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Thanks 😊

If you want to become pupil then you should practice problems with rating 1200, 1300, and 1400... And there is a sheet CP 31 where you can find the best problems to practice. I'm going to start a playlist on that soon...

@@UnfilteredProgrammer thanks a lot!!! I am already doing 1000 rated from ACD ladder

I upload solutions for codeforces also...

Sure I will make a playlist for that.

@@roshankawale720 sure... I will do that.

Here output will be 1.

#include <bits/stdc++.h> using namespace std; int main() { int t; cin>>t; while(t--){ int n,temp,odd=0; cin>>n; for(int i=0; i<n; i++){ cin>>temp; //Count odd elements... (n-odd will be the even elements) if(temp&1) odd++; } //Checking if even elements zero... then output will be odd-1 if(odd==n){ cout<<odd-1<<endl; continue; } //if there are some even elements... then output will be odd+1 cout<<odd+1<<endl; } }

Thanks for the feedback! 😃

@@Vivekrathore6294 yes you are right... I missed one point while explaining... We leave k-1 elements from left and now if remaining elements are a1 to ai.... Then we need to check from a2 to a(i+1) if any of them is not equal to 1 then output will be one otherwise it will be 2. Ex: 1 1 1 1 1 1 2 3 4 n=9 and k=4 So we skip k-1(i.e 3) elements from left [2,3,4]... Now we need to check for these elements a2 to a7.... Because a7 can be the first element of the even index subarray here! Thanks for pointing out this mistake! 😄 We are leaving the k-1 elements but checking from a2 to first element of the k-1 elements... I hope this helps!

@@UnfilteredProgrammer yes sir got it

Uploaded

Last Testcase messed entire logic

Uploaded

please respond

If you want to store only unique keys then we use set and if you want to store the data in key-value pair then we use a map. Ex: Given an array print all unique elements... In this case we can use a set... Because it will store all the unique elements. Ex-2: Given an array... Print the frequency of all the elements.. In this case we need to store both elements and their frequency... That means we need a key-value pair... So we will use a map here... I hope this helps... Thanks!

@@NitinSharma-xy6fx Bro I have a plan for that... To make playlist for DSA and Competitive programming.... But currently I don't have that much subscribers.... So that's my future plan... Once we have a significant amount of subscribers...

Sure... I will do that... Thanks for your feedback...

Thanks 😊

​@@UnfilteredProgrammer Your channel will grow soon....Just wait and watch 👍 Keep doing... Also please provide A and B also from further contest 👍

@@catspeaking7 Thanks for your support 😊

@@aniruddhadabade4198 just by following the given two conditions..

@@UnfilteredProgrammer aap contest ke problems kaise solve kr sakte ho muze codechef contest mai sirf first 2 hi ate hai baki ke nahi ate kuch suggest kr sakte ho kya ki mai kaise improve kru? please help krdo?

@@aniruddhadabade4198 brother you need to solve more problems for practice.... Try to solve problems with the Rating=your current rating + (100, 200, and 300)... Firstly try to solve it by yourself with atleast 2 or 3 attempts... If you are not able to solve the problem then watch the video tutorial.... And even if you are able to solve the go and analyze the others code...

@@UnfilteredProgrammer ok bro

Thanks bro

To prove the second condition that it is impossible to win if we have n=8 and given M moves like this: SSPSP

@UnfilteredProgrammer but if we take n=7 it is also possible to lose the game

@@AnujKumar-y5q9o yes you are right... But we are considering the edge case here.

Uploaded video

#include <bits/stdc++.h> using namespace std; #define int long long signed main() { int t; cin >> t; while (t--) { int n; cin >> n; vector<int> v(n); map<int, int> mp; for (int i = 0; i < n; i++) { cin >> v[i]; mp[v[i]]++; } vector<int> d; int flag = 0; for (auto it : mp) { if (it.second > 3) { cout << it.first << " " << it.first << " " << it.first << " " << it.first << endl; flag = 1; break; } if (it.second >= 2) { d.push_back(it.first); } } if (flag) continue; if (d.size() == 0) { cout << -1 << endl; continue; } if (d.size() > 1) { cout << d[0] << " " << d[0] << " " << d[1] << " " << d[1] << endl; continue; } vector<int> ans; int duplicate = d[0]; int count = 0; sort(v.begin(), v.end()); for (int i = 0; i < n; i++) { if (v[i] == duplicate) { if (count >= 2) ans.push_back(v[i]); count++; } else { ans.push_back(v[i]); } } for (int i = 1; i < ans.size(); i++) { if (ans[i] - ans[i - 1] < 2 * duplicate) { cout << duplicate << " " << duplicate << " " << ans[i - 1] << " " << ans[i] << endl; flag = 1; break; } } if (!flag) { cout << -1 << endl; } } return 0; }

We are using zero based indexing here.... So it will be i.... in this case here...

@@Mukul-k4o bro code is added in the comment section!

Sure bro!

You need practice... Try to solve problems with rating = Your Current Rating + (100, 200, 300)....

Because they are moving away. Suppose i =2 and j=6 then in the next case i=1 and j =7 so initially it is 6-2=4 and now it is 7-1=6

@@tanujagoli3625 if i is 2 and j is 6 shouldn't lower index move right and higher index move left i.e. i =3 and j = 5

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