C++ Program To Find Element That Appears More Than N/2 Times.

Problem :- A Majority Element In An Array A[] of Size N Is An Element That Appears More Than N/2 (and hence there is at most one such element). Times. Write a Function Which Takes An Array And Emits The Majority Element (if it exists), Otherwise Prints NONE .

Logic :- We Need to print the Element Which repeat more than N/2 time's example if the size of array is 9 than number should be repeated at-least 5 times
Let Take A Array
Size =9
Array Element is 1 1 2 3 1 5 1 1 6

Now N/2 times should be 5 ,So here only 1 is a element only repeated more than N/2.
So Candidate Key is 1.

Solution :-

#include<bits/stdc++.h>
using namespace std;

int 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];
}
bool ismejority(int a[],int size,int cand)
{
int count=0;

for(int i=0;i<size;i++)
{
if(a[i]==cand)
count++;
}
if(count>size/2)
return 1;
else
return 0;
}
int main()
{
int a[100],i,n;

cout<<"Enter The Size of An Array\n";
cin>>n;

cout<<"\nEnter The "<<n<<" Elemnts \n";

for(i=0;i<n;i++)
cin>>a[i];

int cand=majority(a,n);

if(ismejority(a,n,cand))
cout<<"Candidate key \n"<<cand;
else