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...
Entire playlist: • Two Pointer and Slidin...
Follow us on our other social media handles: linktr.ee/take...

[@muhammadfarhaan6951]

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

[@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 ?

[@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

[@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

[@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

[@KashishIsOn786]

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

[@rushidesai2836]

Classic problem.. thanks Striver!

[@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)

[@stith_pragya]

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

[@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)

[@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..... 😌

[@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

[@parth_3856]

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

[@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

[@torishi82]

Understood bhai. Thank you.

[@hashcodez757]

"UNDERSTOOD BHAIYA!!"

[@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.

[@ishanaggarwal7265]

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

[@salehaafreen4309]

share it please

[@ted_mosby]

feels a little happy to solve this on my own

[@tlasyashree7590]

Super easily understood

[@ShauryaPundir]

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

[@lavanyaan2158]

Thank you so much for the video bro.

[@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)

[@hareshnayak7302]

Thanks striver for this amazing video.

[@LinhHoang-ml1qo]

understood!Thank you Striver

[@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

[@jayateerthag.h5372]

Thankyou

[@sarangkumarsingh7901]

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

[@ganeshjaggineni4097]

NICE SUPER EXCELLENT MOTIVATED

[@oyeesharme]

understood bhaiya

[@aditya_raj7827]

Understood 😊😊

[@mr_weird3680]

Thanks Brother❤

[@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; } };

[@subee128]

Thank you very much

[@kshitijraj1320]

great video 😇

[@iamnottech8918]

Mja AAGeYa!!

[@codeman3828]

Understood. Thanks

[@naveenau143]

Understood ❤

[@aloklaha8694]

Understood

[@MJBZG]

understood

[@Pushpraj-sf1nz]

it was helpful

[@ok-jg9jb]

Thanks❤

[@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

[@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..

[@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

[@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; } ```

[@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;

[@thoughtsofkrishna8963]

We want Strings playlist Striver

[@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

[@prasannasahoo0806]

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

[@mihiradarsh7604]

You are only allowed to take

[@ManishKumar-dk8hl]

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

[@jivanmainali1742]

Where is it mentoned to keep consecutive number

[@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

[@sandeepgmore1483]

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

[@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

[@angeldeveloper]

🙌🏻

[@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; } ----------------------------------------

[@jitghosh3923]

🙌

[@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

[@DeorajMaharaj]

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

[@adityaroychowdhury3709]

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

[@dhruvmishra9781]

bro konse year mein ho??

[@aryankumar3018]

US

[@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; } }

[@StudyYuv]

😀😀

[@theskyraturi]

hil bhi ni rhe ye questions mujhse

[@manasandmohit]

Are bhaiya itni fast fast

[@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

[@raghavmanish24]

understood

[@adityapandey23]

Understood

[@AkashJaiswal-w4q]

understood

[@ramakrishnakcr4417]

understood

[@shototodoroki4719]

understood