Majority Element, algorithm design big job 1.py

Majority Element

Find majority element;

Input:an array a[1 to n] of elements;

Output:the majority element if it exists;otherwise none;

1.Majority Element:

A majority element in an array a[] of size n was an element, appears more than N/2 times (and hence there are at e such element).

The majority element is the element, which occurs more than half of the size of the array. This means is the majority element occurs more than all other elements combined or if you count the number of times, Maj ority element appears, and subtract the number of all other elements, you'll get a positive number.

So if you count the number of some element, and subtract the number of any other elements and get the number 0, then your Original element can ' t be a majority element.

2,algorithm:

There are variables, counter and possible element. Iterate the stream, if the counter is 0 your overwrite the possible element and initialize of the counter, if the number is T He same as possible element-increase the counter, otherwise decrease it.

3, Pseudocode:

X<--candidate (1);

count<--0;

for J<--1 to N;

If A[j]=x then Count<--count + 1;

End for;

If COUNT>[N/2] then return x;

Else return none;

Candidate (m):

j<--m,x<--a[m],count<--1;

While J<n and count>0:

j<--j+1;

If a[j]= X then count<--count+1;

else Count <--count+1;

End while;

If J=n then return x;

else return candidate (j+1);

Source Code:

#Python Code

def majority_element (arr):

Counter, possible_element = 0, None

For I in Arr:

If counter = = 0:

Possible_element, counter = i, 1

elif i = = possible_element:

Counter + = 1

Else

Counter-= 1

Return possible_element

eg:arr=[2,5,3,5,6,8,5,5]

Start:possible_element=none

Counter=0

When i=2 counter=1 possible_element=2

When I=5 counter=0 possible_element=2 # i!=possible_element, counter-1

When I=3 counter=1 possible_element=3

When I=5 counter=0 possible_element=3 # i!=possible_element, counter-1

When I=6 counter=1 possible_element=6

When I=8 counter=0 possible_element=6 # i!=possible_element, counter-1

When I=5 counter=1 possible_element=5

When I=5 counter=2 possible_element=5

Return possible_element

Majority Element, algorithm design big job 1.py