Majority Element LeetCode Solution

Problem – Majority Element

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.

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

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

Majority Element LeetCode Solution in C++

class Solution {
public:
    int majorityElement(vector<int>& nums) {
        unordered_map<int, int> counter;
        for (int num : nums) {
            if (++counter[num] > nums.size() / 2) {
                return num;
            }
        }
        return 0;
    }
};

Majority Element LeetCode Solution in Java

public int majorityElement2(int[] nums) {
    Map<Integer, Integer> myMap = new HashMap<Integer, Integer>();
    //Hashtable<Integer, Integer> myMap = new Hashtable<Integer, Integer>();
    int ret=0;
    for (int num: nums) {
        if (!myMap.containsKey(num))
            myMap.put(num, 1);
        else
            myMap.put(num, myMap.get(num)+1);
        if (myMap.get(num)>nums.length/2) {
            ret = num;
            break;
        }
    }
    return ret;
}

Majority Element LeetCode Solution in Python

class Solution(object):
    def majorityElement1(self, nums):
        nums.sort()
        return nums[len(nums)//2]
    
    def majorityElement2(self, nums):
        m = {}
        for n in nums:
            m[n] = m.get(n, 0) + 1
            if m[n] > len(nums)//2:
                return n
            
    def majorityElement(self, nums):
        candidate, count = nums[0], 0
        for num in nums:
            if num == candidate:
                count += 1
            elif count == 0:
                candidate, count = num, 1
            else:
                count -= 1
        return candidate

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