Find All Elements in an Array which appears more than N/K times, N is Array Size and k is a Number

By | March 27, 2017

Similarly, original Boyer-Moore Majority Vote Algorithm defined “Majority element of an array A[1 … n]: An array is said to have a majority element if more than half of its entries are the same.” In other words, the majority element is the element that appears more than ⌊ n/2 ⌋. One difference is finding more than one majority element.

Some History Solutions.

Suppose that A array: int [] A = {3, 2, 2, 4, 4, 1, 7, 4, 2, 4}; Therefore, output should be: 4

1. Runtime: O(n), Space: O(n) — Hash table.

Maintain a hash table of the counts of each element,then find the most common one.

public List<Integer> majorityElement(int[] nums) {
    HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
    for(int i: nums){
            map.put(i, map.get(i)+1);
            map.put(i, 1);
    List<Integer> result = new ArrayList<Integer>();
    for(Map.Entry<Integer, Integer> entry: map.entrySet()){
        if(entry.getValue() > nums.length/2){
    return result;

2. Linear Time Majority Vote Algorithm.

Runtime O(n) time O(1) space.

public int majorityElement(int[] nums) {
    int result = 0, count = 0;
    for(int i = 0; i<nums.length; i++ ) {
        if(count == 0){
            result = nums[ i ];
            count = 1;
        }else if(result == nums[i]){
    return result;

3. Runtime O(n log n).

Divide and conquer, Sorting array first solution to find a majority element are best implementation.

Tentative solution.

Thanks to finding the majority element, we temporarily pick up HashMap solution to resolve the problem. N/K= threshold to filter all elements that occur more than N/K times. It is simple to implement but take much space.
int [] A = {3, 2, 2, 4, 4, 1, 7, 4, 2, 4}; Therefore, output should be: {2, 4} if K = 2;

Moreover, an useful reference from improved with datastruct and performance.


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