We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. Algorithms
  3. Implementation
  4. Fair Rations
  5. Discussions

Fair Rations

Sort by

recency

|

747 Discussions

|

  • afzal_saiyed931
     + 0 comments
    def fairRations(B):
    # Write your code here
    c = 0
    n = len(B)
    for i in range(n - 1):
        if B[i] % 2 != 0:
            B[i + 1] += 1
            c += 2
            
    return "NO" if B[-1] % 2 != 0 else str(c)

  • yashdeliwala10
     + 0 comments

    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);
}

    }

  • spriyamalar
     + 0 comments
    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;
}

  • yashdeora98294
     + 1 comment

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

  • alban_tyrex
     + 0 comments

    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";
}