Пікірлер

• 
  @nandanimadhukar7 сағат бұрын

  very good explanation and technique

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  @techdose4u4 сағат бұрын

  Thanks

• 
  @HeloHi-dp6up9 сағат бұрын

  Bro if the question ask for 5 mark.. How much mark will get if i write this method🫠

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  @techdose4u4 сағат бұрын

  10. extra marks for making my life easy in checking 😂

• 
  @AyushKumar-kh8oh11 сағат бұрын

  nice explanation

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  @boomer_mcghee20 сағат бұрын

  i love you

• 
  @DeepakSingh-uf5vv20 сағат бұрын

  What if the array is made up of floats something like 0.76546, 2, 0.65, ....

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  @souravroychowdhury652521 сағат бұрын

  amazing explanation !!

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  @techdose4u4 сағат бұрын

  Thanks

• 
  @tanzimchowdhury320Күн бұрын

  Amazing, just amazing, after 4/5 videos, this is the only guy that makes sense!

• 
  @rohanrajput36940Күн бұрын

  hates(mom,bua). 1:25

• 
  @rey22122Күн бұрын

  makasih bang, besok aku exam jadi mengerti

• 
  @divinity8683Күн бұрын

  ❤

• 
  @harshitsahu2704Күн бұрын

  just awsome made my code just after first 3 min of viedo .

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  @love_for_cars__bikesКүн бұрын

  Man you blew my mind 🤯🤯 Is this always this easy or am I the one who's tripping??

• 
  @abid3801Күн бұрын

  you are Robot.

• 
  @thenepalinerdКүн бұрын

  if you r to merge the sorted chunk and again sort it, why the heck would you sort the chunk first... just merge everything and sort it there... your step is taking double the time. geez youtube ka 50 IQ ka instructor

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  @techdose4uКүн бұрын

  You can choose any inplace algorithm like heapsort. Then find sorted array with O(1) space and then write back the chunks in contiguous memory in the drive. That will effective be done in O(1) extra space and O(NlogN) time complexity.

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  @gayiii1911Күн бұрын

  great explaination!

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  @techdose4uКүн бұрын

  thanks :)

• 
  @dharanish58162 күн бұрын

  Today I have attended Zoho on campus drive.This question is asked in basic programming round💯.

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  @techdose4uКүн бұрын

  nice :)

• 
  @rahuldey55642 күн бұрын

  what is this algorithm called?

• 
  @techdose4u2 күн бұрын

  dutch national flag maybe :)

• 
  @Polly101893 күн бұрын

  Now I will not forget this, thanks

• 
  @techdose4u2 күн бұрын

  great

• 
  @tanishachoudhary25883 күн бұрын

  this is the best video on articulation point, hats off to your comendable efforts 👑

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  @techdose4u2 күн бұрын

  thanks

• 
  @kushagrashekhawat82273 күн бұрын

  I've watched many of your videos. Your explanations are always very easy to understand. Thank you for creating these videos; you are doing so much good.

• 
  @techdose4u2 күн бұрын

  thanks :)

• 
  @user-wg5vc9ik1s4 күн бұрын

  Why we are not moving i back to previous + 1 position?

• 
  @NishagAppu4 күн бұрын

  thanku for explaining pre in and post order in such a easy way my 5 month class converted into 7mints🫡

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  @techdose4u2 күн бұрын

  woww :)

• 
  @sanoferniswan59014 күн бұрын

  Superb explanation ❤

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  @techdose4u2 күн бұрын

  thanks :)

• 
  @user-lj1ny6ts1f4 күн бұрын

  Thank you 🎉🎉

• 
  @techdose4u2 күн бұрын

  welcome

• 
  @mathewsjose19905 күн бұрын

  Nice explanation!

• 
  @techdose4u2 күн бұрын

  thanks

• 
  @user-wt2dq5yt9h5 күн бұрын

  good explanation but I need to implement it using for loops not recursive functions (it is more complex)

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  @IdrisNirah5 күн бұрын

  Thanks

• 
  @techdose4u2 күн бұрын

  welcome

• 
  @pasumarthiashik10995 күн бұрын

  perfect explanation

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  @techdose4u2 күн бұрын

  thanks

• 
  @anandpandey9185 күн бұрын

  // Bruteforce Approach class Solution { public int subarraySum(int[] arr, int k) { int[] prefixSum = new int[arr.length]; for (int i = 0; i < arr.length; i++) { if (i == 0) { prefixSum[i] = arr[i]; } else { prefixSum[i] = prefixSum[i - 1] + arr[i]; } } int count = 0; for (int start = 0; start < arr.length; start++) { for (int end = start; end < arr.length; end++) { if (findSubarraySum(prefixSum, start, end) == k) { count++; } } } return count; } private int findSubarraySum(int[] prefixSum, int startIndex, int endIndex) { if (startIndex == 0) { return prefixSum[endIndex]; } return prefixSum[endIndex] - prefixSum[startIndex - 1]; } } // Better Approach class Solution { public int subarraySum(int[] arr, int k) { int count = 0; for (int start = 0; start < arr.length; start++) { int subArraySum = 0; for (int end = start; end < arr.length; end++) { subArraySum += arr[end]; if (subArraySum == k) { count++; } } } return count; } } //Optimal Approach class Solution { public int subarraySum(int[] arr, int k) { Map<Integer, Integer> sumFrequencyMap = new HashMap<>(); int count = 0, prefixSum = 0; for (int val : arr) { prefixSum += val; if (prefixSum == k) { count++; } if (sumFrequencyMap.containsKey(prefixSum - k)) { count += sumFrequencyMap.get(prefixSum - k); } sumFrequencyMap.put(prefixSum, sumFrequencyMap.getOrDefault(prefixSum, 0) + 1); } return count; } } /* int[] arr = {3, 0, 7 ,2 ,-3 ,1 ,4 ,2}; and k = 7 Let's break down the method's implicit check for all subarrays: Possible Subarrays and Their Sums: Subarrays starting from index 0: [3] : Sum = 3 [3, 0] : Sum = 3 [3, 0, 7] : Sum = 10 [3, 0, 7, 2] : Sum = 12 [3, 0, 7, 2, -3] : Sum= 9 [3, 0, 7, 2, -3 ,1] : Sum= 10 [3, 0, 7, 2, -3, 1, 4]: Sum= 14 [3, 0, 7, 2, -3, 1, 4 , 2]: Sum= 16 Subarrays starting from index 1: [0] : Sum = 0 [0, 7] : Sum = 7 [0, 7, 2] : Sum = 9 [0, 7, 2, -3] : Sum = 6 [0, 7, 2, -3, 1] : Sum = 7 [0, 7, 2, -3, 1, 4] : Sum = 11 [0, 7, 2, -3, 1, 4, 2] : Sum = 13 Subarrays starting from index 2: [7] : Sum = 7 [7, 2] : Sum = 9 [7, 2, -3] : Sum = 6 [7, 2, -3, 1] : Sum = 7 [7, 2, -3, 1, 4] : Sum = 15 [7, 2, -3, 1, 4 , 2] : Sum = 17 Subarrays starting from index 3: [2] : Sum = 2 [2, -3] : Sum = -1 [2, -3, 1] : Sum = 0 [2, -3, 1, 4] : Sum = 4 [2, -3, 1, 4, 2] : Sum = 6 Subarrays starting from index 4: [-3] : Sum = -3 [-3, 1] : Sum = -2 [-3, 1, 4] : Sum = 2 [-3, 1, 4, 2] : Sum = 4 Subarrays starting from index 5: [1] : Sum = 1 [1, 4] : Sum = 5 [1, 4, 2] : Sum = 7 Subarrays starting from index 6: [4] : Sum = 4 [4, 2] : Sum = 6 Subarrays starting from index 7: [2] : Sum = 2 Observation : In first iteration the direct lookup(for prefixSum= 3) is for subarray [3] (doesnot sum to 7) and map lookup(for repetition of (prefixSum-7) i.e -4) is for [] subarray, which does not sum to 7 so ignored. (Hence count=0 and Map:{3:1}) In second iteration the direct lookup(for prefixSum= 3) is for subarray [3 ,0] (doesnot sum to 7) and the map lookup (for repetition of (prefixSum-7) i.e -4) is for subarray [0] , which does not sum to 7 so ignored. (Hence count =0 and Map: {3:2}) In third iteration the direct lookup(for prefixSum= 10) is for subarray[3 ,0 ,7] (doesnot sum to 7) and the map lookup(for repetition of (prefixSum-7)= 3) is for subarrays [0, 7] and [7] , both of these subarrays sum to 7 so include it. (i.e count=0+1+1= 2 i.e [0, 7] and [7] and Map: {3:2,10:1}) In fourth iteration the the direct lookup(for prefixSum= 12) is for subarray [3 , 0, 7, 2] (doesnot sum to 7) and the map lookup (for repetition of (prefixSum-7)= 5) is for subarrays [0, 7, 2] , [7, 2] and [2] but none of these subarrays sum to 7 so ignored (i.e count=2 i.e i.e [0, 7] and [7] and Map: {3:2,10:1,12:1}) In fifth iteration the the direct lookup(for prefixSum= 9) is for subarray [3, 0, 7, 2, -3] (doesnot sum to 7) and the map lookup (for repetition of (prefixSum-7)= 2) is for subarrays [0, 7, 2, -3] , [7, 2, -3] and [2, -3] and [-3] but none of these subarray sum to 7 so ignored (i.e count=2 i.e [0, 7] and [7] and Map: {3:2,10:1,12:1,9:1}) In sixth iteration the the direct lookup(for prefixSum= 10) is for subarray [3, 0, 7, 2, -3 ,1] (doesnot sum to 7) and the map lookup (for repetition of (prefixSum-7)= 3) is for subarrays [0, 7, 2, -3, 1], [7, 2, -3, 1] and [2, -3, 1] and [-3, 1] , two of these subarray sum to 7 (i.e count=2+1+1=4 i.e [0, 7], [7], [0, 7, 2, -3, 1] and [7, 2, -3, 1] and Map: {3:2,10:2,12:1,9:1}) In seventh iteration the the direct lookup(for prefixSum= 14) is for subarray [3, 0, 7, 2, -3, 1, 4] (doesnot sum to 7) and the map lookup (for repetition of (prefixSum-7)= 7) is for subarrays [0, 7, 2, -3, 1, 4] , [7, 2, -3, 1, 4] , [2, -3, 1, 4] , [-3, 1, 4] , [1, 4] and [4] , None of these subarray sum to 7 (i.e count=4 i.e [0, 7] ,[7], [0, 7, 2, -3, 1] and [7, 2, -3, 1] and Map: {3:2,10:2,12:1,9:1,14:1}) In eight iteration the the direct lookup(for prefixSum= 16) is for subarray [3, 0, 7, 2, -3, 1, 4 , 2] (doesnot sum to 7) and the map lookup (for repetition of (prefixSum-7)= 9) is for subarrays [0, 7, 2, -3, 1, 4, 2] , [7, 2, -3, 1, 4 , 2] , [2, -3, 1, 4 , 2] , [-3, 1, 4, 2] , [1, 4, 2] , [4, 2] and [2] ,one of these subarray sum to 7 (i.e count=4 + 1 = 5 i.e [0, 7] ,[7], [0, 7, 2, -3, 1], [7, 2, -3, 1] and [1, 4, 2] and Map: {3:2,10:2,12:1,9:1,14:1,16:1}) */

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  @andresdallarizza15266 күн бұрын

  fucking legend lov u adore u

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  @techdose4u2 күн бұрын

  :)

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  @srivishnudeepika80976 күн бұрын

  Such clear explanation and visual drawings are the reason why u r ny go to person for DSA , you are #1 man

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  @techdose4u6 күн бұрын

  thanks :)

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  @sanatani92737 күн бұрын

  Well explained dude! 😂❤

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  @techdose4u2 күн бұрын

  thanks :)

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  @channadissanayaka64507 күн бұрын

  very clear explaination.

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  @techdose4u2 күн бұрын

  thanks :)

• 
  @n.h.son19027 күн бұрын

  I'm satisfied with the proof on the time complexity. What about the space complexity? May I ask why it is O(logN)?

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  @techdose4u7 күн бұрын

  log N is the height of the tree. we are building heap inplace

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  @amgadshrief72158 күн бұрын

  you are awesome keep going bro

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  @techdose4u2 күн бұрын

  thanks :)

• 
  @channadissanayaka64508 күн бұрын

  Thank you ... This is very clear

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  @techdose4u2 күн бұрын

  thanks :)

• 
  @channadissanayaka64508 күн бұрын

  Thank you very much

• 
  @techdose4u2 күн бұрын

  welcome

• 
  @channadissanayaka64508 күн бұрын

  thank you very much

• 
  @techdose4u2 күн бұрын

  welcome

• 
  @channadissanayaka64508 күн бұрын

  thank you very much

• 
  @techdose4u2 күн бұрын

  welcome

• 
  @umarmasud24628 күн бұрын

  Thanks 😢❤

• 
  @techdose4u2 күн бұрын

  welcome :)

• 
  @greenhawk2719 күн бұрын

  bhai ye to pta hi hota hai you need to find the ans without using inorder aise to bst ke saare que hote hi hai inorder se interview mai aise allowed nai krte

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  @QuynhNhu-gu5ql9 күн бұрын

  How to contact for you. I come to VIỆT NAM

• 
  @abhilashpatel30369 күн бұрын

  didn't notice new intro look. nice!

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  @melwinalm9 күн бұрын

  Such a great explanation

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  @techdose4u2 күн бұрын

  thanks :)

• 
  @DennisSimplifies9 күн бұрын

  😂😂😂 bro simply too good 👍

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  @techdose4u2 күн бұрын

  :)

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  @space_ace77109 күн бұрын

  Best explanation I found!

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  @techdose4u2 күн бұрын

  thanks :)

• 
  @MP-ny3ep9 күн бұрын

  Great explanation. Thank you

• 
  @ventacode9 күн бұрын

  here for daily solution bud !1

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  @techdose4u2 күн бұрын

  welcome :)

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  @jeevatrolls61849 күн бұрын

  Sir, where i can find the problems

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  @prasannakatabattini19749 күн бұрын

  can u send the code for this problem..i got wrong an"Could you please provide the code for this problem? I received an incorrect answer." swer

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  @karthikeyank746710 күн бұрын

  Best solution is still O(N)