In this Leetcode The minimum cost problem solution you are given a binary array (array consists of 0's and 1's) A that contains N elements. You can perform the following operation as many times as you like:
1.  Choose any index 1 <= i <= N and if it is currently 0, then convert it to 1 for C01 coins.
2.  Choose any index 1 <= i <= N and if it is currently 1, then convert it to 0 for C10 coins.
Your task is to transform the given array into a special array that for every index 1 <= i < N, Ai intersection Ai+1 = 1.

## HackerEarth The minimum cost problem solution.

```#include<bits/stdc++.h>
using namespace std;
#define FIO ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define mod 1000000007
#define endl "\n"
#define test ll t; cin>>t; while(t--)
typedef long long int ll;
ll solve(vector<ll>&a,ll c01,ll c10,ll st){
ll ans=0;
for(auto it:a){
if(it!=st){
ans+=(it==0?c01:c10);
}
st^=1;
}
return ans;
}
int main() {
FIO;
test
{
ll n,c01,c10;
cin>>n>>c01>>c10;
vector<ll>a(n);
for(auto &it:a) cin>>it;
ll ans=1e18;
ans=min(ans,solve(a,c01,c10,0ll));
ans=min(ans,solve(a,c01,c10,1ll));
cout<<ans<<endl;
}
return 0;
}```

### Second solution

```t = int(input())
while t > 0:
t -= 1
c = [0, 0]
n, c, c = map(int, input().split())
a = list(map(int, input().split()))
cost = [0, 0]
for i in range(n):
for j in range(2):
cost[j] += c[a[i]] * ((a[i] ^ i) & 1 != j)
print(min(cost))```