I tried to implement the same approach as mentioned in the official editorial of the problem, but I wasn’t able to debug it for WA. Here is my solution.
Please provide me a TC which fails my code. I just didn’t sum the elements initially to check if it is possible to make the array beautiful, I checked it in the later part of my code.
You are complicating your life after taking input.
Just do the following.
if count is our answer
count += two/2;
two = two%2; // Handles numbers with remainder 2.
int min = min(one, three)
one -= min;
three -= min; // after this, either variable one, or variable three will be zero.
count += min;
int rem = max(one,three);
if(two == 1){
if(rem>1){
count += 2;
two–;
rem -= 2;
}else beau = false;
} //Handle merging of 2 with two 1s or two 3s, in two steps.
count += (rem/4)*3; //merges four 1s or four 3s in 3 steps.
if(rem%4 != 0)beau = false;
if(beau)print count;
else print -1;
Your solution is too complicated bro!!!
Using too many if and else statements.
I know my code is cumbersome, but I can’t figure out what is wrong. I think I did what you explained, just the hard way. xD
It’s just that it’s too cumbersome to find bug in a cumbersome code.
Hope you understand this cumbersome problem.
I would recommend to simplify and feel free to ask if any doubt @vatsalsura
u all can use random test cases with some AC solution
In the else part where three > one, you are first setting one=0 and then subtracting it from three ie three -= one,so the value of three never changes. That is what I fell is wrong. You should subtracting one from three and then set one to 0
@omkar__halikar Thank you. Indeed that was the only problem in my code, got ac after inter-changing both the lines.