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 βn / 2β times. You may assume that the majority element always exists in the array.
Example 1:
Input: nums = [3,2,3]
Output: 3
Example 2:
Input: nums = [2,2,1,1,1,2,2]
Output: 2
Constraints:
n == nums.length1 <= 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?
Hint 1
How to solve the problem in O(1) space?
Please search Boyer-Moore majority vote algorithm.
Intuition
Webmaster (Zhang Jian): π
Hi everyone! I am Zhang Jian.
I know the challenge of transitioning from mastering algorithms to actually landing a great job. That's why, in addition to this resource, I personally developed
leader.me!
π leader.me is the ultimate all-in-one platform for programmers to build their personal brand, featuring portfolio hosting, resume builders, and integrated blogs.
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.
Complexity
Time complexity
O(N)
Space complexity
O(N)
Python #
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
Ruby #
# @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
Java #
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
}
}
Other languages
Welcome to contribute code to LeetCode.blog GitHub -> 169. Majority Element. Thanks!Level Up Your Developer Identity
π While mastering algorithms is key, showcasing your talent is what gets you hired.
We recommend leader.me β the ultimate all-in-one personal branding platform for programmers.The All-In-One Career Powerhouse:
- π Resume, Portfolio & Blog: Integrate your skills, projects, and writing into one stunning site.
- π Free Custom Domain: Bind your own personal domain for freeβforever.
- β¨ Premium Subdomains: Stand out with elite tech handle like
name.leader.me.