Відео

bro please do 3 videos per week ❤

2 pointer sort code bro&explain is good

class Solution { public int lengthOfLongestSubstring(String s) { int max_len=0; int i=0; Hashset<Character> set=new Hashset<Character>(); for(int j=0;j<s.length();j++) { while (set.contains(s.CharAt(s(j)))) { set.remove(s.CharAt(s(i))); i+=1; } set.add(s.CharAt(s(j))); max_len=Math.max(max_len,j-i+1); } return max_len ; } } Main.java:13: error: cannot find symbol Hashset<Character> set=new Hashset<Character>(); ^ symbol: class Hashset location: class Main.Solution anna ela error vasthundhi artham kaledhu yendhuku vasthundhi ee error ani

Scholarship

sliding window approach class Solution: def lengthOfLongestSubstring(self, s: str) -> int: i=0 max_length=0 hashset=set() for j in range(0,len(s)): while s[j] in hashset: hashset.remove(s[i]) i+=1 hashset.add(s[j]) max_length=max(max_length,j-i+1) return max_length

Brute force approach lass Solution: def lengthOfLongestSubstring(self, s: str) -> int: max_length=0 for i in range(0,len(s)): for j in range(i,len(s)): hashset=set() flag=True for k in range(i,j+1): if s[k] in hashset: flag=False break hashset.add(s[k]) if flag==True: max_length=max(max_length,j-i+1) return max_length

Anna explanation neat,clarity ga undi. Elantivi Inka pettu Anna

love u bro .... telugu community ki DSA kastalu teerinatte mee chaneel tho....... plz don't stop until this is enough to reach DSA advance

Super bro

Adhaina chesina problem ekuva days gurtu vundali ante ela Tips evu anna plz

Nice explanation with real time examples Brother. and Thanks for forgetting Laptop in Kitchen, that brought the whole explanation for HASHMAP/HASHSET Thanks for the video and keep posting

class Solution: def maxProfit(self, prices: List[int]) -> int: prefix=[0]*len(prices) prefix[0]=prices[0] for i in range(1,len(prices)): res=min(prices[i],prefix[i-1]) prefix[i]=res maz=0 for i in range(1,len(prices)): if prices[i]-prefix[i-1]>maz: maz=prices[i]-prefix[i-1] return maz this is my another code by own thinking

def isValidSudoku(self, board: List[List[str]]) -> bool: for i in range(len(board)): row_check=[] col_check=[] for j in range(len(board)): if board[i][j]!=".": if board[i][j] in row_check: return False row_check.append(board[i][j]) if board[j][i]!=".": if board[j][i] in col_check: return False col_check.append(board[j][i]) start_row=start=0 row_end=col_end=3 for i in range(3): for _ in range(3): temp=[] for j in range(start_row,row_end): for k in range(start,col_end): if board[j][k]!=".": if board[j][k] in temp: return False temp.append(board[j][k]) start+=3 col_end+=3 start_row+=3 row_end+=3 start=0 col_end=3 return True Bro ee code ki leetcode lo O(1) time complexity ani istundi , is it true?

bro...........binary search video cheyandhi

anna first question start chesemundhu meru aa question ni neetcode platform lo chupinchi tharvatha white board medha chepandi appudu memu aa question ni artham chesukonni memu try chese motham video chustham

def decode(self, s: str) -> List[str]: i=0 t='' result=[] while (i < len(s)): flag=0 j=i if s[j]=="#": length_ss=int(t) cur=0 s1='' while(cur<length_ss): j+=1 s1+=s[j] cur+=1 result.append(s1) flag=1 if flag==1: t='' else: t+=s[j] j+=1 i=j return result Bro dini time complexity O(n**2) or O (n * l)

Bro expalination peaks but code tho step by step explain cheyy 🙏

Bharat, love babbar 450 questions kuda solve cheyava, this would be great help for future generations.

Well explained on two sum problem statement, tq Bharat

Hi Bharath Chandra, 2 unavailable videos hidden.

My doubt In for loop, syntax is * for i in range (start, end, step): -In this end = 10 then it gives 9 is end th element . -In this problem y we use len(nums) instead of len(nums)+1

superb explanation anna👍

Next videos fast ga pettu bro

Day -3 ✅ java brute force : class Solution { public int[] twoSum(int[] nums, int target) { for ( int i = 0 ; i < nums.length ; i ++) { for ( int j = i + 1 ; j < nums.length ; j ++) { if ( nums[i] + nums [j] == target ) { return new int[] { i, j }; } } } return null ; } }

Linked list explain plz😢

very good way of explanation thank you bro .

#optimal approach prices = [7, 1, 5, 3, 6, 4] n = len(prices) maxprofit = 0 i = 0 j = 1 while j < n: if prices[j] - prices[i] > maxprofit: maxprofit = prices[j] - prices[i] if prices[j] < prices[i]: i = j j += 1 print(maxprofit) # Output: 5

#brute force approach arr=[7 , 1, 5, 3, 6, 4] n=len(arr) maxprofit=0 for i in range(0,n-1): #loop for buying day for j in range(i+1,n): #loop for selling down profit=arr[j]-arr[i] #profit=selling price- buying price maxprofit=max(maxprofit,profit) #updating profit print(maxprofit)

int mini=INT_MAX; for(int i=0;i<n;i++) { ans=max(ans,arr[i]-mini); mini=min(arr[i],mini); } return ans;

a=[3,1,0,2,1,0,1,3,2,1,2,3] m=0 #level by level water for i in range(1,max(a)+1): k=i #check every unit area for j in range(len(a)): if(i<=a[j]): if(j!=a.index(i) and j-k!=1): if j>=k: m=m+j-k-1 else: m=m+k-j-1 k=j else: k=j print(m) Water filling level by level genaral approach 😅

Hi Anna iam Java developer looking for freelancing jobs as Java developer could you please share info

Hi Anna o(n3 ) undi kada 9:13 could you please explain in detail

Hi Anna, Meeku telisina Python dsa youtube channel suggest cheyara

bro last approach ardam kaaledu bro🙄😥

hi anna rough ga enii days padhuthundiii dsa complete cheyadaniki and knch ma kosam daily vedios upload chey anna.....alage 3rd year vallaki oka guidance or roadmap la pettandiii anna...

efficient approach def trap(height): if not height: return 0 n = len(height) left_max = [0] * n right_max = [0] * n total_water_stored = 0 # Fill left_max array left_max[0] = height[0] for i in range(1, n): left_max[i] = max(left_max[i - 1], height[i]) # Fill right_max array right_max[n - 1] = height[n - 1] for i in range(n - 2, -1, -1): right_max[i] = max(right_max[i + 1], height[i]) # Calculate total water stored with explicit check for i in range(n): water_level = min(left_max[i], right_max[i]) water_stored = water_level - height[i] if water_stored > 0: total_water_stored += water_stored return total_water_stored

brute force approach height=[0,1,0,2,1,0,1,3,2,1,2,1] totalwaterstored=0 for i in range(0,len(height)): if i>0: leftmax=max(height[:i]) else: leftmax=0 if i<len(height)-1: rightmax=max(height[i+1:]) else: rightmax=0 waterlevel=min(leftmax,rightmax) waterstored=waterlevel-height[i] if waterstored>0: totalwaterstored+=waterstored return totalwaterstored

Brother,topic wise possible patterns like arrays,strings ala problems solve cheyandhi.My small request kindly explain the zero duplicates problem with in place approach.Thanks in advance brother

Sorry for the delay guys, I know you all are waiting! Chaala effort padindhi ee video lo, reshoot cheyalsi vachindi. Next time nunchi videos evaina chaala delay avthunte playlist lo next videos release cheyadaaniki try chestha parellel ga

anna c++ loni cheyppu anna youtube loni anni akkuva channels java and python loni unnayi codes but c++ ki vachinapitiki sudo code cheypaysi vadileystunaru anna c++ loni cheyyu anna

Good explanation brother😊

Bro code is not visible

Thank you very much brother...!!!

Superr bro thanku

anna waiting for videos

Bro Fast Ga upload chyandi

Bro ee video chala bagundhi but vere videos anni time chala ekkuva vunde next videos kuda duration ila ga vunte chala baguntadhi bro

Where u are working my brother

Hi bharath, this swaroopa u r explanation was so good and impressive. I feel that when i cleared my mca i didn't get the guidence like this and moved to govt sector with a lot of hard work. Now i just want to prove my self in coading like DSA. Is it possible now.

Hai anna, nenu 1st year complete chesa,c lang basics vachayi anthey 2nd year nunchi ey language choose chesukovalo cheppunanna, That it should be help for getting job or any other skills to prepare from now itself.Plz anna reply 🙏