Majority Element Problem from gfg is added

Author: Ishitamukherjee2004

dsa-solutions/gfg-solutions/Easy problems/Majority-Element.md

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+---
+id: majority-element
+title: Majority Element
+sidebar_label: Majority-Element
+tags:
+  - Beginner
+  - Geeks for Geeks
+  - CPP
+  - Python
+  - DSA
+description: "This is a solution to the Majority Element Problem on Geeks for Geeks."
+---
+
+This tutorial contains a complete walk-through of the Majority Element problem from the Geeks for Geeks website. It features the implementation of the solution code in two programming languages: Python and C++.
+
+## Problem Description
+Given an array `A` of `N` elements. Find the majority element in the array. A majority element in an array `A` of size `N` is an element that appears strictly more than `N/2` times in the array.
+
+## Examples
+
+**Example 1:**
+
+```
+Input:
+N = 3 
+A[] = {1,2,3} 
+Output:
+-1
+Explanation:
+Since, each element in 
+{1,2,3} appears only once so there 
+is no majority element.
+```
+
+**Example 2:**
+
+```
+Input:
+N = 5 
+A[] = {3,1,3,3,2} 
+Output:
+3
+Explanation:
+Since, 3 is present more
+than N/2 times, so it is 
+the majority element.
+```
+
+## Your Task
+The task is to complete the function majorityElement() which returns the majority element in the array. If no majority exists, return -1.
+
+Expected Time Complexity: $O(n)$
+Expected Auxiliary Space: $O(1)$
+
+## Constraints
+
+`100 <= N <= 2*10^9`
+
+
+## Code Implementation
+
+<Tabs>
+  <TabItem value="Python" label="Python" default>
+  <SolutionAuthor name="@ishiitamukherjee2004"/>
+  ```py
+  class Solution:
+  
+  	
+def majority_element(A):
+    count = 0
+    candidate = None
+    for num in A:
+        if count == 0:
+            candidate = num
+            count = 1
+        elif candidate == num:
+            count += 1
+        else:
+            count -= 1
+    return candidate
+
+	```
+
+  </TabItem>
+  <TabItem value="C++" label="C++">
+  <SolutionAuthor name="@ishitamukherjee2004"/>
+
+  ```cpp
+  int majorityElement(vector<int>& A) {
+    int count = 0;
+    int candidate;
+    for (int num : A) {
+        if (count == 0) {
+            candidate = num;
+            count = 1;
+        } else if (candidate == num) {
+            count++;
+        } else {
+            count--;
+        }
+    }
+    return candidate;
+}
+
+```
+
+  </TabItem>  
+
+    <TabItem value="Java" label="Java" default>
+  <SolutionAuthor name="@ishiitamukherjee2004"/>
+  ```java
+  public int majorityElement(int[] A) {
+    int count = 0;
+    int candidate = 0;
+    for (int num : A) {
+        if (count == 0) {
+            candidate = num;
+            count = 1;
+        } else if (candidate == num) {
+            count++;
+        } else {
+            count--;
+        }
+    }
+    return candidate;
+}
+
+
+	```
+
+  </TabItem>
+
+    <TabItem value="Javascript" label="Javascript" default>
+  <SolutionAuthor name="@ishiitamukherjee2004"/>
+  ```js
+ 
+function majorityElement(A) {
+    let count = 0;
+    let candidate;
+    for (let num of A) {
+        if (count === 0) {
+            candidate = num;
+            count = 1;
+        } else if (candidate === num) {
+            count++;
+        } else {
+            count--;
+        }
+    }
+    return candidate;
+}
+
+
+	```
+
+  </TabItem>
+</Tabs>
+
+
+
+
+## Time Complexity
+
+The time complexity is $O(n)$ because the operations involve a fixed number of steps regardless of the size of N:
+
+## Space Complexity
+
+The space complexity is $O(1)$ as well since the operations use a constant amount of extra space for storing the products and concatenated strings.