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leetcode/note/easy/169_majority_element.md

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+---
+id: 169
+title: Majority Element
+difficulty: Easy
+tags:
+  - array
+  - hash-table
+  - sorting
+  - counting
+status: Solved
+date_solved: 2026-05-26
+leetcode_url: https://leetcode.com/problems/majority-element/
+review_needed: false
+---
+
+# 169. Majority Element
+
+> [!info] **Problem Link**: [LeetCode - Majority Element](https://leetcode.com/problems/majority-element/)
+
+## 📝 Problem Description
+
+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`
+
+
+---
+
+## 💡 Approaches & Explanations
+have cont, if cont  is equal to 0 the res become the highest amount 
+if i equal to the res than add one to count anything else cont -1
+return res at the end
+
+## 💻 Code Implementations
+
+### Python3
+```python
+class Solution:
+	def majorityElement(self, nums: List[int]) -> int:
+		cont = 0
+		res = None
+		for i in nums:
+			if cont == 0:
+				res = i
+			cont +=1 if res == i else -1
+		return res
+```