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

Fair Rations

  • 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)