Notes/Codes/Problem links under step 10 of A2Z DSA Course: takeuforward.o... Entire playlist: • Two Pointer and Slidin... Follow us on our other social media handles: linktr.ee/take...
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@ArjunAR062 ай бұрын
I'm visiting this problem after a week (I solved this by watching striver's solution). I was not able to solve a single problem on my own thinking. I saw few comments stating that they were able to solve this by themselves and I was like "Am I doing something wrong. Why am I not able to solve on my own?" To my surprise, I completed the entire sliding window playlist and when I revisited after like a week, I'm able to solve every single sliding window problem in like 20 minutes. You deserve a huge salute, Striver!
@CleancodeCentralАй бұрын
I used to have the Same problem manhh. Thats Normal. Complete the whole series or half of the problems. then take a problem and try to solve the problem on paper and pen. For some it takes only 50 leetcode problems But for some it takes More than 200 . " Consistency is the Key". keep improving...
@aryanjain943223 күн бұрын
@@CleancodeCentral Tiru
@Sandeep-tn8mz5 ай бұрын
This was asked to me in AMAZON
@himanshu_0474 ай бұрын
You got selected??
@Sandeep-tn8mz4 ай бұрын
@@himanshu_047 NO bro .. Im not from CS background so, my CS fundamentasl are not strong.. especially OSI, System design.. I was able to solve the current problem but messed up in system design problem.
@oppie3354 ай бұрын
@@himanshu_047 he's here so , you know..
@Shivam-ux8mn4 ай бұрын
Bhaiya mai abhi class 12 pass kiya haii aur abhi btech cse krne ka decide kiya hu and in jee mains meri 73 percentile bani aur abhi mai drop lu ya na lu ke doubt mw hu , ek point ye kehta hai ki college se zada skills matter krti hai agar tumhare pass skills hai to tumhari placement off campus bhi ho jayegi but ek taraf man me ye bhi hai ki kya ek baar try krna chhaiye , is taking a drop is worth it?
@Sandeep-tn8mz4 ай бұрын
Hey bro , At the end of the day your skills matter more than any college bro. Yes , the top tier colleges will give you that early career boost but its you who has to push yourself. Irrespective of college , if you keep yourself pushing you can do it from any college. yes you will have more difficulties , but they will definitely teach you a good lesson bro... ALL THE BEST (And if you are confident that you can do better by taking drop, go for it. You are still young, don't think much about outside noise. At the end of the day Its you who has to feed your family. )
@AkankshaGupta-zn7he5 ай бұрын
Just by watching & understanding the previous 4 videos, I was able to solve it on my own. First with brute then the sliding window technique. That's the magic of Striver😃❤
@jimit27952 ай бұрын
how did you come to know that you have to use map or set in it.
@darkistheknight2 ай бұрын
@@jimit2795 Look I am a noob myslef, but consider this, wherever we have to maintain something in terms of numbers like k distinct, at most k, you should know it is gonna be map. Here also we have two baskets i.e. we can store maximum of two types and we have to maximize the numbers of those 2 types of fruits. So at most K type questions. Hence map/set. That's all there's to it. Don't worry keep grinding. You'll get it.
@TheAditya-zt9pbАй бұрын
@@darkistheknight exactly your intuition is similar to mine
@selviraghavi-jp4ep3 ай бұрын
This question was asked in infosys specialist programmer role and bro made a video thankfully :)
@user-bf7yz3qd2i4 ай бұрын
This was asked to me in Adobe.
@shankarrakh7744Ай бұрын
(best solution without using map) int totalFruits(int N, vector &fruits) { int f = -1; // first fruit type int s = -1; // second fruit type int l = 0; // left pointer int r = 0; // right pointer int ans = 0; while (r < N) { if (f == -1 || fruits[r] == fruits[f]) { f = r; } else if (s == -1 || fruits[r] == fruits[s]) { s = r; } else { // When a new fruit type is encountered if (f < s) { l = f + 1; f = r; } else { l = s + 1; s = r; } } ans = max(ans, r - l + 1); r++; } return ans; }
@raghavmanish242 ай бұрын
hey striver you make things so simple , some time i got jealous that some others also can know this method from this videos ........thanku again striver
@rajharsh37395 ай бұрын
I just solved before seeing this video by just copy paste the last code from L4 and did some tweaks. 🙌🙌
@hashcodez7573 күн бұрын
"UNDERSTOOD BHAIYA!!"
@jagansubramanian52345 ай бұрын
You made my DSA journey easy
@2amCoder5 ай бұрын
althugh i have alredy did all these patterns questions stii here for your support ! thanks a lot for that dp and graph series and others too.
@nishantshrivastava89372 ай бұрын
Theoretically, we have optimised the solution to O(n) but the solution of O(2N) is faster on runtime while programming.
@deepakvajpayee23964 ай бұрын
This solution can also be optimised by in place of storing count of the character in map if we store last index of the character
@AK-nj1jeАй бұрын
Actually the space complexity of the most optimal approach will be almost O(n) Coz just imagine all the elements are different then, the code will not care what's the size of the map, it will just keep adding elements in the map, howeven the maxlen will not be affected. Amazing lecture!!!!
@shubhtandon8315Ай бұрын
Right!!! but suppose the array has half no. of elements as of type 1 and the rest of the array has different types(2,3,4,.....) then while trimming down the left side we will be adding different elements to the map. So instead of O(n), it can said to be near about O(maxlen ) or O(n/2) as at max we will be storing half of the elements. What do you say?
@Aparichi78gft24 күн бұрын
No bro map will not store the let say 1--- element rather it will store the entry of it so it will o(1) not o(n/2) so at max map will be o(3).
@akashwasson42203 ай бұрын
Striver bro you are a pro....Salute!
@watch2-grow5 күн бұрын
Understood thanks bhaiya
@CaptionamericaАй бұрын
Bhai saaahab tussi great ho ❤❤❤❤. A big hearts to you bade bhaiyaaa
@stith_pragya4 ай бұрын
UNDERSTOOD...Thank You So Much for this wonderful video.......🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻
@Ethical-Boy-No-15 ай бұрын
28:58 I have a question bhaiya .. Let's take this example {1,2,1,2,1,2,1,2,3,4,5,6,7,8,9,10,11} .. if you see in case of optimal solution then SC=O(n/2) for map ... so how it is O(1) . And if you make an example of 10^5 length array the same as the upper example then also SC=O(10^5/2)..
@fizzgaming79102 ай бұрын
Pointing out a small error, instead of erasing mpp[arr[l]] in the code, erase arr[l] and the code works flawlessly
@soumeshnayak45462 ай бұрын
I was unable to solve a single question in Sliding Window and two pointers after watching the above concepts I can think of and able to solve this question. This is the solution that I have solved on my own. def totalFruit(self, fruits: List[int]) -> int: maxlen=0 map={} i=0 for j in range(len(fruits)): map[fruits[j]]=map.get(fruits[j],0)+1 while len(map)>2: map[fruits[i]]-=1 if map[fruits[i]]==0: del map[fruits[i]] i+=1 if len(map)
@takagaminggxdАй бұрын
beautiful explanation ❤
@sahulraj953629 күн бұрын
for optimal approach space complexity cant be O(1) right, because map can have more than 2 elements as well for this example {1,1,1,1,1,3,3,2,4,5,6,7,8,9} for the above example map can have almost n/2 elements correct me if im wrong
@Aparichi78gft24 күн бұрын
No bro map will have at max 3 elements in any case
@verma_jay7 сағат бұрын
You are correct
@user-fz1tl2dh9bАй бұрын
lovee you brooo thanksss a lott!!!
@sonakshibajpai64452 ай бұрын
Thank you striver !!
@AtulSingh-dp8wo5 ай бұрын
most awaited ❤thanks a lot!
@RadheShyam334554 күн бұрын
tried using set instead of hashset and directly removing(arr[i]) from set and setting i=j-1; whenever setSize >2 but it failed certain test cases
@anshkothariAK3 күн бұрын
Your code will not work for this testcase: {1, 2, 1, 2, 3, 2, 2, 1}
@oyeesharme16 күн бұрын
understood bhaiya
@Abcd-jt1qs2 ай бұрын
Understood sir! Thank you :)
@augustinradjou39095 ай бұрын
Understood bhaiya ❤
@namannema33494 ай бұрын
thanks striver
@techmateinАй бұрын
i think i solved it in better way : class Solution { public: int totalFruit(vector& fruits) { pairp1={-1,-1}; pairp2={-1,-1}; int l=0; int ans=-1; for(int i=0;i
@shivisingh99754 ай бұрын
Understood sir!
@subee1285 ай бұрын
thanks
@adityapandey23Ай бұрын
Understood
@AJ-ve1yb4 ай бұрын
should i follow sequence of this playlist according to a2z sheet or to watch first which has been uploaded first @striver
@raghavmanish242 ай бұрын
understood
@lucifee94073 ай бұрын
Striver if you are the interviewer ,you ask to optimize it to 2N to N ??
@Harsh-jc2bz3 ай бұрын
mostly dont ask...9 out of 10
@5Q8NAZEER12 күн бұрын
we should erase mp.erase(arr[l]) not mp.erase(mp[arr[l]]) ...... a small correction from my side
@akworld27393 ай бұрын
but i am happy with 0(2N) time complexity😂😂😂😂
@techmateinАй бұрын
after solving the previous question on my own and reading this one, i thought i could not be able to solve this one on my own but gave it a try and solved it on my own and that also in just o(n). quite happy today.
@alessandrocamilleri12394 ай бұрын
Great explanation as usual. However the map is actually not necessary. int findMaxFruits(vector &arr, int n) { int fruit1 = arr[0], fruit2 = -1; int nextLeftPos = -1; int l = 0, r = 1; int count = 1; while (r < n) { if (arr[r] == fruit1 || arr[r] == fruit2) { if(arr[r] != arr[r-1]) nextLeftPos = r; count = max(count, r - l + 1); r++; } else if (fruit2 == -1) { fruit2 = arr[r]; nextLeftPos = r; count = max(count, r - l + 1); r++; } else { l = nextLeftPos; fruit1 = arr[l]; fruit2 = arr[r]; } } return count; }
@saketjaiswal93814 ай бұрын
done
@noone-fl1qb4 ай бұрын
Can anyone tell me why this is 'nt working int findMaxFruits(vector &arr, int n) { int left = 0, right = 0, maxLength = 0; unordered_map map; while (right < arr.size()) { map[arr[right]]++; if (map.size() > n) { map[arr[left]]--; if (map[arr[left]] == 0) { map.erase(arr[left]); } left++; } if(map.size()
@shubhiiii2203 ай бұрын
Add "right++" after closing the bracket of "if(map.size()
@ayushisingh66383 ай бұрын
If this is geeksforgeesks question the varibale n is the size of array. if questio says 2 basket then you need to use 2 instead of n the following will produce correct result: int totalFruits(int n, vector &arr) { int left = 0, right = 0, maxLength = 0; unordered_map map; while (right < n) { map[arr[right]]++; if (map.size() > 2) { map[arr[left]]--; if (map[arr[left]] == 0) { map.erase(arr[left]); } left++; } maxLength = max(maxLength, right - left + 1); right++; } return maxLength; }
@hardiksingh83444 ай бұрын
thanks sir can you update in take you forward site as well
@txbankush46012 ай бұрын
can anyone provide me the full code because my some testcases are not being passed according to this pseudocode
@simarpreetsingh723528 күн бұрын
A better method could be storing {value, last picked position} , since only two values would be there in the map pick the minimum of last picked position of the two, O(N) solution
@shreerangaraju10133 ай бұрын
Can anyone tell me what iPad software is striver using?
@harshit7794 ай бұрын
bahaiya please upload article on the website for this
@user-oj3wh8bx8z5 ай бұрын
Please add C++ code also
@parvahuja76185 ай бұрын
then what will you do
@sangharshpipare6664 ай бұрын
@@parvahuja7618 I will dance
@ramdangi53354 ай бұрын
@@parvahuja7618 nagin dance 😂😂
@lakshsinghania3 ай бұрын
do it yourself bro
@angeldeveloper5 ай бұрын
🙌🏻
@user-gk4zy6bk7l3 ай бұрын
god
@Beeplov23375685 ай бұрын
😍
@angeldeveloper5 ай бұрын
❤🙌🏻👍
@rohanbera622711 күн бұрын
int totalFruits(vector arr) { int n = arr.size(); if (n == 1) return 1; int i = 0, j = 0, maxi = 0; int elem1 = arr[0], elem2 = -1; int ind1 = 0, ind2 = 0; int lastConsec = 0; while (j < n) { if (elem2 != -1 && elem1 != arr[j] && elem2 != arr[j]) { i = lastConsec; elem1 = arr[i]; elem2 = arr[j]; } elem2 = (elem1 != arr[j] && elem2 == -1) ? arr[j] : elem2; if (arr[j] != arr[lastConsec]) lastConsec = j; maxi = max(maxi, j - i + 1); j++; } return maxi; }