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

Question: 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[] ), 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 element at index i + n/2. If 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 */
#include<bits/stdc++.h>
using namespace std;

bool isMajority(int arr[], int n, int x)
{
    int i;

    /* get last index according to n (even or odd) */
    int last_index = n % 2 ? (n / 2 + 1): (n / 2);

    /* 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 1;
    }
    return 0;
}

/* Driver code */
int main()
{
    int arr[] ={1, 2, 3, 4, 4, 4, 4};
    int n = sizeof(arr)/sizeof(arr[0]);
    int x = 4;
    if (isMajority(arr, n, x))
        cout <<    x <<" appears more than "<<
                              n/2 << " times in arr[]"<< endl;
    else
        cout <<x <<" does not appear more than" << n/2 <<"  times in arr[]" << endl;

return 0;
}

// This code is contributed by shivanisinghss2110

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
#include<bits/stdc++.h>
using namespace std;

// If x is present in arr[low...high]
// then returns the index of first
// occurrence of x, otherwise returns -1
int _binarySearch(int arr[], int low,
                  int high, int x);

// This function returns true if the x
// is present more than n/2 times in
// arr[] of size n
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;
}

// If x is present in arr[low...high] then
// returns the index of first occurrence
// of x, otherwise returns -1
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;
}

// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 3, 3, 3, 10 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int x = 3;

    if (isMajority(arr, n, x))
        cout << x << " appears more than "
             << n / 2 << " times in arr[]"
             << endl;
    else
        cout << x << " does not appear more than"
             << n / 2 << "  times in arr[]" << endl;

    return 0;
}

// This code is contributed by shivanisinghss2110

Output: 

3 appears more than 3 times in arr[]

Time Complexity: O(Logn) 

Auxiliary Space: O(1)

As constant extra space is used.
Algorithmic Paradigm: Divide and Conquer

METHOD 3: 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++

#include <iostream>
using namespace std;

bool isMajorityElement(int arr[], int n, int key)
{
    if (arr[n / 2] == key)
        return true;
    else
        return false;
}

int main()
{
    int arr[] = { 1, 2, 3, 3, 3, 3, 10 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int x = 3;
    if (isMajorityElement(arr, n, x))
        cout << x << " appears more than "
             << n / 2 << " times in arr[]"
             << endl;
    else
        cout << x << " does not appear more than"
             << n / 2 << "  times in arr[]" << endl;

    return 0;
}

// This code is contributed by shivanisinghss2110

Output

3 appears more than 3 times in arr[]

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

Please refer complete article on Check for Majority Element in a sorted array for more details!