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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Fair Rations
  5. Discussions

Fair Rations

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733 Discussions

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  • 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";
}

  • dhihanlaksanaaj1
     + 0 comments

    SWIFT

    func fairRations(B: [Int]) -> String {
    // Write your code here
    var oddIndex: [Int] = [Int]()
    var oddDict: [Int: Int] = [:]
    var steps: Int = 0
    
    for index in 0 ... B.count - 1 {
        if B[index] % 2 == 1 {
            oddIndex.append(index)
            oddDict[index] = index
        }
    }
    
    if oddIndex.count % 2 == 1 { return "NO" }    
    
    while(oddIndex.count > 1) {
      var currentOddIndex: Int = oddIndex.first!
      steps += 2
      let nextOddIndex: Int = currentOddIndex + 1
        if oddDict[nextOddIndex] == nil {
            oddIndex[0] = nextOddIndex
            oddDict.removeValue(forKey: currentOddIndex)
            oddDict[nextOddIndex] = nextOddIndex
            continue
        }
          
        while(currentOddIndex <= nextOddIndex) {
            oddIndex.removeFirst()
            oddDict.removeValue(forKey: currentOddIndex)
            currentOddIndex += 1
        }
    }
        
    return String(steps)
}

  • krishnatripathi9
     + 0 comments

    string fairRations(vector B) { int result=0; int n=B.size(); for (int i=0;i 

    }
return B[B.size() - 1] % 2 == 0 ? to_string(result*2) : "NO";

    }

  • suryansh_gupta
     + 0 comments

    string fairRations(vector B) {

     int res = 0;
 // size1 = sizeof(B)/sizeof(B[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";

    }

  • jmcparland
     + 0 comments

    Haskell

    module Main where

solve :: [Int] -> Maybe Int
solve bs = go bs 0
  where
    go [] acc = Just acc
    go [x] acc = if even x then Just acc else Nothing
    go (x : y : xs) acc
        | even x = go (y : xs) acc
        | otherwise = go (y + 1 : xs) (acc + 2)

main :: IO ()
main = do
    n <- readLn :: IO Int
    bs <- map read . words <$> getLine :: IO [Int]
    case solve bs of
        Just acc -> print acc
        Nothing -> putStrLn "NO"