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+---
+id: Majority-Element
+title: Majority Element
+sidebar_label: 0169-Majority-Element
+tags:
+   - Array
+   - Hash Table
+   - Sorting
+   - Counting
+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."
+---
+
+
+
+## Problem Statement
+
+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.
+
+**Follow-up:** Could you solve the problem in linear time and in O(1) space?
+
+
+### Examples
+
+**Example 1:**
+
+```plaintext
+Input: nums = [3,2,3]
+Output: 3
+```
+
+**Example 2:**
+
+```plaintext
+Input: nums = [2,2,1,1,1,2,2]
+Output: 2
+```
+
+### Constraints
+
+- `n == nums.length`
+- `1 <= n <= 5 * 104`
+- `-109 <= nums[i] <= 109`
+  
+
+## Solution
+
+If the array contains a majority element, its occurrence must be greater than the floor(N/2). 
+Now, we can say that the count of minority elements and majority elements is equal up to a certain point in the array. 
+So when we traverse through the array we try to keep track of the count of elements and the element itself for which we are tracking the count. 
+
+After traversing the whole array, we will check the element stored in the variable. 
+The question states that the array must contain a majority element, so the stored element will be that one 
+
+### Approach 
+
+#### Algorithm
+
+1. Initialize 2 variables:
+     - Count –  for tracking the count of element
+     - Element – for which element we are counting
+2. Traverse through the given array.
+     - If Count is 0 then store the current element of the array as Element.
+     - If the current element and Element are the same increase the Count by 1.
+     - If they are different decrease the Count by 1.
+3. The integer present in Element should be the result we are expecting 
+
+
+
+#### Solution
+
+### Java Solution
+
+```Java
+import java.util.*;
+
+public class tUf {
+    public static int majorityElement(int []v) {
+        //size of the given array:
+        int n = v.length;
+        int cnt = 0; // count
+        int el = 0; // Element
+
+        //applying the algorithm:
+        for (int i = 0; i < n; i++) {
+            if (cnt == 0) {
+                cnt = 1;
+                el = v[i];
+            } else if (el == v[i]) cnt++;
+            else cnt--;
+        }
+
+        return el;
+    }
+
+    public static void main(String args[]) {
+        int[] arr = {2, 2, 1, 1, 1, 2, 2};
+        int ans = majorityElement(arr);
+        System.out.println("The majority element is: " + ans);
+
+    }
+
+}
+
+```
+
+### C++ Solution
+
+```cpp
+#include <bits/stdc++.h>
+using namespace std;
+
+int majorityElement(vector<int> v) {
+
+    //size of the given array:
+    int n = v.size();
+    int cnt = 0; // count
+    int el; // Element
+
+    //applying the algorithm:
+    for (int i = 0; i < n; i++) {
+        if (cnt == 0) {
+            cnt = 1;
+            el = v[i];
+        }
+        else if (el == v[i]) cnt++;
+        else cnt--;
+    }
+
+return el;
+}
+
+int main()
+{
+    vector<int> arr = {2, 2, 1, 1, 1, 2, 2};
+    int ans = majorityElement(arr);
+    cout << "The majority element is: " << ans << endl;
+    return 0;
+}
+```
+
+### Python Solution
+
+```python
+def majorityElement(arr):
+    # Size of the given array
+    n = len(arr)
+    cnt = 0  # Count
+    el = None  # Element
+
+    # Applying the algorithm
+    for i in range(n):
+        if cnt == 0:
+            cnt = 1
+            el = arr[i]
+        elif el == arr[i]:
+            cnt += 1
+        else:
+            cnt -= 1
+
+    return el
+    
+
+arr = [2, 2, 1, 1, 1, 2, 2]
+ans = majorityElement(arr)
+print("The majority element is:", ans)
+```
+
+### Complexity Analysis
+
+- **Time complexity**: $O(N)$, where N = size of the given array.
+- **Space complexity**: $O(1)$ as we are not using any extra space.
+