Live Codechef Starters 133 Discussion | Abhinav Awasthi | coding75 Pro

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00:00 Introduction
02:22 Devouring Donuts
03:50 Legs on a Farm
08:06 Maximum Alternating Sum
16:08 Powered Parameters
36:48 ABC Conjecture 3
46:48 coding75 Pro

[@AbhinavAwasthi]

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[@sarathchandra5895]

@AbhinavAwasthi have a look at my approach
int n = s.nextInt() ;
int even_indices = 0 ;
if (n % 2 == 0) even_indices = n/2 ;
else even_indices = n/2 + 1 ;
int arr[] = new int[n] ;
for (int i = 0 ; i < n ; i++) {
arr[i] = s.nextInt() ;
}
int sum = 0 ;
Arrays.sort(arr) ;
for (int i = n - 1 ; i >= 0 ; i--) {
if (even_indices>0) {
sum += arr[i] ;
}
else {
sum -= arr[i] ;
}
even_indices-- ;
}
System.out.println(sum ) ;

[@jitendrakumar-vv8ho]

who knew that brute force would work in powered parameters so i didn't tried it lol

[@anshumaan1024]

bhaiya pls cc ke aur contest ke bhi discussion kia kro na...

[@AbhinavAwasthi]

Live discussion will be here: coding75.com/classroom

[@sdmfslkdm]

bhaiya proof wgera to dia ni ABC conjecture ka ? What was the intution of such solution ?

[@priyanshkumariitd]

Thanks for Powered Parameters time complexity explanation!!!

[@AbhinavAwasthi]

Welcome

[@Dharmik852]

Thank you for making such informative videos!

[@AbhinavAwasthi]

Thanks

[@user-ti7jb6xp4n]

Very Nice explanation thanku bhaiya

[@tarunsingh2480]

Great explanation sir❤️🎉

[@vibhanshusharma9143]

sir,in worst case it will be run for 30 times so the time complexity should be O(30*N) eventualy O(N), but you said O(logn) where do i mistaken.. because take {2,2,2,2,2,2........,(2*10^5)-1 time and,10^9} so in this case every 2 take 30 time ,

[@ritvikreddy3959]

Yeah it is O(30*N).

[@akshayff2016]

string ans =s;
cta=0;
ctb=0;
ctc=0;
int buffer=0;
int fans=0;
int cc=0;
int ca=0;
for(int i=0;i

[@top_g755]

Question d❤

[@AbhinavAwasthi]

Thanks

[@nikhilswami1096]

sir why this solution for ABC conjecture 3 is not right
int a = 0;
int b = 0;
long long ans = 0;
for(int i =0;i0){
b++;
}else if(s[i] == 'c'){
if(a>0 && b>0){
ans++;
a--;
}
}
}
cout

[@AbhinavAwasthi]

You are not handling cases where b is before a, you are just checking that a and b both are there

[@nikhilswami1096]

like i want to ask what is the use of ca why we are not using a directly

[@nikhilswami1096]

@@AbhinavAwasthi sir im checking in else if im checking if s[i] == 'b' and if a is greater than 1 than only im updating it

[@AbhinavAwasthi]

You can join telegram for better discussion

[@nikhilswami1096]

@@AbhinavAwasthi okay sir and yes i understood it like after finding one b we can't take a after that b but I'm taking it like if we have something like abaac in this we can make only one ABC but I will count 3 a and b is also 1 so we will make 3 but its is not correct. thank you sir

[@OnlyAyushAgarwal]

Too good 😊😊❤

[@user-ti7jb6xp4n]

#include
using namespace std;
void solve()
{
int n;
cin >> n;
vector v(n);
for (auto &x : v)
cin >> x;
long long int maxi = *max_element(v.begin(), v.end());
long long count = 0;
for (int i = 0; i < n; i++)
{
if (v[i] == 1)
{
count += n;
}
}
for(int i=0;i

[@ritvikreddy3959]

It is because of overflow.
In the case of 2 , 2^31 ~ 1e9 but if you take 7^31 it overflows as 7^31 is very large due to which wrong value gets stored in your val variable. You can just add an if condition to check if val is greater than 1e9 , if it is you can just break the loop.

[@singhshivam6730]

take continue for case v[i]==1 it will work

[@user-ti7jb6xp4n]

@@ritvikreddy3959 thanku bhai 🙏

[@user-ti7jb6xp4n]

@@singhshivam6730 thanku bhai 🙏