE] = {5,5,5,3,3,1,1,3,3,3,1,3,3};
    printf("majority: %i\n", boyerMoore(numbers));
    return 0;
}

And here’s a fun explanation (more fun than reading the paper, at least): http://userweb.cs.utexas.edu/~moore/best-ideas/mjrty/index.html