222A - Shooshuns and Sequence - CodeForces Solution

brute force implementation *1200

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Python Code: 

n , k = map(int,input().split())

a = list(map(int,input().split()))

l = a[k-1]
for i in a[k:]:
	if i != l:
		print(-1)
		break
else:
	for i,val in enumerate(reversed(a[:k-1])):
		if val != l:
			print(k-i-1)
			break
	else:
		print(0)

C++ Code: 

#include<bits/stdc++.h> 
using namespace std;
int a[100005],n,k,ans;
int main()
{
    cin>>n>>k;
    for(int i=1;i<=n;i++) 
    {
    	cin>>a[i];
	}
    for(int i=k;i<=n;i++)
    {
    	if(a[i]!=a[k])
		{
	        cout<<-1<<endl;
	        return 0;
	    }
	}
    for(int i=1;i<k;i++)
    {
    	if(a[i]!=a[k]) 
    	{
    		ans=i;
		}
	}
  	cout<<ans<<endl;
  	return 0;
}


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