# HackerEarth The minimum cost problem solution

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:
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[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))`