Pre-requisite: Majority Element, Majority Element | Set-2 (Hashing)
Given an array of size N, find the majority element. The majority element is the element that appears more than n/2 times in the given array.
Examples:
Input: {3, 3, 4, 2, 4, 4, 2, 4, 4}
Output: 4
Input: {3, 3, 6, 2, 4, 4, 2, 4}
Output: No Majority Element
Approach:
In this post, we solve the problem with the help of binary representation of the numbers present in the array.
The task is to find the element that appears more than n/2 times. So, it appears more than all other numbers combined.
So, we starting from LSB (least significant bit) of every number of the array, we count in how many numbers of the array it is set. If any bit is set in more than n/2 numbers, then that bit is set in our majority element.
The above approach works because for all other numbers combined the set bit count can’t be more than n/2, as the majority element is present more than n/2 times.
Lets see with the help of example
Input : {3, 3, 4, 2, 4, 4, 2, 4, 4}
Binary representation of the same are:
3 - 0 1 1
3 - 0 1 1
4 - 1 0 0
2 - 0 1 0
4 - 1 0 0
4 - 1 0 0
2 - 0 1 0
4 - 1 0 0
4 - 1 0 0
----------
- 5 4 0
Here n is 9, so n/2 = 4 and an only 3rd bit from right satisfy count>4 and hence set in majority element and all other bits are not set.
So, our majority element is 1 0 0, which is 4
But there more to it, This approach works when the majority element is present in the array. What if it is not present?
Let’s see with the help of this example:
Input : {3, 3, 6, 2, 4, 4, 2, 4}
Binary representation of the same are:
3 - 0 1 1
3 - 0 1 1
6 - 1 1 0
2 - 0 1 0
4 - 1 0 0
4 - 1 0 0
2 - 0 1 0
4 - 1 0 0
----------
- 4 5 0
Here n is 8, so n/2 = 4 and an only 2nd bit from right satisfy count>4 and hence set it should be set in majority element and all other bits not set.
So, our majority element according to this is 0 1 0, which is 2 But actually majority element is not present in the array. So, we do one more pass of the array, to make sure this element is present more than n/2 times.
Here is the implementation of the above idea
C++
#include <iostream>using namespace std;void findMajority(int arr[], int n){ // Number of bits in the integer int len = sizeof(int) * 8; // Variable to calculate majority element int number = 0; // Loop to iterate through all the bits of number for (int i = 0; i < len; i++) { int count = 0; // Loop to iterate through all elements in array // to count the total set bit // at position i from right for (int j = 0; j < n; j++) { if (arr[j] & (1 << i)) count++; } // If the total set bits exceeds n/2, // this bit should be present in majority Element. if (count > (n / 2)) number += (1 << i); } int count = 0; // iterate through array get // the count of candidate majority element for (int i = 0; i < n; i++) if (arr[i] == number) count++; // Verify if the count exceeds n/2 if (count > (n / 2)) cout << number; else cout << "Majority Element Not Present";}// Driver Programint main(){ int arr[] = { 3, 3, 4, 2, 4, 4, 2, 4, 4 }; int n = sizeof(arr) / sizeof(arr[0]); findMajority(arr, n); return 0;} |
Java
class GFG { static void findMajority(int arr[], int n) { // Number of bits in the integer int len = 32; // Variable to calculate majority element int number = 0; // Loop to iterate through all the bits of number for (int i = 0; i < len; i++) { int count = 0; // Loop to iterate through all elements in array // to count the total set bit // at position i from right for (int j = 0; j < n; j++) { if ((arr[j] & (1 << i)) != 0) count++; } // If the total set bits exceeds n/2, // this bit should be present in majority Element. if (count > (n / 2)) number += (1 << i); } int count = 0; // iterate through array get // the count of candidate majority element for (int i = 0; i < n; i++) if (arr[i] == number) count++; // Verify if the count exceeds n/2 if (count > (n / 2)) System.out.println(number); else System.out.println("Majority Element Not Present"); } // Driver Code public static void main (String[] args) { int arr[] = { 3, 3, 4, 2, 4, 4, 2, 4, 4 }; int n = arr.length; findMajority(arr, n); } }// This code is contributed by AnkitRai01 |
Python3
def findMajority(arr, n): # Number of bits in the integer Len = 32 # Variable to calculate majority element number = 0 # Loop to iterate through # all the bits of number for i in range(Len): count = 0 # Loop to iterate through all elements # in array to count the total set bit # at position i from right for j in range(n): if (arr[j] & (1 << i)): count += 1 # If the total set bits exceeds n/2, # this bit should be present in # majority Element. if (count > (n // 2)): number += (1 << i) count = 0 # iterate through array get # the count of candidate majority element for i in range(n): if (arr[i] == number): count += 1 # Verify if the count exceeds n/2 if (count > (n // 2)): print(number) else: print("Majority Element Not Present")# Driver Codearr = [3, 3, 4, 2, 4, 4, 2, 4, 4]n = len(arr)findMajority(arr, n)# This code is contributed by Mohit Kumar |
C#
using System;class GFG{ static void findMajority(int []arr, int n) { // Number of bits in the integer int len = 32; // Variable to calculate majority element int number = 0; // Loop to iterate through all the bits of number for (int i = 0; i < len; i++) { int count = 0; // Loop to iterate through all elements // in array to count the total set bit // at position i from right for (int j = 0; j < n; j++) { if ((arr[j] & (1 << i)) != 0) count++; } // If the total set bits exceeds n/2, // this bit should be present in majority Element. if (countt > (n / 2)) number += (1 << i); } int count = 0; // iterate through array get // the count of candidate majority element for (int i = 0; i < n; i++) if (arr[i] == number) count++; // Verify if the count exceeds n/2 if (count > (n / 2)) Console.Write(number); else Console.Write("Majority Element Not Present"); } // Driver Code static public void Main () { int []arr = { 3, 3, 4, 2, 4, 4, 2, 4, 4 }; int n = arr.Length; findMajority(arr, n); } }// This code is contributed by @Tushi.. |
Javascript
<script>function findMajority(arr, n) { // Number of bits in the integer let len = 32; // Variable to calculate majority element let number = 0; // Loop to iterate through all // the bits of number for(let i = 0; i < len; i++) { let countt = 0; // Loop to iterate through all elements // in array to count the total set bit // at position i from right for(let j = 0; j < n; j++) { if ((arr[j] & (1 << i)) != 0) countt++; } // If the total set bits exceeds n/2, // this bit should be present in // majority Element. if (countt > parseInt(n / 2, 10)) number += (1 << i); } let count = 0; // Iterate through array get // the count of candidate // majority element for(let i = 0; i < n; i++) if (arr[i] == number) count++; // Verify if the count exceeds n/2 if (count > parseInt(n / 2, 10)) document.write(number); else document.write("Majority Element Not Present"); } // Driver Codelet arr = [ 3, 3, 4, 2, 4, 4, 2, 4, 4 ]; let n = arr.length; findMajority(arr, n); // This code is contributed by decode2207</script> |
4
Time complexity: O(N*logN) Where N is the number of elements present in the array, logN time is taken by the number of bits of an integer and, N time is taken to iterate all the elements of the array. So Time Complexity is O(len*N), which can be written in the form of N like this O(NlogN).
Auxiliary Space: O(1)
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