Відео

nice explanation bro keep going

Wrong time complexity explanation. If you solve it using a queue. You're processing every index just once. Whenever you process an index you send another index at the end, but that takes O(1) time. Overall you're only processing O(1) operations n times. Hence O(n) is the time complexity of the while loop on the queue. Correct, final TC is O(nlogn), but it is because it gets dominated via sorting. If you solve the question using two pointers, then the TC would be nlogn of even the while loop.

Where did we update the string?

Got confused More!

Thanks a lot

Excellent explanation thank you

thank you sir

maxi not required to be same while increasing i. ex: [ 3,2,1,3,4 ] k = 2

2 pointer

its better not to use seive for this question instead use sime check function for the prime number

great explaination... can you go over DP please

Thank you! nicely done explanation!

Great explanation !! Thanks

nice brother

Basically you are doing take not take and max-value among them right ??

what if the number gets repeated, then we need to store both the index of the value in the map?

Nice Explanation, hope to see you in greater heights!

i understand the solution but ... i dont know how to come up with the idea. how can we know to return min(value[node], sum)... what is the idea to that specific value ??

Nice Explanation

Bro can you share how you plugin the BigOmega extension to your leetcode, I am just not able to do that

Why 1+search() can you please help

This time i was able to solve this question.Here is my solution class Solution { public: long long f(int ind,int x,vector<int>& a, vector<int>& b,vector<vector<long long>> &dp) { if(x<0) return 0; if(ind<0) return -1e15; if(dp[ind][x]!=-1) return dp[ind][x]; long long notpick=f(ind-1,x,a,b,dp); long long pick=INT_MIN; if(x >=0) pick=(long long)a[x]*b[ind] + f(ind-1,x-1,a,b,dp); return dp[ind][x]=max(pick,notpick); } long long maxScore(vector<int>& a, vector<int>& b) { int n=b.size(); vector<vector<long long>> dp(n+1,vector<long long>(4,-1)); return f(n-1,3,a,b,dp); } };

the code did not work

Is the brutefore time complexity is #columns^(#rows) ?

"Tanya" & the smile though 😆. Well explained though!

Sir at 7:19 we are using set to store all the unique elements present in the vector nums which is taking O(N) space in worst case.

Damn this was the 3rd what will the 4th look like

There is no way I can come up with this. Hopefully, it builds with experience

hello bro, Could you make another video about this problem (3276. Select Cells in Grid With Maximum Score). I wrote a backtracking solution and I'm stuck to transform it to dynamic. make a deep explanation if possible about how to to think in term of dynamic. here ici my backtrack solution : # backtracking approach O(n^m) where n = {nombre of column} and m = {nombre of line} def maxScore(self, grid: List[List[int]]) -> int: ans = {"res" : 0} selected = set() n = len(grid) def aux(i,currentSum): if i >= n : return for j in range(len(grid[i])): el = grid[i][j] if el not in selected : selected.add(el) # select this element ans["res"] = max(ans["res"], el + currentSum) aux(i+1, el + currentSum) # go to next line selected.remove(el) # backtracking aux(i+1, currentSum) # go to next line aux(0,0) return ans["res"]

thank you this awesome content

good explanation man

startup vs service based company(both offering same package)

class Solution { public: int solve(vector<int>&nums, int target, int i, int dp[], int &maxSize, int size){ if(target == 0){ maxSize = max(maxSize, size); return dp[i] = maxSize; } if(i == nums.size()){ return dp[i] = -1; } if(dp[i]!=0) return dp[i]; if(nums[i] <= target){ return dp[i] = solve(nums, target - nums[i], i+1, dp, maxSize, size+1); } return dp[i] = solve(nums, target, i+1,dp, maxSize, size); } int lengthOfLongestSubsequence(vector<int>& nums, int target) { int max_size = 0, size = 0; int i = 0; int dp[1000]; memset(dp, -1, sizeof(dp)); solve(nums, target, i, dp, max_size, size); if(max_size == 0){ return -1; } return max_size; } }; I am applying 1d DP, thinking that i is always changing that is why i chosed 1d DP

How do you get to know which paramater to take in dp? Or whether it is 1d DP or 2d DP?

Hi sir, The explanation of the approach is clear, but in the solve function why are we taking base case as i==n, because i is not representing the row index, instead isn't it representing the ith max value from the t array, so in case when we not taking the ith max value and moving ahead, it can be the case that the value we skipped was from same row(whose max is already included), so with this we will always be checking the top n max values of the grid instead of values specific to the different row? Can you please clear this doubt.

Bro mindBlowing. Actually, what I did in contest is. I iterated over the grid and chose the maximum value. If the maximum value occurred in more than one row, I did backtracking for every row. This solution may potentially lead to TLE but Leet code forgot to add testcases for this soln which gave me a global rank of 1028.

bro ur explanation was too good 🫡 but i have a question why use masking when we can use set or a map for checking index

why we didn't take in example 2: 0,1,2,3 and 6 nodes because after taking this 4 and 3 are present in there leaf node so sum will not be zero

i dont know graphs and DP. can i crack big tech job?

the intro feels like sriver's intro. cool

Great explanation bro! although I was able to solve the problem during the contest, your approach was way more better and clear. If i may ask, what tool are you using for writing on OneNote, some kind of digital writing pad or on ipad?

Thank you

should't time be n*n since length of num of digits is not related to n and is at max 7 digits long.

Best explanation I came seeing your solution on leetcode

Does this gives TLE in the further part II of this question?

explained the question very well

Bro try to open up words, we have to listen very carefully each of your word..they r not clear..be concise, clear. And explain dry run slow..

Why the complexity is not O(n*50*50*50), can you pls tell this?

Thank you, man.

Nice Explanation!! Thanks!!