Boyer–Moore majority vote algorithm in C

March 23, 2023

Boyer-Moore majority vote algorithm

Boyer–Moore majority vote algorithm in C is one of the popular optimal algorithms which is used to find the majority element among the given elements of the array, that have more than N/ 2 occurrences. Boyer–Moore’s majority vote algorithm 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.

Boyer-Moore Majority Vote Algorithm in C

The Boyer-Moore Majority Vote Algorithm is an efficient algorithm for finding the majority element in an array, which is an element that appears more than n/2 times, where n is the length of the array.

Note The algorithm works in O(n) time complexity and O(1) space complexity, which makes it very efficient for large arrays.

Example:

Input: k = 6,  arr = [ 1, 2, 5, 2, 2, 2 ]

Output: Majority Element is 2

Size of sub-array with max sum

Sub-array with given sum

Triplet that sum to a given value

Segregate 0’s and 1’s in array

Sort elements in array by frequency

Algorithm :

Take the size of the array and store it in variable n.
Take n elements of the array from the user.
Declare two variables one is maj_index and initialize it with 0, and one variable count with value 0.
Run a loop from index 1 to n-1
Check if a[i]==a[maj_index] then increase the value of count by 1, else decrease it by 1.
If count becomes zero then set maj_index=i and count =1.
Now create one function check that will check whether the maj_index value presents n/2 times or not.
If it is present then return 1, otherwise 0(means no elements present more than n/2 times).

C code

Run

#include<stdio.h>

int find_majority (int a[], int size)
{
  int maj_index = 0, count = 1;

  for (int i = 1; i < size; i++)
    {
      if (a[i] == a[maj_index])
	count++;
      else
	count--;

      if (count == 0)
	{
	  maj_index = i;
	  count = 1;
	}
    }
  return a[maj_index];
}

int check_maj (int a[], int n, int k)
{
  int count = 0;

  for (int i = 0; i < n; i++) { if (a[i] == k) count++; } if (count > n / 2)
    return 1;
  else
    return 0;
}

int main ()
{
  int k;
  int a[] = { 3, 3, 4, 2, 4, 4, 2, 4, 4 };
  int n = sizeof (a) / sizeof (a[0]);
  k = find_majority (a, n);

  if (check_maj (a, n, k))
    printf ("Majority element is %d", k);
  else
    printf ("No Majority element\n");

  return 0;
}

Output:

 Majority Element is 4

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