*refered from:-http://www.geeksforgeeks.org/archives/503
basically want to know that whether voting algo is corret???
*2,2,2,3,3,4,4
in this case it is giving majority element as 4 but it should be 3???plz
explain.
*
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"
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