very simple and uderstable solution with cpp

string fairRations(vector B) {

int key=0;

for(int i = 0;i < B.size()-1;i++){ // because lin last element we have to do other operation
    if(i == B.size()-1){  //this is for last element only
        if(B[i] % 2!=0){
             B[i]++;
             B[i-1]++;
        }

    }
    if( B[i] % 2 !=0){
        B[i]++;
        B[i+1]++;
        key+=2;

    }
}
bool flag = 1;

for(int i = 0;i < B.size() ; i++){
    if(B[i] % 2!=0){
        flag=0;
        break;
    }
}


if( flag ==0){
    return "NO";
}
else{
    return to_string(key);
}

}

public static String fairRations(List<Integer> B) {
    List<Integer> ration = new ArrayList<>(B);
    int count = 0;
    for(int i=0 ;i< ration.size()-1; i++) {
        if(ration.get(i) % 2 == 0) continue;
        if(ration.get(i) % 2 != 0) {
            ration.set(i, ration.get(i)+1);
            ration.set(i+1, ration.get(i+1)+1);
            count += 2;
        }
    }
    if(ration.get(ration.size()-1)%2 != 0) return "NO";
    return ""+count;
}    

Here is problem solution in PYthon, Java, C++, C and Javascript - https://programmingoneonone.com/hackerrank-fair-rations-problem-solution.html

Here is my c++ solution: you can watch the explanation here : https://youtu.be/pAzUgM52d60

string fairRations(vector<int> B) {
    int res = 0;
    for(int i = 0; i < B.size() - 1; i++){
        if(B[i] % 2 == 1){
            B[i+1]++;
            res+=2;
        }
    }
    return B[B.size() - 1] % 2 == 0 ? to_string(res) : "NO";
}

Whithout the if

string fairRations(vector<int> B) {
    int res = 0;
    for(int i = 0; i < B.size() - 1; i++){
        B[i+1] += B[i] % 2;
        res += B[i] % 2;
    }
    return B[B.size() - 1] % 2 == 0 ? to_string(res*2) : "NO";
}