L2. Maximum Points You Can Obtain from Cards | 2 Pointers and Sliding Window Playlist

Notes/Codes/Problem links under step 10 of A2Z DSA Course: takeuforward.o...
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[@Gottastudyhard-m3b]

Implementing it yourself makes it much more clearer.

[@4444-c4s]

East or West, Striver bhaiya sabse Best KZbin pe esa koi bhi nahi hoga jo job ke saath saath itna content banata ho aur apni company bhi chalata ho...🙏🙏🙏

[@Coder_Buzz07]

codewithmik bhi hai bro

[@Amanh729]

@@Coder_Buzz07 yup

[@adarshnegi8906]

This was a great problem, i hope one day i can come up with such approach all by myself. For those who might of two pointers i at 0 and j at n-1 and compare instances , this method will fail on TC such as [11,49,100,20,86,29,72], because at first it looks 72 should be chosen but all the elements after 11 are large and they will compensate and produce maximum score

[@sruthimajhi5610]

Thank you... I was dying to understand the problem statement. Now this example makes it easier why a sliding window approach should be used and not the two pointer approach.

[@ritik_deswal77]

my approach using while loop in java - int[] ar= {6,2,3,4,7,2,1,7,1}; int k=4; int sum=0; for(int i=0;iar.length-k-1) { sum+=ar[l]; sum-=ar[r]; maxx=Math.max(maxx,sum); l--;r--; } System.out.println(maxx);

[@mayanksaurabhmayanksaurabh9271]

thanks for so easy and intuitive solution. Lot of solutions for this problem are problem which have made it look really complex

[@unknown2698]

public static int maxScore(int[] cardPoints, int k) { int lsum =0, rsum =0, max =0,sum =0; int n = cardPoints.length; for(int i=0;imax){ max = sum; } } return max; } same approach but easier to understand

[@VarshaSingh-hi2sb]

In this case with above solution in video we can't take 7 i.e 7,2,1,7 which will result in more sum . or is it important to take the last card ?

[@aakanshavishwakarma8235]

another approach? we can take a consecutive window of size (n-k) and find the minimum sum window , that yields us the maximum sum of the remaining 4 elements from the first or last

[@akshaysharmaBSG]

Yes...that also makes sense...but what will be time complexity for getting min sum of n-k array size ?

[@AryanGupta-2024]

@@akshaysharmaBSG O(N) time and O(1) space

[@sjain6320]

@@AryanGupta-2024 yeah i did this but ig his solution is better as it is O(2k)

[@time-barbaad]

@@sjain6320 the above said solution would be better in the cases where k > n/2

[@109_debjitacharjee9]

u are the best bhaiya🤩 eagerly waiting for your string playlist

[@KashishIsOn786]

Thank you @striver , for making DSA Easy for us . Hatts of to you .

[@rushidesai2836]

Classic problem.. thanks Striver!

[@parth_3856]

OTHER CREATORS! be like "BHAI SAANS THO LENE DE......".

[@stith_pragya]

Understood...........Thank You So Much for this wonderful video..........🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻

[@someThingisFishy5]

One more approach: As the qsn asking us to pickup the maximum points , and the points can be pickup either front or back side, (which leaves the least sum of points in the array) so we can find the minimum sum ,with window size n-k and subtract the result from total sum of the array ,we can use sliding window to solve this Time Complexity is O(N) Space Complexity is O(1)

[@IstiakGametube]

I have been following your a2z DSA course. I want to do strings but there is no videos and problems in your sheet. Please make videos on them and upload the problems

[@mrinceptionist7038]

bro what are you talking about.....the a2z dsa sheet has 2 dedicated section for strings....step-5 and step-18, which covers all basics,medium and hard string questions !!!

[@IstiakGametube]

@@mrinceptionist7038 He has questions but there is no solve videos for them

[@bhaiinnaa6535]

You can take n-k as windowlength , totalsum=sum(arr) windowlength=n-k currwindowsum=sum(arr[:windowlength) maxwin=curwin , for l in range(windowlen,n): curwin +=arr[l]-arr[l-windowlen] netsum=total-curwin maxwin=max(maxwin,netsum)

[@torishi82]

Understood bhai. Thank you.

[@subhasishdas3011]

One more approach is finding min_sum of all windows of size array_length - k , array_total_sum - min_sum will be the ans.

[@hashcodez757]

"UNDERSTOOD BHAIYA!!"

[@bharath3387]

Do we need seperate variables for right and left sum, can we not just maintain a single variable and 2 pointers and remove left pointer value and increase right bponter value

[@ShauryaPundir]

thanks vvvvv much sir,i really wanted a playlist like this

[@jivanmainali1742]

With modulo , its much easier class Solution { public: int maxScore(vector& cardPoints, int k) { int n = cardPoints.size(); int left = n-k; int right = n-k; int ans = 0; int sum = 0; while (left < n) { if ((right - left + 1)

[@lavanyaan2158]

Thank you so much for the video bro.

[@ishanaggarwal7265]

Hi striver, I knew a better solution that solves in single pass.

[@salehaafreen4309]

share it please

[@tlasyashree7590]

Super easily understood

[@sandeeppp9040]

class Solution { public int maxScore(int[] cardPoints, int k) { int lsum=0,rsum=0,maxSum=0; for(int i=0;i=0;i--){ lsum-=cardPoints[i]; rsum+=cardPoints[rindex]; rindex--; maxSum=Math.max(maxSum,(lsum+rsum)); } return maxSum; } } i was here and anyone can have this java code !!!! and also initial loop will go to end k because of correct calculation of lsum

[@Codingforugeek]

class Solution { public: int maxScore(vector& nums, int k) { int leftSum=0,rightSum=0,maxSum=0; for(int i=0;i=0;i--){ leftSum-=nums[i]; rightSum+=nums[rightIndex]; rightIndex--; maxSum=max(maxSum,leftSum+rightSum); } return maxSum; } }; here's the working code..... 😌

[@hareshnayak7302]

Thanks striver for this amazing video.

[@ted_mosby]

feels a little happy to solve this on my own

[@LinhHoang-ml1qo]

understood!Thank you Striver

[@raghavmanish24]

hey striver ... i'm the one of those follower who always give time for like and comment whenever i watch your videos.....thanku again

[@shivanshchaturvedi2601]

More of a two pointer approach rather than sliding window....as sliding window format is generally Diffrent.

[@8daudio672]

class Solution { public int maxScore(int[] nums, int k) { //int left=0,right=0,maxi=0,sum=0; int n=nums.length,leftsum=0,rightsum=0,maxi=0; for(int i=0;i=0;i--){ leftsum-=nums[i];//contract //if(j>n-k-1) rightsum+=nums[j];//expand j--; maxi=Math.max(maxi,leftsum+rightsum); } return maxi; } } whats wong with it

[@kisnagoyal9694]

I think you are ignoring value of leftsum from (idx = 0 to idx=k-1) in maxi... So it doesn't take value of leftsum for first kth elements. So after the first loop try maxi = leftsum before going for second loop in which you are decreasing value of leftsum....try it...may help..

[@sarangkumarsingh7901]

Wow......nice approach......

[@ganeshjaggineni4097]

NICE SUPER EXCELLENT MOTIVATED

[@iamnottech8918]

Mja AAGeYa!!

[@AmanPandey-bd1sj]

thanks you bhaiya, Understood😊 import java.util.*; public class main{ public static void main(String args[]){ int a[] = {6, 2, 3, 4, 7, 2, 1, 7, 1}; int k = 4; System.out.println("MAX SUM IS: "+ findMaxPoints(a, k)); } public static int findMaxPoints(int a[], int k){ int lsum = 0; int rsum = 0; int maxsum = 0; for(int i=0;i

[@mr_weird3680]

Thanks Brother❤

[@aditya_raj7827]

Understood 😊😊

[@oyeesharme]

understood bhaiya

[@kshitijraj1320]

great video 😇

[@vaibhavmurarka5179]

my approach is similar to urs what i did is take k last elements followed by k first then i took sum of last k and then kept removing the front and adding the latest element in the sum and took max of these steps: int n=arr.size(); int ans=0; int sum=0; int ret=0; for(int i=n-k;i

[@jayateerthag.h5372]

Thankyou

[@dipanshuraj7868]

can anyone guide me why the two pointer approach is not working here that I am place i in the 0 and the J = n-1;

[@shreyxnsh.14]

this method will fail on TC such as [11,49,100,20,86,29,72], because at first it looks 72 should be chosen but all the elements after 11 are large and they will compensate and produce maximum score copied from another comment

[@naveenau143]

Understood ❤

[@subee128]

Thank you very much

[@Pushpraj-sf1nz]

it was helpful

[@sandeepgmore1483]

@striver where can i get the source code solution for content

[@MuralidharanSrec]

here that you have included all the elements as 4 size but the index 4 element 7 is not at all included if its included the output will be 17 am i correct

[@codeman3828]

Understood. Thanks

[@ok-jg9jb]

Thanks❤

[@prasannasahoo0806]

what about the elements in the middle ? how do we reach them ?

[@mihiradarsh7604]

You are only allowed to take

[@vamsilanka5639]

optimal solution in another way ------------------------------------------------------ public int maxScore(int[] arr, int k) { int n = arr.length; int i = 1, j = 1; int sum = 0,maxsum = -1; int end = 2 * k; /// to consider only first k and last k elements while (j k) { sum -= arr[(n + k - i) % n]; i++; } if ((j - i) + 1 == k) maxsum = Math.max(sum, maxsum); j++; } return maxsum; } ----------------------------------------

[@ShahNawaz-nk3po]

Another approach with is exactly same as one of the very standard Sliding window question. class Solution { public: int maxScore(vector& cardPoints, int k) { // we need to find substring of size (n-k) and minimum sum // final answer = is total sum of length n - minimum sum of substring of length (n-k) // = maximum sum of length k taken from extreme left/right int n = cardPoints.size(); int l = 0; int r = 0; int ans = INT_MAX; int sum = 0; int val = 0; for(auto it:cardPoints) val+=it; while(r(n-k)){ sum-=cardPoints[l]; l++; } if(r-l+1==(n-k)) ans = min(ans, sum); r++; } return val-ans; } };

[@jivanmainali1742]

Where is it mentoned to keep consecutive number

[@adityapandey23]

Understood

[@kuldeeprawat-zp7od]

Hello everyone, another approach we can think of is We can take sum of all the elements and as according to example we want sum of 4 maximum elements with given conditions and the size of array is 9 so actually we can calculate the sum of 5 consecutive elements which is minimum among all and then we can subtract it from the total sum of all the elements of the array

[@sanukyadav]

Understood, implementing with pseudocode is not able to pass leetcode test cases! did i miss something! def (nums, k): lsum = 0 maxsum = 0 n = len(nums) for i in range(k-1): lsum = lsum + nums[i] maxsum = lsum r_idx = n-1 maxsum = 0 rsum = 0 for i in range(k-1, -1, -1): lsum = lsum - nums[i] rsum = rsum + nums[r_idx-i] r_idx = r_idx - 1 maxsum = max(maxsum, lsum+rsum) return maxsum

[@vatsalpoddar6660]

int maxScore(vector& cardPoints, int k) { int n = cardPoints.size(); int left = k-1; int right = n-1; int maxSum = INT_MIN; int sum = 0; for(int i = 0; i < k; i++){ sum += cardPoints[i]; } maxSum = max(maxSum, sum); while(left >= 0){ sum = sum - cardPoints[left]; sum = sum + cardPoints[right]; maxSum = max(maxSum, sum); left--; right--; } return maxSum; }

[@Josuke217]

You set maxsum to 0 and r_idx-i is wrong

[@Prakash-jh7hp]

same approach but implemented in a simpler way class Solution { private static int sumupto(int k , int[]arr){ int res=0; for(int i =0;i

[@rishabsharma5307]

Started from the end and circularly rotated sliding window ``` int maxScore(vector& cardPoints, int k) { int i, j, maxSum, sum, n = cardPoints.size(); bool flag = false; i = j = n - k; maxSum = sum = 0; while(true) { sum += cardPoints[j]; if(j-i+1 == k || flag) { flag = true; maxSum = max(maxSum, sum); if(i == 0) break; sum -= cardPoints[i]; i = (i + 1) % n; } j = (j + 1) % n; } return maxSum; } ```

[@shototodoroki4719]

understood

[@karthik-varma-1579]

class Solution { public int maxScore(int[] cardPoints, int k) { long start = System.nanoTime(); int maxy = 0; int lsum = 0; int rsum = 0; for(int i=0;i=0;j--){ rsum = rsum + cardPoints[rightIndex]; rightIndex--; lsum = lsum - cardPoints[j]; maxy = Math.max(maxy,lsum+rsum); } long end = System.nanoTime(); System.out.println(end-start); return maxy; } }

[@ManishKumar-dk8hl]

JAVA BOLNE WALO K LIYE : -- class Solution { public int maxScore(int[] arr, int k) { int sum=0; for(int i=0;i

[@thoughtsofkrishna8963]

We want Strings playlist Striver

[@siddiqabr7110]

I thought of this approach at the very beginning and then thought of how to optimise this for an hour then came here to see the optimised approach but 😂😂😂😂😂😂 after coming here i realised I just wasted an hour

[@THAKURPRAJWALSINGH-o7o]

class Solution { public: int maxScore(vector& nums, int k) { int lsum=0; int rsum = 0; int maxSum = 0; int n = nums.size(); for(int i=0;i=0;i--){ lsum=lsum-nums[i]; rsum=rsum+nums[rindex]; rindex=rindex-1; maxSum= max(maxSum,lsum+rsum); } return maxSum; } };this code not passing testcases in leetcode

[@ManishLakkavatri]

the condition in the first loop will be i < k or i

[@ShobhitVerma-j3f]

i think everyone is providing solution when N=2k+1, what if k is lesser than that??

[@paulbarsha1740]

Why 2*k and not 2+k? Isn't 2 separate loop takes adding and nested takes multiplication?

[@sohaildarwajkar9979]

Bhai tu first year se wapas repeat kr

[@aryankumar3018]

US

[@angeldeveloper]

🙌🏻

[@theskyraturi]

hil bhi ni rhe ye questions mujhse

[@jitghosh3923]

🙌

[@StudyYuv]

😀😀

[@DeorajMaharaj]

bhaiya aap thora sick ya kaafi stressed lg rhe ho. Wo phle jaisa energy kahi na kahi missing laga.

[@manasandmohit]

Are bhaiya itni fast fast

[@adityaroychowdhury3709]

Subah uthe, aapke darshan hogye. Ab saare contest badia jayenge, sare hard question solve hone lag jayenge 😂😂

[@dhruvmishra9781]

bro konse year mein ho??

[@saketjaiswalSJ]

class Solution { public: int maxScore(vector& a, int k) { int n=a.size(); int s1=0; int s2=0; int i=0; int j=n-k; while(j k. tk loop chala k times o(k) { s2=s2+a[j]; j++; } int maxi = s2; j=n-k; while(i

[@SibiRanganathL]

Understood

[@ramakrishnakcr4417]

understood

[@aloklaha8694]

Understood

[@raghavmanish24]

understood

[@anshsaxena7297]

Understood

[@AkashJaiswal-w4q]

understood

[@riaa55]

Understood

[@MJBZG]

understood