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# 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
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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