Check for Majority Element in a sorted array

2.3

Question: Write a C 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 <stdio.h>
# include <stdbool.h>

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 program to check above function */
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))
        printf("%d appears more than %d times in arr[]",
               x, n/2);
     else
        printf("%d does not appear more than %d times in arr[]",
                x, n/2);

   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.

C

/* Program to check for majority element in a sorted array */
# include <stdio.h>
# include <stdbool.h>

/* 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 program to check above functions */
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))
        printf("%d appears more than %d times in arr[]",
               x, n/2);
    else
        printf("%d does not appear more than %d times in arr[]",
               x, n/2);
    return 0;
}

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

Please write comments if you find any bug in the above program/algorithm or a better way to solve the same problem.

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