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


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))