Majority Element: A majority element in an array A[] of size n is an element that appears more than n/2 times (and hence there is at most one such element).
Write a function which takes an array and emits the majority element (if it exists), otherwise prints NONE as follows:
I/P : 3 3 4 2 4 4 2 4 4
O/P : 4
I/P : 3 3 4 2 4 4 2 4
O/P : NONE
METHOD 1 (Basic)
The basic solution is to have two loops and keep track of maximum count for all different elements. If maximum count becomes greater than n/2 then break the loops and return the element having maximum count. If maximum count doesn’t become more than n/2 then majority element doesn’t exist.
Time Complexity: O(n*n).
Auxiliary Space : O(1).
METHOD 2 (Using Binary Search Tree)
Thanks to Sachin Midha for suggesting this solution.
Node of the Binary Search Tree (used in this approach) will be as follows.
struct
tree
{
int
element;
int
count;
}BST;
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Insert elements in BST one by one and if an element is already present then increment the count of the node. At any stage, if count of a node becomes more than n/2 then return.
The method works well for the cases where n/2+1 occurrences of the majority element is present in the starting of the array, for example {1, 1, 1, 1, 1, 2, 3, 4}.
Time Complexity: If a binary search tree is used then time complexity will be O(n^2). If a self-balancing-binary-search tree is used then O(nlogn)
Auxiliary Space: O(n)
METHOD 3 (Using Moore’s Voting Algorithm)
This is a two step process.
1. Get an element occurring most of the time in the array. This phase will make sure that if there is a majority element then it will return that only.
2. Check if the element obtained from above step is majority element.
1. Finding a Candidate:
The algorithm for first phase that works in O(n) is known as Moore’s Voting Algorithm. Basic idea of the algorithm is if we cancel out each occurrence of an element e with all the other elements that are different from e then e will exist till end if it is a majority element.
findCandidate(a[], size)
1. Initialize index and count of majority element
maj_index = 0, count = 1
2. Loop for i = 1 to size – 1
(a)If a[maj_index] == a[i]
count++
(b)Else
count--;
(c)If count == 0
maj_index = i;
count = 1
3. Return a[maj_index]
Above algorithm loops through each element and maintains a count of a[maj_index], If next element is same then increments the count, if next element is not same then decrements the count, and if the count reaches 0 then changes the maj_index to the current element and sets count to 1.
First Phase algorithm gives us a candidate element. In second phase we need to check if the candidate is really a majority element. Second phase is simple and can be easily done in O(n). We just need to check if count of the candidate element is greater than n/2.
Example:
A[] = 2, 2, 3, 5, 2, 2, 6
Initialize:
maj_index = 0, count = 1 –> candidate ‘2?
2, 2, 3, 5, 2, 2, 6
Same as a[maj_index] => count = 2
2, 2, 3, 5, 2, 2, 6
Different from a[maj_index] => count = 1
2, 2, 3, 5, 2, 2, 6
Different from a[maj_index] => count = 0
Since count = 0, change candidate for majority element to 5 => maj_index = 3, count = 1
2, 2, 3, 5, 2, 2, 6
Different from a[maj_index] => count = 0
Since count = 0, change candidate for majority element to 2 => maj_index = 4
2, 2, 3, 5, 2, 2, 6
Same as a[maj_index] => count = 2
2, 2, 3, 5, 2, 2, 6
Different from a[maj_index] => count = 1
Finally candidate for majority element is 2.
First step uses Moore’s Voting Algorithm to get a candidate for majority element.
2. Check if the element obtained in step 1 is majority
printMajority (a[], size)
1. Find the candidate for majority
2. If candidate is majority. i.e., appears more than n/2 times.
Print the candidate
3. Else
Print "NONE"
Implementation of method 3:
/* Program for finding out majority element in an array */
# include<stdio.h>
# define bool int
int
findCandidate(
int
*,
int
);
bool
isMajority(
int
*,
int
,
int
);
/* Function to print Majority Element */
void
printMajority(
int
a[],
int
size)
{
/* Find the candidate for Majority*/
int
cand = findCandidate(a, size);
/* Print the candidate if it is Majority*/
if
(isMajority(a, size, cand))
printf
(
" %d "
, cand);
else
printf
(
"NO Majority Element"
);
}
/* Function to find the candidate for Majority */
int
findCandidate(
int
a[],
int
size)
{
int
maj_index = 0, count = 1;
int
i;
for
(i = 1; i < size; i++)
{
if
(a[maj_index] == a[i])
count++;
else
count--;
if
(count == 0)
{
maj_index = i;
count = 1;
}
}
return
a[maj_index];
}
/* Function to check if the candidate occurs more than n/2 times */
bool
isMajority(
int
a[],
int
size,
int
cand)
{
int
i, count = 0;
for
(i = 0; i < size; i++)
if
(a[i] == cand)
count++;
if
(count > size/2)
return
1;
else
return
0;
}
/* Driver function to test above functions */
int
main()
{
int
a[] = {1, 3, 3, 1, 2};
printMajority(a, 5);
getchar
();
return
0;
}
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Time Complexity: O(n)
Auxiliary Space : O(1)
Now give a try to below question
Given an array of 2n elements of which n elements are same and the remaining n elements are all different. Write a C program to find out the value which is present n times in the array. There is no restriction on the elements in the array. They are random (In particular they not sequential).