Here is problem solution in python, java, c++, c and javascript programming - https://programmingoneonone.com/hackerrank-fair-rations-problem-solution.html

This code will timeout

public static string fairRations(List B) { int count=0;
bool found=false;

    do{ 

        found=false;

        if(B.Count < 2 || (B.Count == 2 && (B[0]%2 != B[1]%2)))
        {
            break;
        }

        for(int i=0; i<B.Count; i++)
        {
            if(B[i]%2 != 0)
            {
                found=true;

                B[i]+=1;

                if(i+1 < B.Count)
                {
                    B[i+1]+=1;
                }
                else if(i-1 >=0)
                {
                    B[i-1]+=1;
                }

                count+=2;

                break;
            }                
        }

    }while(found);

    return count>0?count.ToString():"NO";
}

}

include

include

include

using namespace std;

string fairRations(vector B) { int loaves = 0; int n = B.size();

for (int i = 0; i < n - 1; ++i) {
    if (B[i] % 2 != 0) {
        B[i]++;
        B[i + 1]++;
        loaves += 2;
    }
}

if (B[n - 1] % 2 != 0) {
    return "NO";
} else {
    return to_string(loaves);
}

}

int main() { ios::sync_with_stdio(false); cin.tie(NULL);

int n; cin >> n;
vector<int> B(n);
for (int i = 0; i < n; ++i) cin >> B[i];

cout << fairRations(B) << "\n";

return 0;

}

Solution in go

func fairRations(B []int32) string {
    oddIndex := -1
    totalBreads := int32(0)
    
    for i := 0; i < len(B); i++ {
        if B[i]%2 > 0 {
            if oddIndex < 0 {
                oddIndex = i
                continue
            }
            
            totalBreads += int32((i - oddIndex) * 2)
            oddIndex = -1
        }
    }
    if oddIndex >= 0 {
        return "NO"
    }
    
    return fmt.Sprintf("%d", totalBreads)
}

My Rust solution: -

fn is_odd(number: &i32) -> bool {
    number % 2 != 0
}


fn fairRations(B: &[i32]) -> String {
    let mut b_vec = B.to_vec();
    let mut counter = 0;
    for i in 0..(b_vec.len() - 1) {
        if is_odd(&b_vec[i]) {
            b_vec[i + 1] += 1;
            counter += 2;
        }
    }
    
    if b_vec.last().unwrap() % 2 == 0 && b_vec.len() > 2 {
        return counter.to_string() 
    }
    return "NO".to_string()
}