In this

**HackerEarth 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:- Choose any index 1 <= i <= N and if it is currently 0, then convert it to 1 for C01 coins.
- 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[0], c[1] = 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))

## 0 Comments