please bring string series as soon as possible

Please try to make and upload string, stacks n queues and heaps playlist as soon as possible. I understand you must be very busy, but still you are making time for us and uploading videos and playlists at regular intervals. Thanks a lot❤❤

You are next level in explaining, hands up 🙌🙌

Thank you so much💯.....please bring stacks and queue playlist

Hope you are doing extremely well.

please consider bringing a playlist on stacks and queues as soon as possible. I am totally unable to figure out the intuition by just seeing the question in an interview

Please bring strings series ASAP bhaiya ❤ lots of love and thanks for your content ❤️🙏🏻

Thank u so much for this playlist

Thankyou Please bring a playlist on strings

Please bring the string series as soon as possible.

Thankyou so much Striver for all you efforts throughout in delivering us so much valuable content. Any student / working professional can now be able to transition their career without paying money for courses. Would also like your insights on the point : While preparing for interviews most of the aspirants are going through the videos solely and solving the question after completely watching the video. And also are feeling lazy trying to solve the question on our own. What is the best way to complete any topic without being lazy and how should an aspirant approach any topic/playlist?

"Striver, your DSA Sheet is absolutely phenomenal! It's been an invaluable resource for mastering data structures and algorithms. Looking forward to the remaining topics, especially the much-anticipated sections on strings and heaps. Thanks for all your hard work!"

@ok-jg9jb

Ай бұрын

Why are you spamming in every video?

Never thought, there would be a linear solution for this question!

Would also recommend solving Minimum Jumps problem in gfg. Same as above but with a little caveat. Amazing solution btw

I had solved this long back using 1D Dp. Just took the index as state. Below is the recurrence- int func(int index, vector& arr) { if(index >= arr.size()-1) return 0; //1 is already added while reaching this. if(index + arr[index] return INT_MAX; //impossible to reach int mini = 0; for(int i = 1; i

@jaydabhi5539

Ай бұрын

I think time complexity will be O(n*maxjump) I had also solved this using 1D DP.

@rishabhagarwal6057

22 күн бұрын

same

awesome content... please make string playlist soon

love your tutorials till now can you pls add string series also

Thanks Great Content!

Sir please do playlist in strings Really it is needed 🙏

Bhaiya please make a series on strings badly need it it's a humble request

Can we expect Stack and Queue playlist by end of this month or next month ?

you are best❤

Clowns in the comments demand everything but not once appreciate the guy for uploading all these lectures, lol

@iamnoob7593

Ай бұрын

in india if u see lot of people want everything for free.

got it bro!!!

Hey raj, can you bring the string series soon???

Bhaiyya, please start heap series after this one

sir please fix the saved notes issue of striver sheet after the new update i am facing a problem that notes saved for question A gets saved to notes of question B(happens when you restart the website and go to saved notes navbar section to check your notes)

thankyou sir

Wow ! what a solution

I have solved using one for loop only int jump(vector& nums) { int jumps = 0; int left = 0; int right = 0; for (int i = 0; i right = max(right, nums.at(i) + i); if (left == i) { jumps++; left = right; } } return jumps; }

Understood 💯

love it

Bhaiya pattern wise recursion prr bhi daal do

awesome

Striver brilliant solution man , I had done this problem using dp only , No wonder u r in GOOGLE

ty sir

Isn't that i+arr[i] inside the for loop? Why striver has written i+arr[ind]? Won't that be different?

Best!

Understood

Bhaiya please start sde sheet challange 2024

String please

class Solution { public int jump(int[] nums) { if (nums.length == 1) return 0; int n = nums.length; int l = 0, r = 0, jumps = 0, farthest = 0; while (r

@tanyacharanpahadi158

28 күн бұрын

If you remove = n - 1) ....)

Started your playlist a week ago, didn't know there are more videos in the making. What else is remaining in the course?

@brainmosquito7

18 күн бұрын

all major portions are covered strings is just remaining i recommend you to go to TUF wesite and start following A2Z sheet

Why code studio is gone

please bring the string video first .A humble request from us

HOW IS IT 2 JUMPS FOR ALL INDEXES FROM F(1,1) IN TREE @3:09 ?

@AlokTripathi

27 күн бұрын

it's wrong computation.

hey striver here in my O(n) time complexity solution int jump(vector& nums) { int final=nums.size()-1; int i=0; int count=0; while(final!=0){ if(nums[i]>=final-i){ final=i; count++; i=0; } else i++; } return count; }

was the greedy solution intuitive or not ?coz i dont find it intuitive!!!!

Please bring heaps bro

Waiting for strings playlist

yehi too chahiye tha 😭

Striver, there is no need for 2D DP here. It can be solved using 1D DP.

US

brother ye toh DP ka question hai then put it there why in greedy playlist :)

@Rahul_Mongia

Ай бұрын

question has different ways to solve, this soln has greedy as optimal approach

the dp solution is 1d why use 2d:class Solution { public: int solve(vector& nums,int id,vector& dp){ if(id>=nums.size()-1)return 0; if(dp[id]!=-1)return dp[id]; int reach=id+nums[id]; int n=nums.size(); int mini=1e9; for(int k=id+1;k

There can one more simple greedy solution #Java class Solution { public int jump(int[] nums) { int z; int smallest[]=new int[nums.length]; smallest[nums.length-1]=0; for(int i=nums.length-2;i>=0;i--){ if(i+nums[i]>=nums.length-1) smallest[i]=1; else{ z=getsmallest(smallest, i+1, i+nums[i]); smallest[i]=1+z; } } return smallest[0]; } int getsmallest(int ary[], int a, int b){ int small=10000000; for(int j=a;j

Data Structures & Algorithm ❌ Data STRIVERS & Algorithm ✅

bro, why is your voice very low in this greedy series, can't here u properly

00:04 Finding minimum number of jumps to reach the end 02:01 Using recursion to find the minimum number of jumps in a smaller example 04:04 Return the number of jumps when index is greater than or equal to n - 1 06:19 Optimizing dynamic programming solution using a quadratic state approach 08:37 Understanding jump range in the context of Greedy Algorithm 10:31 Optimizing jump game II algorithm by carrying a range instead of individual recursive calls 12:42 Determine farthest jump for each range and update jumps array 14:47 Implementing non-recursive range based solution for jump game with linear time complexity.