Відео

how to know which on of this to use to solve any problem

Really great explanation!

Extraordinary bro, continue the Playlist. It is very rare to see this type of high quality content.

Sir currently I am in my 3 sem and doing dsa but not able to solve questions on codeforces...any advice regarding dsa and cp

Bro, Bahut hi sahi explanation... Especially loved the part, Where you, showed the dp relationship, and how the map helps replace it.... (You not only explained the question, But the way you showed the trick, IT sinks down in the brain, to help me further solve other questions....) Straight to the point solution....

Bhaiya How to thik directly about tabulation? Means i first need to think about recursive then i need to convert it into tabulation

is atcoder dp educational and cses dp question sufficient to solve such dp problem?

from where we can learn dp for CP?

Ist question ke lia map ka use tha bhi

Can you explain in English 😢

Please explain in english sir❤

Nicely Explained✌️

Bro provide code in java too

well explained !!

I am here after seeing your LinkedIn post. Hope we get a new video on Div 2

Hindi me maja aya sir keep it up!

Keep them coming

Was giving first contest. 4 ke idea aaye. 3 mai hidden me gya😭

Abhishek Bhai i have a query i know Bhai aapne m.tech ka exam nhi diya hai but Bhai I'm preparing for Gate cse exam brother suggest me best Book of c programming and algorithm in depth

is it possible to do the same for Max deque ?

Hi bhaiya, For the past 2 months, I have been trying Competitive Programming , but most of the time I am unable to handle math-related problems. Since the majority of CP problems are math-related, please suggest how I can improve my math skills to better handle CP problems. Also, please suggest some sources.

class Solution { public int maximumLength(int[] nums, int k) { int n = nums.length; int[][][] dp = new int[n][k + 1][n + 1]; for (int[][] d : dp) { for (int[] row : d) { Arrays.fill(row, -1); } } return fxn(n - 1, nums, k, n, dp); } int fxn(int idx, int[] arr, int k, int prevIdx, int[][][] dp) { if (idx < 0) { return 0; } if (dp[idx][k][prevIdx] != -1) { return dp[idx][k][prevIdx]; } int notTake = fxn(idx - 1, arr, k, prevIdx, dp); int take = 0; if (prevIdx == arr.length || arr[idx] == arr[prevIdx]) { take = 1 + fxn(idx - 1, arr, k, idx, dp); } else if (arr[idx] != arr[prevIdx] && k > 0) { take = 1 + fxn(idx - 1, arr, k - 1, idx, dp); } dp[idx][k][prevIdx] = Math.max(notTake, take); return dp[idx][k][prevIdx]; } }what optimization can be done here like it passed 3rd one but not 4th one ??!

Bhai shorts ki DSA series upload Karo then tier 2-3 wale relate kar payenge .. shorts video bnao nhi toh time waste hai on UA-cam this is my opinion 😊.. best of luck Bhai

Bahiya ive written its rec code its showing tle 3d dp

Great explanation my man 🤗

Thanks a ton bhaiya for the help.....🙏🏻🙏🏻🙏🏻

class Solution: def maximumLength(self, nums: List[int], k: int) -> int: n = len(nums) c = Counter(nums) if k == 0: return max(c.values()) if max(c.values()) == 1: return min(k + 1, n) runs = {} best = [0] * (k + 1) for x in nums: if x not in runs: runs[x] = [1] * (k + 1) else: runs[x] = [z + 1 for z in runs[x]] runs[x][1:] = [max(lo + 1, xrun) for lo, xrun in zip(best, runs[x][1:])] best[1:] = [max(lo + 1, b) for lo, b in pairwise(best)] best = [max(b, xrun) for b, xrun in zip(best, runs[x])] return max(best) that's Python easy solution

i have a doubt in second optimization (was able to think of first one on own). The doubt is why prev index doesn't matter. condition says nums[i] != nums[prev] . Like assume nums[i] =1 say at some i we and trying to calculate for j=3 and there was a sequence let say 1 2 2 1 so obv a length of 4 is already precalculated for some prev index also ending at 1 and j=2 as u can see . Now if you will calculate ans for current i you will say 1 + dp[p][2] as you ans => 1 + 4 but actually this is wrong na coz you made a sequence of length 5 saying there are 3 indices in 1 2 2 2 1 1

where to study dp from, I always write the recursive solutions but can't write the iterative ones, even here my solution to the problem with smaller constraints got accepted but couldn't solve the 4th one, something similar happened in one of the problems of recent Amazon hack-on OA. Please suggest some resource

If someone solves this using DP with O(n*n*k) TC and SC, is it good enough to get Strong Hire or Hire verdict at Google for L4 and above?

ya i understood this quiz but we could have done one more thing....iterate from (0--->k) and use a temp array to store the updated values for overallMaxForK array throughout the j loop and then do overallMaxForK = temp bcoz we require values of overallMaxForK array only for p = (0--->i-1) so we can update the array once the j loop is over Am I right ??

I was using only a map to store the length of the max sequence ending at that particular value...so had problem while updating the map, and the second array which represent the length of the longest subseq with atmost k different pairs simultaneously....implementation issues...but glad i could think in somewhat the same direction as yours !!

can anyone explain why it is giving MLE bool check(vector<int>&ele,int k){ int n=ele.size(); int count=0; for(int i=0;i<=n-2;i++){ if(ele[i]!=ele[i+1]){ count++; } } return count<=k; } vector<vector<int>>ans; void getsub(vector<int>&nums,int i,int n,vector<int>&ele){ if(i>=n){ ans.push_back(ele); return; } ele.push_back(nums[i]); getsub(nums,i+1,n,ele); ele.pop_back(); getsub(nums,i+1,n,ele); } int maximumLength(vector<int>& nums, int k) { int n=nums.size(); vector<int>ele; getsub(nums,0,n,ele); int MAX=0; for(int i=0;i<ans.size();i++){ if(check(ans[i],k)){ int m=ans[i].size(); MAX=max(MAX,m); } } return MAX; }

Still not understood

thanks bhaiya 🙏🙏 quite helpful video .... best thing for me is I was trying this from last 1hr but was unable to come up with a soln and i found this video.

Answer to quiz: The reverse might result in wrong answer because we don't wanna use the result of something[j - 1] or something[j] computed at index i to be used for something[j] at the same index.

Nice Thank you... C felt kind of standard but for D i thought similar approach but end up storing for each K using vector<map> and then finding maxx for each k for every iteration.. Took O(n*k*k) . It got accepted 😅 tho I was expecting it to be MLE.

sir the lecture was godly.

Sorry brother, you lost me at white theme. BTW, keep up the good work.

In the problem E suppose i declare both the array inside Int main then i get some random output why ?? rest mine logic was same , also i have matched from your submission

Today in Codechef 134 B i solved 3rd problem using it. It was amazing as you Sir.

Time complexity(fun part was truly fun) until i saw the last ques, that one's a menace. Need to grow a lot. Thanks for the video!

For problem 'F' Let's say we are calculating for x = 6 and i = 6 in that case cnt = 0 here we didn't count any extra points, then why would we subtract - 1 from the count

int lastElementLessThanD(vector<int> arr, int d) { auto it = std::lower_bound(arr.begin(), arr.end(), d); if (it == arr.begin()) { // If lower_bound returns the beginning of the array, // it means all elements are greater than or equal to d. return -1; // Indicating no such index found } else { // Return the index of the last element less than d return std::distance(arr.begin(), --it); } } int32_t main() { // your code goes here jaldi_kar_kal_panvel_nikalna_hein int t = 1; cin >> t; while(t--){ int n, k, q; cin >> n >> k >> q; vi a(k); scan(a,k); vi b(k); scan(b,k); while(q--){ int d; cin >> d; if(d <= a[0]){ cout << (d*b[0])/a[0] << " "; continue; } if((k-2) >= 0 && d >= a[k-2]){ int x = d - a[k-2]; int tot_d = n - a[k-2]; int tot_t = b[k-1] - b[k-2]; int prin = b[k-2] + (x * tot_t) / tot_d; cout << prin << " "; continue; } int idx = lastElementLessThanD(a,d); // cout << endl << d << " : " << idx << endl; int x = d - a[idx]; int tot_d = a[idx+1] - a[idx]; int tot_t = b[idx+1] - b[idx]; int ans = b[idx] + (x * tot_t) / tot_d; cout << ans << " "; } cout << endl; } return 0; } Sir, can you plz tell me where is the problem in this solution for Problem E?? I have used BS on array "a" only ..but it have me TLE on 7th case.

Time complexity +1

Amazing as always! Can you suggest some resources for bit manipulation. I always feel very underconfident while solving them.