CHFPROF- Editorial

editorial
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CHFPROF- Editorial:

Practice
Contest

Author: Setter’s name
Tester: Tester’s name

DIFFICULTY:

CAKEWALK, EASY

PREREQUISITES:

Greedy, Math, etc.

PROBLEM:

In this problem, a list is given which consists of only 1’s or 0’s elements. We have to count the number of times 1’s and 0’s appear and display the count that is greater compared to other

QUICK EXPLANATION:

First count the number of 1’s and 0’s present in the list, compare both and display the greater number.

EXPLANATION:

First create variables n,count_1,count_0 and initialize all to 0.
Take the number of element in a list as input and store in n.
Then while taking input for list check if input is 1 or 0 and accordingly increment the values of count_1 and count_0.
Finally check which count i.e count_1 or count_0 is greater and display the greater number

SOLUTIONS:

Setter's Solution 
#include <bits/stdc++.h>
using namespace std;
int main()
{
   int n,count0=0,count1=0;
   cin>>n;
   int a[n];
   for(int i = 0;i<n;i++){
	   cin>>a[i];
	   if (a[i]==0){
		   count0+=1;
	   }
	   else{
		   count1+=1;
	   }
   }
   cout<<max(count0,count1);

   return 0;
}
Tester's Solution 
n = int(input())
lis = list(map(int,input().split()))
c1=0
c2=0
for x in lis:
	if x==1:
		c1+=1
	else:
		c2+=1
print(max(c1,c2))