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Visit original link: 169. Majority Element - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions for a better experience!

LeetCode link: 169. Majority Element, difficulty: Easy.

Given an array nums of size n, return the majority element.

The majority element is the element that appears more than ⌊n / 2⌋ times. You may assume that the majority element always exists in the array.

Input: nums = [3,2,3]

Output: 3

Input: nums = [2,2,1,1,1,2,2]

Output: 2

• n == nums.length
• 1 <= n <= 5 * 10^4
• -10^9 <= nums[i] <= 10^9

Follow-up: Could you solve the problem in linear time and in O(1) space?

The key to solving this problem is to use a hash table to store the occurrence count of each num. The key is the num, and the value is the number of times it appears.

• Time complexity: O(N).
• Space complexity: O(N).

class Solution:
    def majorityElement(self, nums: List[int]) -> int:
        num_to_count = defaultdict(int)

        for num in nums:
            num_to_count[num] += 1
            if num_to_count[num] >= len(nums) / 2:
                return num

# @param {Integer[]} nums
# @return {Integer}
def majority_element(nums)
  num_to_count = Hash.new(0)

  nums.each do |num|
    num_to_count[num] += 1

    if num_to_count[num] > nums.size / 2
      return num
    end
  end
end

class Solution {
    public int majorityElement(int[] nums) {
        Map<Integer, Integer> numToCount = new HashMap<>();

        for (int num : nums) {
            numToCount.put(num, numToCount.getOrDefault(num, 0) + 1);

            if (numToCount.get(num) > nums.length / 2) {
                return num;
            }
        }

        return -1; // This line won't be reached due to problem constraints
    }
}

// Welcome to create a PR to complete the code of this language, thanks!

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Visit original link: 169. Majority Element - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions for a better experience!

GitHub repository: leetcode-python-java.