229. Majority Element II LeetCode Solution

Last updated: February 15, 2025

In this guide, you will get 229. Majority Element II LeetCode Solution with the best time and space complexity. The solution to Majority Element II problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Majority Element II solution in C++
Majority Element II solution in Java
Majority Element II solution in Python
Additional Resources

229. Majority Element II LeetCode Solution image

Problem Statement of Majority Element II

Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.

Example 1:

Input: nums = [3,2,3]
Output: [3]

Example 2:

Input: nums = [1]
Output: [1]

Example 3:

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

Constraints:

1 <= nums.length <= 5 * 104
-109 <= nums[i] <= 109

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

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(1)

229. Majority Element II LeetCode Solution in C++

class Solution {
 public:
  vector<int> majorityElement(vector<int>& nums) {
    vector<int> ans;
    int candidate1 = 0;
    int candidate2 = 1;   // any number different from candidate1
    int countSoFar1 = 0;  // the number of candidate1 so far
    int countSoFar2 = 0;  // the number of candidate2 so far

    for (const int num : nums)
      if (num == candidate1) {
        ++countSoFar1;
      } else if (num == candidate2) {
        ++countSoFar2;
      } else if (countSoFar1 == 0) {  // Assign the new candidate.
        candidate1 = num;
        ++countSoFar1;
      } else if (countSoFar2 == 0) {  // Assign the new candidate.
        candidate2 = num;
        ++countSoFar2;
      } else {  // Meet a new number, so pair with the previous counts.
        --countSoFar1;
        --countSoFar2;
      }

    const int count1 = ranges::count(nums, candidate1);
    const int count2 = ranges::count(nums, candidate2);

    if (count1 > nums.size() / 3)
      ans.push_back(candidate1);
    if (count2 > nums.size() / 3)
      ans.push_back(candidate2);
    return ans;
  }
};
/* code provided by PROGIEZ */

229. Majority Element II LeetCode Solution in Java

class Solution {
  public List<Integer> majorityElement(int[] nums) {
    List<Integer> ans = new ArrayList<>();
    int candidate1 = 0;
    int candidate2 = 1;  // any number different from candidate1
    int countSoFar1 = 0; // the number of candidate1 so far
    int countSoFar2 = 0; // the number of candidate2 so far

    for (final int num : nums)
      if (num == candidate1) {
        ++countSoFar1;
      } else if (num == candidate2) {
        ++countSoFar2;
      } else if (countSoFar1 == 0) { // Assign the new candidate.
        candidate1 = num;
        ++countSoFar1;
      } else if (countSoFar2 == 0) { // Assign the new candidate.
        candidate2 = num;
        ++countSoFar2;
      } else { // Meet a new number, so pair with the previous counts.
        --countSoFar1;
        --countSoFar2;
      }

    int count1 = 0;
    int count2 = 0;

    for (final int num : nums)
      if (num == candidate1)
        ++count1;
      else if (num == candidate2)
        ++count2;

    if (count1 > nums.length / 3)
      ans.add(candidate1);
    if (count2 > nums.length / 3)
      ans.add(candidate2);
    return ans;
  }
}
// code provided by PROGIEZ

229. Majority Element II LeetCode Solution in Python

class Solution:
  def majorityElement(self, nums: list[int]) -> list[int]:
    ans1 = 0
    ans2 = 1
    count1 = 0
    count2 = 0

    for num in nums:
      if num == ans1:
        count1 += 1
      elif num == ans2:
        count2 += 1
      elif count1 == 0:
        ans1 = num
        count1 = 1
      elif count2 == 0:
        ans2 = num
        count2 = 1
      else:
        count1 -= 1
        count2 -= 1

    return [ans for ans in (ans1, ans2) if nums.count(ans) > len(nums) // 3]
 # code by PROGIEZ

Additional Resources

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