Find the majority element using bit logic

The problem at hand is to find the majority element in a given array. A majority element is an element that appears more than n/2 times in an array of n elements, where n is the size of the array.

Example

Let's consider an example to understand the problem better. Suppose we have an array arr[] = {1, 7, 3, 7, 4, 3, 7, 7, 51, 7, 7}. In this array, the number 7 appears 6 times out of the total 11 elements, which is more than n/2 (11/2 = 5). Hence, the majority element in this array is 7.

Pseudocode

Before diving into the algorithm, let's summarize the approach in pseudocode:

findMajorityElement(arr[], n):
    bits = 32
    count = 0
    x = 0

    for i in range(0, bits):
        for j in range(0, n):
            if (arr[j] & (1 << i)) != 0:
                count++

        if count > (n / 2):
            x += (1 << i)

        count = 0

    count = 0
    for i in range(0, n):
        if arr[i] == x:
            count++

    if count > n / 2:
        print("Majority element is", x)
    else:
        print("No Majority element in this array")

Algorithm Explanation

1. The function findMajorityElement takes the array arr[] and its size n as input.
2. We use a variable bits to denote the number of bits in an integer (typically 32 for most systems).
3. We initialize a variable count to keep track of the number of active bits at a specific position (ith position) across all array elements.
4. We use a variable x to find the candidate majority element by setting the bits at positions where the count of set bits is greater than n/2.
5. We loop through each bit position (from 0 to bits-1) and count the number of set bits (1) at that position across all elements of the array.
6. If the count of set bits at the current position is greater than n/2, we set the corresponding bit of x.
7. After obtaining x, we count its frequency in the array to verify if it indeed appears more than n/2 times.
8. If x satisfies the majority condition, we print it as the majority element; otherwise, we print that there is no majority element.

Code Solution

Here given code implementation process.

• C
• Java
• C++
• C#
• Go
• Php
• Node Js
• Python
• Ruby
• Scala
• Swift 4
• Kotlin

// C Program for
// Find the majority element using bit logic
#include <stdio.h>

void findMajorityElement(int arr[], int n)
{
	// Auxiliary variable
	int bits = 32;
	int count = 0;
	int x = 0;
	// This loop are work on range (1...31) bit number
	for (int i = 0; i < bits; ++i)
	{
		// Count active bits in ith position in every array element
		for (int j = 0; j < n; ++j)
		{
			if ((arr[j] & (1 << i)) != 0)
			{
				count++;
			}
		}
		if (count > (n / 2))
		{
			// Add value of i-th bit
			x += (1 << i);
		}
		count = 0;
	}
	// Count frequency of x element
	for (int i = 0; i < n; ++i)
	{
		if (arr[i] == x)
		{
			count++;
		}
	}
	if (count > n / 2)
	{
		// x appears more than n/2 times in the given array
		printf("\n Majority element is %d", x);
	}
	else
	{
		printf("\n No Majority element in this array ");
	}
}
int main()
{
	int arr[] = {
		1 , 7 , 3 , 7 , 4 , 3 , 7 , 7 , 51 , 7 , 7
	};
	// Get the length of array
	int n = sizeof(arr) / sizeof(arr[0]);
	// Test
	findMajorityElement(arr, n);
	return 0;
}

 Majority element is 7

// Java program for
// Find the majority element using bit logic
public class Searching
{
	public void findMajorityElement(int[] arr, int n)
	{
		// Auxiliary variable
		int bits = 32;
		int count = 0;
		int x = 0;
		// This loop are work on range (1...31) bit number
		for (int i = 0; i < bits; ++i)
		{
			// Count active bits in ith position in every array element
			for (int j = 0; j < n; ++j)
			{
				if ((arr[j] & (1 << i)) != 0)
				{
					count++;
				}
			}
			if (count > (n / 2))
			{
				// Add value of i-th bit
				x += (1 << i);
			}
			count = 0;
		}
		// Count frequency of x element
		for (int i = 0; i < n; ++i)
		{
			if (arr[i] == x)
			{
				count++;
			}
		}
		if (count > n / 2)
		{
			// x appears more than n/2 times in the given array
			System.out.print("\n Majority element is " + x);
		}
		else
		{
			System.out.print("\n No Majority element in this array ");
		}
	}
	public static void main(String[] args)
	{
		Searching task = new Searching();
		int[] arr = {
			1 , 7 , 3 , 7 , 4 , 3 , 7 , 7 , 51 , 7 , 7
		};
		// Get the length of array
		int n = arr.length;
		// Test
		task.findMajorityElement(arr, n);
	}
}

 Majority element is 7

// Include header file
#include <iostream>
using namespace std;
// C++ program for
// Find the majority element using bit logic
class Searching
{
	public: void findMajorityElement(int arr[], int n)
	{
		// Auxiliary variable
		int bits = 32;
		int count = 0;
		int x = 0;
		// This loop are work on range (1...31) bit number
		for (int i = 0; i < bits; ++i)
		{
			// Count active bits in ith position in every array element
			for (int j = 0; j < n; ++j)
			{
				if ((arr[j] &(1 << i)) != 0)
				{
					count++;
				}
			}
			if (count > (n / 2))
			{
				// Add value of i-th bit
				x += (1 << i);
			}
			count = 0;
		}
		// Count frequency of x element
		for (int i = 0; i < n; ++i)
		{
			if (arr[i] == x)
			{
				count++;
			}
		}
		if (count > n / 2)
		{
			// x appears more than n/2 times in the given array
			cout << "\n Majority element is " << x;
		}
		else
		{
			cout << "\n No Majority element in this array ";
		}
	}
};
int main()
{
	Searching *task = new Searching();
	int arr[] = {
		1 , 7 , 3 , 7 , 4 , 3 , 7 , 7 , 51 , 7 , 7
	};
	// Get the length of array
	int n = sizeof(arr) / sizeof(arr[0]);
	// Test
	task->findMajorityElement(arr, n);
	return 0;
}

 Majority element is 7

// Include namespace system
using System;
// Csharp program for
// Find the majority element using bit logic
public class Searching
{
	public void findMajorityElement(int[] arr, int n)
	{
		// Auxiliary variable
		int bits = 32;
		int count = 0;
		int x = 0;
		// This loop are work on range (1...31) bit number
		for (int i = 0; i < bits; ++i)
		{
			// Count active bits in ith position in every array element
			for (int j = 0; j < n; ++j)
			{
				if ((arr[j] & (1 << i)) != 0)
				{
					count++;
				}
			}
			if (count > (n / 2))
			{
				// Add value of i-th bit
				x += (1 << i);
			}
			count = 0;
		}
		// Count frequency of x element
		for (int i = 0; i < n; ++i)
		{
			if (arr[i] == x)
			{
				count++;
			}
		}
		if (count > n / 2)
		{
			// x appears more than n/2 times in the given array
			Console.Write("\n Majority element is " + x);
		}
		else
		{
			Console.Write("\n No Majority element in this array ");
		}
	}
	public static void Main(String[] args)
	{
		Searching task = new Searching();
		int[] arr = {
			1 , 7 , 3 , 7 , 4 , 3 , 7 , 7 , 51 , 7 , 7
		};
		// Get the length of array
		int n = arr.Length;
		// Test
		task.findMajorityElement(arr, n);
	}
}

 Majority element is 7

package main
import "fmt"
// Go program for
// Find the majority element using bit logic
type Searching struct {}
func getSearching() * Searching {
	var me *Searching = &Searching {}
	return me
}
func(this Searching) findMajorityElement(arr[] int, n int) {
	// Auxiliary variable
	var bits int = 32
	var count int = 0
	var x int = 0
	// This loop are work on range (1...31) bit number
	for i := 0 ; i < bits ; i++ {
		// Count active bits in ith position in every array element
		for j := 0 ; j < n ; j++ {
			if (arr[j] & (1 << i)) != 0 {
				count++
			}
		}
		if count > (n / 2) {
			// Add value of i-th bit
			x += (1 << i)
		}
		count = 0
	}
	// Count frequency of x element
	for i := 0 ; i < n ; i++ {
		if arr[i] == x {
			count++
		}
	}
	if count > n / 2 {
		// x appears more than n/2 times in the given array
		fmt.Print("\n Majority element is ", x)
	} else {
		fmt.Print("\n No Majority element in this array ")
	}
}
func main() {
	var task * Searching = getSearching()
	var arr = [] int {1 , 7 , 3 , 7 , 4 , 3 , 7 , 7 , 51 , 7 , 7}
	// Get the length of array
	var n int = len(arr)
	// Test
	task.findMajorityElement(arr, n)
}

 Majority element is 7

<?php
// Php program for
// Find the majority element using bit logic
class Searching
{
	public	function findMajorityElement($arr, $n)
	{
		// Auxiliary variable
		$bits = 32;
		$count = 0;
		$x = 0;
		// This loop are work on range (1...31) bit number
		for ($i = 0; $i < $bits; ++$i)
		{
			// Count active bits in ith position in every array element
			for ($j = 0; $j < $n; ++$j)
			{
				if (($arr[$j] & (1 << $i)) != 0)
				{
					$count++;
				}
			}
			if ($count > ((int)($n / 2)))
			{
				// Add value of i-th bit
				$x += (1 << $i);
			}
			$count = 0;
		}
		// Count frequency of x element
		for ($i = 0; $i < $n; ++$i)
		{
			if ($arr[$i] == $x)
			{
				$count++;
			}
		}
		if ($count > (int)($n / 2))
		{
			// x appears more than n/2 times in the given array
			echo("\n Majority element is ".$x);
		}
		else
		{
			echo("\n No Majority element in this array ");
		}
	}
}

function main()
{
	$task = new Searching();
	$arr = array(1, 7, 3, 7, 4, 3, 7, 7, 51, 7, 7);
	// Get the length of array
	$n = count($arr);
	// Test
	$task->findMajorityElement($arr, $n);
}
main();

 Majority element is 7

// Node JS program for
// Find the majority element using bit logic
class Searching
{
	findMajorityElement(arr, n)
	{
		// Auxiliary variable
		var bits = 32;
		var count = 0;
		var x = 0;
		// This loop are work on range (1...31) bit number
		for (var i = 0; i < bits; ++i)
		{
			// Count active bits in ith position in every array element
			for (var j = 0; j < n; ++j)
			{
				if ((arr[j] & (1 << i)) != 0)
				{
					count++;
				}
			}
			if (count > (parseInt(n / 2)))
			{
				// Add value of i-th bit
				x += (1 << i);
			}
			count = 0;
		}
		// Count frequency of x element
		for (var i = 0; i < n; ++i)
		{
			if (arr[i] == x)
			{
				count++;
			}
		}
		if (count > parseInt(n / 2))
		{
			// x appears more than n/2 times in the given array
			process.stdout.write("\n Majority element is " + x);
		}
		else
		{
			process.stdout.write("\n No Majority element in this array ");
		}
	}
}

function main()
{
	var task = new Searching();
	var arr = [1, 7, 3, 7, 4, 3, 7, 7, 51, 7, 7];
	// Get the length of array
	var n = arr.length;
	// Test
	task.findMajorityElement(arr, n);
}
main();

 Majority element is 7

#  Python 3 program for
#  Find the majority element using bit logic
class Searching :
	def findMajorityElement(self, arr, n) :
		#  Auxiliary variable
		bits = 32
		count = 0
		x = 0
		i = 0
		#  This loop are work on range (1...31) bit number
		while (i < bits) :
			j = 0
			#  Count active bits in ith position in every list element
			while (j < n) :
				if ((arr[j] & (1 << i)) != 0) :
					count += 1
				
				j += 1
			
			if (count > (int(n / 2))) :
				#  Add value of i-th bit
				x += (1 << i)
			
			count = 0
			i += 1
		
		i = 0
		#  Count frequency of x element
		while (i < n) :
			if (arr[i] == x) :
				count += 1
			
			i += 1
		
		if (count > int(n / 2)) :
			#  x appears more than n/2 times in the given list
			print("\n Majority element is ", x, end = "")
		else :
			print("\n No Majority element in this array ", end = "")
		
	

def main() :
	task = Searching()
	arr = [1, 7, 3, 7, 4, 3, 7, 7, 51, 7, 7]
	#  Get the length of list
	n = len(arr)
	#  Test
	task.findMajorityElement(arr, n)

if __name__ == "__main__": main()

 Majority element is  7

#  Ruby program for
#  Find the majority element using bit logic
class Searching 
	def findMajorityElement(arr, n) 
		#  Auxiliary variable
		bits = 32
		count = 0
		x = 0
		i = 0
		#  This loop are work on range (1...31) bit number
		while (i < bits) 
			j = 0
			#  Count active bits in ith position in every array element
			while (j < n) 
				if ((arr[j] & (1 << i)) != 0) 
					count += 1
				end

				j += 1
			end

			if (count > (n / 2)) 
				#  Add value of i-th bit
				x += (1 << i)
			end

			count = 0
			i += 1
		end

		i = 0
		#  Count frequency of x element
		while (i < n) 
			if (arr[i] == x) 
				count += 1
			end

			i += 1
		end

		if (count > n / 2) 
			#  x appears more than n/2 times in the given array
			print("\n Majority element is ", x)
		else
 
			print("\n No Majority element in this array ")
		end

	end

end

def main() 
	task = Searching.new()
	arr = [1, 7, 3, 7, 4, 3, 7, 7, 51, 7, 7]
	#  Get the length of array
	n = arr.length
	#  Test
	task.findMajorityElement(arr, n)
end

main()

 Majority element is 7

// Scala program for
// Find the majority element using bit logic
class Searching()
{
	def findMajorityElement(arr: Array[Int], n: Int): Unit = {
		// Auxiliary variable
		var bits: Int = 32;
		var count: Int = 0;
		var x: Int = 0;
		var i: Int = 0;
		// This loop are work on range (1...31) bit number
		while (i < bits)
		{
			var j: Int = 0;
			// Count active bits in ith position in every array element
			while (j < n)
			{
				if ((arr(j) & (1 << i)) != 0)
				{
					count += 1;
				}
				j += 1;
			}
			if (count > (n / 2))
			{
				// Add value of i-th bit
				x += (1 << i);
			}
			count = 0;
			i += 1;
		}
		i = 0;
		// Count frequency of x element
		while (i < n)
		{
			if (arr(i) == x)
			{
				count += 1;
			}
			i += 1;
		}
		if (count > n / 2)
		{
			// x appears more than n/2 times in the given array
			print("\n Majority element is " + x);
		}
		else
		{
			print("\n No Majority element in this array ");
		}
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: Searching = new Searching();
		var arr: Array[Int] = Array(1, 7, 3, 7, 4, 3, 7, 7, 51, 7, 7);
		// Get the length of array
		var n: Int = arr.length;
		// Test
		task.findMajorityElement(arr, n);
	}
}

 Majority element is 7

import Foundation;
// Swift 4 program for
// Find the majority element using bit logic
class Searching
{
	func findMajorityElement(_ arr: [Int], _ n: Int)
	{
		// Auxiliary variable
		let bits: Int = 32;
		var count: Int = 0;
		var x: Int = 0;
		var i: Int = 0;
		// This loop are work on range (1...31) bit number
		while (i < bits)
		{
			var j: Int = 0;
			// Count active bits in ith position in every array element
			while (j < n)
			{
				if ((arr[j] & (1 << i))  != 0)
				{
					count += 1;
				}
				j += 1;
			}
			if (count > (n / 2))
			{
				// Add value of i-th bit
				x += (1 << i);
			}
			count = 0;
			i += 1;
		}
		i = 0;
		// Count frequency of x element
		while (i < n)
		{
			if (arr[i] == x)
			{
				count += 1;
			}
			i += 1;
		}
		if (count > n / 2)
		{
			// x appears more than n/2 times in the given array
			print("\n Majority element is ",
                  x, terminator: "");
		}
		else
		{
			print("\n No Majority element in this array ", 
                  terminator: "");
		}
	}
}
func main()
{
	let task: Searching = Searching();
	let arr: [Int] = [1, 7, 3, 7, 4, 3, 7, 7, 51, 7, 7];
	// Get the length of array
	let n: Int = arr.count;
	// Test
	task.findMajorityElement(arr, n);
}
main();

 Majority element is  7

// Kotlin program for
// Find the majority element using bit logic
class Searching
{
	fun findMajorityElement(arr: Array < Int > , n: Int): Unit
	{
		// Auxiliary variable
		val bits: Int = 32;
		var count: Int = 0;
		var x: Int = 0;
		var i: Int = 0;
		// This loop are work on range (1...31) bit number
		while (i < bits)
		{
			var j: Int = 0;
			// Count active bits in ith position in every array element
			while (j < n)
			{
				if ((arr[j] and(1 shl i)) != 0)
				{
					count += 1;
				}
				j += 1;
			}
			if (count > (n / 2))
			{
				// Add value of i-th bit
				x += (1 shl i);
			}
			count = 0;
			i += 1;
		}
		i = 0;
		// Count frequency of x element
		while (i < n)
		{
			if (arr[i] == x)
			{
				count += 1;
			}
			i += 1;
		}
		if (count > n / 2)
		{
			// x appears more than n/2 times in the given array
			print("\n Majority element is " + x);
		}
		else
		{
			print("\n No Majority element in this array ");
		}
	}
}
fun main(args: Array < String > ): Unit
{
	val task: Searching = Searching();
	val arr: Array < Int > = arrayOf(1, 7, 3, 7, 4, 3, 7, 7, 51, 7, 7);
	// Get the length of array
	val n: Int = arr.count();
	// Test
	task.findMajorityElement(arr, n);
}

 Majority element is 7

Explanation of Result

After running the given code with the provided example array, the output is "Majority element is 7", which matches our expectations.

Time Complexity

Let's analyze the time complexity of the algorithm. The outer loop runs bits times, which is a constant value (32 in this case). The inner loop runs n times. Hence, the time complexity of the algorithm is O(bits * n), which is equivalent to O(n). In most practical cases, bits is a small constant, making the algorithm linear in the size of the input array.