# Majority Element

## Appearing more than N / 2 times

This type of majority element problem is a classical problem. There are two algorithm is know can solve this problem in a time cost of .

The first is that we find the median number in the array, it must be the majority element. The simplest way to get a median number is to sort the array, but its time cost is . There is a way to find the median number in based on array partition.

public class Solution {
private void swap(int[] nums, int i, int j) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}

private int partition(int[] nums, int l, int r) {
int pivot = nums[l];
int i = r + 1;
int m = r + 1;
do {
while (nums[--i] < pivot);
swap(nums, --m, i);
} while (i > l);
return m;
}

private int median(int[] nums, int l, int r, int k) {
int p = partition(nums, l, r);
if (p == k) return nums[k];
else if (p < k) return median(nums, p + 1, r, k);
else return median(nums, l, p - 1, k);
}

public int majorityElement(int[] nums) {
return median(nums, 0, nums.length - 1, nums.length / 2);
}
}


Although its time complexity is , but it uses a space of . It is also not fast in enough on LeetCode.

There are also several other solutions such as Using a HashTable to count element numbers, divide and conquer and so on. But the best solusion is commonly thought to be Moore Voting algorithm. Moore voting algorithm's time cost is and space cost is .

public class Solution {
public int majorityElement(int[] nums) {
int majority = nums[0];
int sameCount = 1;

for (int i = 1; i < nums.length; i++) {

if (nums[i] != majority) {
sameCount--;
if (sameCount == 0) {
majority = nums[i];
sameCount++;
}
} else {
sameCount++;
}

}
return majority;
}
}


More description on Moore voting algorithm and other algorithms please refer to the solution page on LeetCode.

## Appearing more than N / 3 times

One step harder, how to find elements that appears more than times? At first we know, there are at most two element which can appear more than times and as to make use of the Moore voting algorithm described above we need a tweak that let one type of elements appears more than times.

Thus we maintain a set which contains two items, apply Moore algorithm on elements in the set against all other elements. That is, if a element is in the set, increase its vote count by 1. Otherwise, we decrease vote count of every element by 1. The total vote of all elements in the set will exceed so that Moore voting algorithm will give a result that contains all such majority elements, but it is possible that there are numbers not are mojority elements left in the set. So we have to check each element to make sure it appears more than times.

Obviously, we can easily generalize this method to cases.

public class MajorityElementII {
private int count(int[] nums, int value) {
int count = 0;
for (int e : nums) {
count += (e == value) ? 1 : 0;
}
return count;
}

public List<Integer> majorityElement(int[] nums) {
ArrayList<Integer> result = new ArrayList<Integer>(2);
int firstCount = 0;
int secondCount = 0;
int first = 0;
int second = 0;

for (int e : nums) {
if (firstCount == 0 || e == first) {
firstCount++;
first = e;
} else if (secondCount == 0 || e == second) {
secondCount++;
second = e;
} else {
firstCount--;
secondCount--;
}
}

if (count(nums, first) > nums.length / 3) result.add(first);
if (count(nums, second) > nums.length / 3) {
if (result.size() == 0 || (result.size() > 0 && result.get(0) != second)) {