The Boyer-Moore voting algorithm is one of the popular optimal algorithms which is used to find the majority element among the given elements that have more than N/ 2 occurrences. This works perfectly fine for finding the majority element which takes 2 traversals over the given elements, which works in O(N) time complexity and O(1) space complexity.
Let us see the algorithm and intuition behind its working, by taking an example –
Input :{1,1,1,1,2,3,5}
Output : 1
Explanation : 1 occurs more than 3 times.
Input : {1,2,3}
Output : -1This algorithm works on the fact that if an element occurs more than N/2 times, it means that the remaining elements other than this would definitely be less than N/2. So let us check the proceeding of the algorithm.
- First, choose a candidate from the given set of elements if it is the same as the candidate element, increase the votes. Otherwise, decrease the votes if votes become 0, select another new element as the new candidate.
Intuition Behind Working :
When the elements are the same as the candidate element, votes are incremented whereas when some other element is found (not equal to the candidate element), we decreased the count. This actually means that we are decreasing the priority of winning ability of the selected candidate, since we know that if the candidate is in majority it occurs more than N/2 times and the remaining elements are less than N/2. We keep decreasing the votes since we found some different element(s) than the candidate element. When votes become 0, this actually means that there are the equal number of votes for different elements, which should not be the case for the element to be the majority element. So the candidate element cannot be the majority and hence we choose the present element as the candidate and continue the same till all the elements get finished. The final candidate would be our majority element. We check using the 2nd traversal to see whether its count is greater than N/2. If it is true, we consider it as the majority element.
Steps to implement the algorithm :
Step 1 – Find a candidate with the majority –
- Initialize a variable say i ,votes = 0, candidate =-1
- Traverse through the array using for loop
- If votes = 0, choose the candidate = arr[i] , make votes=1.
- else if the current element is the same as the candidate increment votes
- else decrement votes.
Step 2 – Check if the candidate has more than N/2 votes –
- Initialize a variable count =0 and increment count if it is the same as the candidate.
- If the count is >N/2, return the candidate.
- else return -1.
Dry run for the above example:
Given :
arr[]= 1 1 1 1 2 3 5
votes =0 1 2 3 4 3 2 1
candidate = -1 1 1 1 1 1 1 1
candidate = 1 after first traversal
1 1 1 1 2 3 5
count =0 1 2 3 4 4 4 4
candidate = 1
Hence count > 7/2 =3
So 1 is the majority element.
// C++ implementation for the above approach#include <iostream>using namespace std;
// Function to find majority elementint findMajority(int arr[], int n)
{ int i, candidate = -1, votes = 0;
// Finding majority candidate
for (i = 0; i < n; i++) {
if (votes == 0) {
candidate = arr[i];
votes = 1;
}
else {
if (arr[i] == candidate)
votes++;
else
votes--;
}
}
int count = 0;
// Checking if majority candidate occurs more than n/2
// times
for (i = 0; i < n; i++) {
if (arr[i] == candidate)
count++;
}
if (count > n / 2)
return candidate;
return -1;
}int main()
{ int arr[] = { 1, 1, 1, 1, 2, 3, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
int majority = findMajority(arr, n);
cout << " The majority element is : " << majority;
return 0;
} |
import java.io.*;
class GFG
{ // Function to find majority element
public static int findMajority(int[] nums)
{
int count = 0, candidate = -1;
// Finding majority candidate
for (int index = 0; index < nums.length; index++) {
if (count == 0) {
candidate = nums[index];
count = 1;
}
else {
if (nums[index] == candidate)
count++;
else
count--;
}
}
// Checking if majority candidate occurs more than
// n/2 times
count = 0;
for (int index = 0; index < nums.length; index++) {
if (nums[index] == candidate)
count++;
}
if (count > (nums.length / 2))
return candidate;
return -1;
// The last for loop and the if statement step can
// be skip if a majority element is confirmed to
// be present in an array just return candidate
// in that case
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 1, 1, 1, 2, 3, 4 };
int majority = findMajority(arr);
System.out.println(" The majority element is : "
+ majority);
}
}// This code is contributed by Arnav Sharma |
# Python implementation for the above approach# Function to find majority elementdef findMajority(arr, n):
candidate = -1
votes = 0
# Finding majority candidate
for i in range (n):
if (votes == 0):
candidate = arr[i]
votes = 1
else:
if (arr[i] == candidate):
votes += 1
else:
votes -= 1
count = 0
# Checking if majority candidate occurs more than n/2
# times
for i in range (n):
if (arr[i] == candidate):
count += 1
if (count > n // 2):
return candidate
else:
return -1
# Driver Code arr = [ 1, 1, 1, 1, 2, 3, 4 ]
n = len(arr)
majority = findMajority(arr, n)
print(" The majority element is :" ,majority)
# This code is contributed by shivanisinghss2110 |
using System;
class GFG
{ // Function to find majority element
public static int findMajority(int[] nums)
{
int count = 0, candidate = -1;
// Finding majority candidate
for (int index = 0; index < nums.Length; index++) {
if (count == 0) {
candidate = nums[index];
count = 1;
}
else {
if (nums[index] == candidate)
count++;
else
count--;
}
}
// Checking if majority candidate occurs more than
// n/2 times
count = 0;
for (int index = 0; index < nums.Length; index++) {
if (nums[index] == candidate)
count++;
}
if (count > (nums.Length / 2))
return candidate;
return -1;
// The last for loop and the if statement step can
// be skip if a majority element is confirmed to
// be present in an array just return candidate
// in that case
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 1, 1, 1, 1, 2, 3, 4};
int majority = findMajority(arr);
Console.Write(" The majority element is : "
+ majority);
}
}// This code is contributed by shivanisinghss2110 |
<script>// Function to find majority elementfunction findMajority(nums)
{
var count = 0, candidate = -1;
// Finding majority candidate
for (var index = 0; index < nums.length; index++) {
if (count == 0) {
candidate = nums[index];
count = 1;
}
else {
if (nums[index] == candidate)
count++;
else
count--;
}
}
// Checking if majority candidate occurs more than
// n/2 times
count = 0;
for (var index = 0; index < nums.length; index++) {
if (nums[index] == candidate)
count++;
}
if (count > (nums.length / 2))
return candidate;
return -1;
// The last for loop and the if statement step can
// be skip if a majority element is confirmed to
// be present in an array just return candidate
// in that case
}
// Driver code var arr = [ 1, 1, 1, 1, 2, 3, 4 ];
var majority = findMajority(arr);
document.write(" The majority element is : "
+ majority);
// This code is contributed by shivanisinghss2110.</script> |
The majority element is : 1
Time Complexity: O(n) ( For two passes over the array )
Space Complexity: O(1)