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/

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