Majority Element

Array

Easy

Given an array of sizen, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋times.

You may assume that the array is non-empty and the majority element always exist in the array.

Example 1:

Input:
[3,2,3]

Output:
3

Example 2:

Input:
[2,2,1,1,1,2,2]

Output:
2

Solution

Approach 1: HashMap

We can use a HashMap that maps elements to counts in order to count occurrences in linear time by looping over nums. Then, we simply return the key with maximum value.

• Time complexity: O(n)
• Space complexity : O(n)
class Solution {
private Map<Integer, Integer> countNums(int[] nums) {
Map<Integer, Integer> counts = new HashMap<Integer, Integer>();
for (int num : nums) {
if (!counts.containsKey(num)) {
counts.put(num, 1);
}
else {
counts.put(num, counts.get(num)+1);
}
}
return counts;
}

public int majorityElement(int[] nums) {
Map<Integer, Integer> counts = countNums(nums);

Map.Entry<Integer, Integer> majorityEntry = null;
for (Map.Entry<Integer, Integer> entry : counts.entrySet()) {
if (majorityEntry == null || entry.getValue() > majorityEntry.getValue()) {
majorityEntry = entry;
}
}

return majorityEntry.getKey();
}
}

Approach 2: Sorting

If the elements are sorted in monotonically increasing (or decreasing) order, the majority element can be found at index ⌊ n / 2 ⌋ (and ⌊ n / 2 ⌋ + 1, incidentally, if n is even).

• Time complexity : O(nlgn) Sorting the array costs O(nlgn) time in Python and Java, so it dominates the overall runtime.
• Space complexity : O(1) or (O(n)) We sorted nums in place here - if that is not allowed, then we must spend linear additional space on a copy of nums and sort the copy instead.
class Solution {
public int majorityElement(int[] nums) {
Arrays.sort(nums);
return nums[nums.length/2];
}
}

Approach 3: Boyer-Moore Voting Algorithm

If we had some way of counting instances of the majority element as +1 and instances of any other element as −1, summing them would make it obvious that the majority element is indeed the majority element.

Complexity Analysis

• Time complexity :O(n)

Boyer-Moore performs constant work exactlynntimes, so the algorithm runs in linear time.

• Space complexity :O(1)

Boyer-Moore allocates only constant additional memory.

class Solution {
public int majorityElement(int[] nums) {
int count = 0;
Integer candidate = null;

for (int num : nums) {
if (count == 0) {
candidate = num;
}
count += (num == candidate) ? 1 : -1;
}

return candidate;
}
}

Reference

https://leetcode.com/problems/majority-element/solution/