Java Program for Check for Majority Element in a sorted array

Last Updated : 13 Dec, 2021

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.

Java

/* Program to check for majority element in a sorted array */
import java.io.*; 
  
class Majority { 
  
    static boolean 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 function to check for above functions*/
    public static void main (String[] args) { 
        int arr[] = {1, 2, 3, 4, 4, 4, 4}; 
        int n = arr.length; 
        int x = 4; 
        if (isMajority(arr, n, x)==true) 
           System.out.println(x+" appears more than "+ 
                              n/2+" times in arr[]"); 
        else
           System.out.println(x+" does not appear more than "+ 
                              n/2+" times in arr[]"); 
    } 
} 
/*This article is contributed by Devesh Agrawal*/

Output:

4 appears more than 3 times in arr[]

Time Complexity: O(n)

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.

Java

/* Java Program to check for majority element in a sorted array */
import java.io.*; 
  
class Majority { 
  
    /* 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 boolean 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 function to check for above functions*/
    public static void main (String[] args)  { 
  
        int arr[] = {1, 2, 3, 3, 3, 3, 10}; 
        int n = arr.length; 
        int x = 3; 
        if (isMajority(arr, n, x)==true) 
            System.out.println(x + " appears more than "+ 
                              n/2 + " times in arr[]"); 
        else
            System.out.println(x + " does not appear more than " + 
                              n/2 + " times in arr[]"); 
    } 
} 
/*This code is contributed by Devesh Agrawal*/

Output:

3 appears more than 3 times in arr[]

Time Complexity: O(Logn)
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.

Java

import java.util.*; 
  
class GFG{ 
  
static boolean 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) 
{ 
    int arr[] = { 1, 2, 3, 3, 3, 3, 10 }; 
    int n = arr.length; 
    int x = 3; 
      
    if (isMajorityElement(arr, n, x)) 
        System.out.printf("%d appears more than %d " + 
                          "times in arr[]", x, n / 2); 
    else
        System.out.printf("%d does not appear more " + 
                          "than %d times in " + "arr[]", 
                          x, n / 2); 
} 
} 
  
// This code is contributed by aashish1995

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!

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