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

Fair Rations

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737 Discussions

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  • chauhansatish978
     + 0 comments
    int count = 0;
    if(B.size() <= 1) return "NO";
    else {
        for(int i = 0; i< B.size(); i++){
            if(B[i]%2 == 0) continue;
            if (i + 1 < B.size()) {
                B[i]++;
                B[i + 1]++;
                count += 2;
            }
        }
    }
    
    return B[B.size()-1]%2 == 0? to_string(count) : "NO";

  • kunalkawate424
     + 0 comments

    Solution from my side

    count = 0
    for i in range(len(B)-1):
        if(B[i]%2 ==1 ):
            B[i]+= 1
            B[i+1]+=1
            count +=2
    for j in range(len(B)):
        if(B[j]%2 == 1):
            return 'NO'
    return str(count)

  • 220416d_CSE22
     + 0 comments

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

    for (int i=0;i<n;i++){
    if(i==n-1 && B[i]%2!=0){
        ans="NO";
        return ans;


    }
    if(B[i]%2!=0){

       B[i]++;
       B[i+1]++;
       cnt+=2;

    }
}
ans=to_string(cnt);
return ans; ;

    }

  • 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";
}

  • voronin_ilia
     + 0 comments

    C++ Solution:

    string fairRations(vector<int> B) {
  int res = 0;
  char pattern[3];

  for(size_t i = 0; i < B.size() - 1; i++) {
    pattern[0] = B[i] % 2 == 0 ? 'E' : 'O';
    pattern[1] = B[i + 1] % 2 == 0 ? 'E' : 'O';
    pattern[2] = 0;

    if(((i + 1) == (B.size() - 1)) &&
       (strcmp(pattern, "OE") == 0 ||
        strcmp(pattern, "EO") == 0)) {
      return "NO";
    }

    if (strcmp(pattern, "OO") == 0 || 
        strcmp(pattern, "OE") == 0) {
      B[i] += 1;
      B[i + 1] += 1;
      res+=2;
    } 
  }
  
  return to_string(res);
}