C# Program to Check for Majority Element in a sorted array

Problem Statement: Write a function to find if a given integer x appears more than n/2 times in a sorted array of n integers.

Basically, we need to write a function say isMajority() that takes an array (arr[] ), the array’s size (n) and a number to be searched (x) as parameters and returns true if x is a majority element (present more than n/2 times).

Examples:

Input: arr[] = {1, 2, 3, 3, 3, 3, 10}, x = 3
Output: True (x appears more than n/2 times in the given array)

Input: arr[] = {1, 1, 2, 4, 4, 4, 6, 6}, x = 4
Output: False (x doesn't appear more than n/2 times in the given array)

Input: arr[] = {1, 1, 1, 2, 2}, x = 1
Output: True (x appears more than n/2 times in the given array)

Method 1: Using linear search

Linearly search for the first occurrence of the element, once you find it (let at index i), check the element at index i + n/2. If the element is present at i+n/2 then return 1 else return 0.

C#

// C# Program to check for majority
// element in a sorted array
using System;
 
class GFG {
    static bool isMajority(int[] arr,
                            int n, int x)
    {
        int i, last_index = 0;
 
        // Get last index according to
        // n (even or odd)
        last_index = (n % 2 == 0) ? n / 2 :
                                    n / 2 + 1;
 
        // Search for first occurrence
        // of x in arr[]
        for (i = 0; i < last_index; i++) {
            // Check if x is present and
            // is present more than n/2 times
            if (arr[i] == x && arr[i + n / 2] == x)
                return true;
        }
        return false;
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = { 1, 2, 3, 4, 4, 4, 4 };
        int n = arr.Length;
        int x = 4;
        if (isMajority(arr, n, x) == true)
            Console.Write(x + " appears more than " +
                            n / 2 + " times in arr[]");
        else
            Console.Write(x + " does not appear more than " +
                             n / 2 + " times in arr[]");
    }
}
 
// This code is contributed by Sam007

Output:

4 appears more than 3 times in arr[]

Time Complexity: O(n)
Auxiliary Space: O(1), as constant extra space is used.

Method 2: Using Binary Search

Use binary search methodology to find the first occurrence of the given number. The criteria for binary search is important here.

C#

// C# Program to check for majority
// element in a sorted array */
using System;
 
class GFG {
 
    // If x is present in arr[low...high]
    // then returns the index of first
    // occurrence of x, otherwise returns -1
    static int _binarySearch(int[] arr, int low, int high,
                             int x)
    {
        if (high >= low) {
            int mid = (low + high) / 2;
            // low + (high - low)/2;
 
            // Check if arr[mid] is the first
            // occurrence of x.    arr[mid] is
            // first occurrence if x is one of
            // the following is true:
            // (i) mid == 0 and arr[mid] == x
            // (ii) arr[mid-1] < x and arr[mid] == x
 
            if ((mid == 0 || x > arr[mid - 1])
                && (arr[mid] == x))
                return mid;
            else if (x > arr[mid])
                return _binarySearch(arr, (mid + 1), high,
                                     x);
            else
                return _binarySearch(arr, low, (mid - 1),
                                     x);
        }
 
        return -1;
    }
 
    // This function returns true if the x is
    // present more than n/2 times in arr[]
    // of size n
    static bool isMajority(int[] arr, int n, int x)
    {
 
        // Find the index of first occurrence
        // of x in arr[]
        int i = _binarySearch(arr, 0, n - 1, x);
 
        // If element is not present at all,
        // return false
        if (i == -1)
            return false;
 
        // check if the element is present
        // more than n/2 times
        if (((i + n / 2) <= (n - 1)) && arr[i + n / 2] == x)
            return true;
        else
            return false;
    }
 
    // Driver code
    public static void Main()
    {
 
        int[] arr = { 1, 2, 3, 3, 3, 3, 10 };
        int n = arr.Length;
        int x = 3;
        if (isMajority(arr, n, x) == true)
            Console.Write(x + " appears more than " + n / 2
                          + " times in arr[]");
        else
            Console.Write(x + " does not appear more than "
                          + n / 2 + " times in arr[]");
    }
}
 
// This code is contributed by Sam007

Output:

3 appears more than 3 times in arr[]

Time Complexity: O(Logn)
Auxiliary Space: O(1), as constant extra space is used.

Method 3: Divide and Conquer

If it is already given that the array is sorted and there exists a majority element, checking if a particular element is as easy as checking if the middle element of the array is the number we are checking against.

Since a majority element occurs more than n/2 times in an array, it will always be the middle element. We can use this logic to check if the given number is the majority element.

C#

using System;
 
// Class
class GFG {
 
    // Method to find majority element
    static bool isMajorityElement(int[] arr, int n, int key)
    {
        if (arr[n / 2] == key)
            return true;
        else
            return false;
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
 
        // Custom input array
        int[] arr = { 1, 2, 3, 3, 3, 3, 10 };
 
        int n = arr.Length;
        int x = 3;
 
        if (isMajorityElement(arr, n, x))
            Console.Write(x + " appears more than " + n / 2
                          + " times in []arr");
        else
            Console.Write(x + " does not appear more "
                          + "than " + n / 2
                          + " times in arr[]");
    }
}
 
// This code is contributed by aashish1995

Output

3 appears more than 3 times in arr[]

Time complexity: O(1)
Auxiliary Space: O(1)

Method 4: Using Moore’s Voting Algorithm

Initialize the candidate element as the first element of the array setting count to 1.
Traverse through the array from the second element to the last.
- If the current element is the same as the candidate element, increment the count otherwise decrement the count.
- If the count becomes 0, set the current element as the candidate element and reset the count to 1.
Check if the candidate element is the majority element by traversing the array again and counting the number of occurrences of the candidate element.
If the number of occurrences of the candidate element is greater than n/2, return the candidate element. Otherwise, return -1.

C#

using System;
 
// Class
class Program {
 
    // Method to find majority
    static int FindMajority(int[] arr, int n)
    {
        // Initialize candidate as first element
        int candidate = arr[0];
        int count = 1;
 
        // Find majority element
        for (int i = 1; i < n; i++) {
            if (arr[i] == candidate)
                count++;
            else
                count--;
 
            if (count == 0) {
                candidate = arr[i];
                count = 1;
            }
        }
 
        // Checking if candidate is the majority element
        int countCandidate = 0;
 
        for (int i = 0; i < n; i++) {
            if (arr[i] == candidate)
                countCandidate++;
        }
 
        if (countCandidate > n / 2)
            return candidate;
        else
            return -1;
    }
 
    // Driver method
    static void Main()
    {
        // Custom input array
        int[] arr = { 1, 2, 3, 3, 3, 3, 10 };
 
        int n = arr.Length;
        int majorityElement = FindMajority(arr, n);
 
        // Printing the majority element
        // it is found
        if (majorityElement != -1)
            Console.Write(majorityElement
                          + " is the majority element");
        else
            // Display message there exists no majority
            // element
            Console.Write("No majority element");
    }
}