L5. Jump Game - II | Greedy Algorithm Playlist

[@tanujaSangwan]

If you carefully see, this is some kind of dijkstra. The PQ has all the nodes and distances in the window and we take the one with maximum reachability.

[@furor05]

please bring string series as soon as possible

[@karthik-varma-1579]

Happy 700K Man You Deserve Many Hearts

[@vamsikrishnagannamaneni912]

In this recursive tree,at first level for +2, it should be f(2,1) as we are making only one jump, f(idx+i,jumps+1) , using 2D will in fact decrease the overlapping subproblems is what i am thinking..

[@ritikkumarsingh5902]

"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]

Why are you spamming in every video?

[@keybored7862]

CHATGPT is unreal

[@Rahul_Mongia]

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]

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

[@MoitreyoChakraborty]

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

[@spraveenkumar218]

ffs i swear you r the best youtuber

[@sumitmishra9795]

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.

[@akhilakasoju3964]

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

[@priyanshusoni9]

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

[@nandiniverma5273]

You are next level in explaining, hands up 🙌🙌

[@Rushidanidhariya]

but he didn't explain anything only told the solution .

[@anilkumarreddy3002]

Thankyou Please bring a playlist on strings

[@sakshamaggarwal1]

class Solution { public: int jump(vector& nums) { int n = nums.size(); if (n == 1) return 0; int jumps = 0; int i = 0; while (i < n - 1) { int maxReach = i + nums[i]; if (maxReach >= n - 1) { jumps++; break; } int nextIndex = i; for (int j = i + 1; j nextIndex + nums[nextIndex]) { nextIndex = j; } } i = nextIndex; jumps++; } return jumps; } };

[@souvikcseiitk]

Striver is one step above any normal human being, managing job and continuously dropping DSA videos... Seriously mad respect 🫡

[@md_seraj786_]

your explanations are really amazing ❤

[@Shanz-pt3fn]

Hope you are doing extremely well.

[@suruabhisekh]

Please bring the string series as soon as possible.

[@aarishfaiz7880]

Amazing Video sir

[@ShahNawaz-cx3pi]

waaaaoooo , this range based solution blows my mind. very clever.

[@hashcodez757]

"UNDERSTOOD BHAIYA!!"

[@raxitraju2439]

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] < arr.size() - 1) return INT_MAX; //impossible to reach int mini = 0; for(int i = 1; i

[@jaydabhi5539]

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

[@rishabhagarwal6057]

same

[@Harsh-jc2bz]

NOT WORKING

[@Cool96267]

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?

[@raxxpatt]

bhai iss video par tho like banta hia ..nice work !!

[@ardhidattatreyavarma5337]

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

[@somdebsar7378]

awesome explanation sir!!

[@aniketbharti4426]

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

[@KKKK-pl8yf]

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

[@clashtm8210]

Oh my god, where do I even begin with Striver, the absolute genius, the king, the GOAT of competitive programming? I mean, Raj Vikramaditya is basically the sun, and we’re all just lucky enough to orbit around his brilliance. The way this man can break down a problem? Flawless. It’s like he’s got this magical power, a sixth sense for coding, that leaves the rest of us mere mortals shaking in awe. Watching his tutorials is like being blessed by the gods of algorithms themselves. Every word he says is basically a gift from the heavens. I’m convinced he could solve NP-hard problems in his sleep and then write a blog about it that makes it sound like child’s play. Honestly, Striver isn’t just a role model; he’s the role model, and if you’re not trying to be even a fraction of what this guy is, are you even living right? Legend. Absolute legend.

[@siddharthbanga7301]

aaram se bhai aaram se

[@sudo_ayush]

@@siddharthbanga7301 lagta hai bhai naya hai 😅

[@subhajitdey135]

One question : How to think of the range intuition u wrote ? I tried to solve by taking the maximal value of arr[i], as the question asks minimum number of jumps, so I thought that the arr[i] values should be maximum to get the minimum jumps. Btw thanks Striver for uploading Greedy playlist !!

[@iamnoob7593]

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

[@kavyahegde3586]

Sir please do playlist in strings Really it is needed 🙏

[@Flash-qr5oh]

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

[@AlokTripathi]

it's wrong computation.

[@sksadiruddin4191]

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 < nums.size() - 1; ++i) { right = max(right, nums.at(i) + i); if (left == i) { jumps++; left = right; } } return jumps; }

[@cswaale4498]

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

[@AyushKumar-hi5uy]

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❤❤

[@aniketbansod8136]

Bhaiyya, please start heap series after this one

[@jotsinghbindra8317]

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)

[@deepanshuchaudhary9299]

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

[@monicaaahaha]

basically you should give disclaimer that please watch DP series before this - (or create a separate playlist which include all the questions which needs to covered after covering all concepts) - idk what m saying

[@lakshyadalal8296]

The last approach is very easy to understand and also has linear TC and SC = O(1) Then why do we even need a recursive sonl which is so diff to understand 😂

[@MAX_RITIK]

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

[@aeshwer]

awesome content... please make string playlist soon

[@ashian09]

Why here we need to take minimum as we can only return jump at the if we reached to the index which is => n-1 by using void function because at same level jump value would be the same why to take minimum of all...anyone?

[@hajeeramohamad7641]

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

[@shwetanshu13]

Thank u so much for this playlist

[@bruvhellnah]

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

[@iamnoob7593]

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

[@Soum1947]

for ppl who are watching first time, there is a error in the title.. dynamic programming soln, not greedy soln

[@apmotivationakashparmar722]

Thank you so much bhaiya

[@UECAshutoshKumar]

Thank you

[@Rahul-jy9wg]

word of advice, interviewer should not expect this solution from you because this is not at all intuitive unless you have solved this question in the past, this does not mean you should not give this solution in the interview, but most probably interview might already know that you have already solved this question.

[@satyadev_k]

int minimumJumps(vector& arr, int n) { int l = 0; int r = 0; int farthest = 0; int cnt = 0; while (r < n - 1) { for (int i = l; i

[@MayankPareek]

Bhaiya pattern wise recursion prr bhi daal do

[@devanshshroff1820]

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

[@brainmosquito7]

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

[@aakashjha8393]

Why is R always L+1?

[@sagarsm23]

Why code studio is gone

[@the_random_youtuber12]

Is Recursion playlist completed?

[@prateek4279]

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

[@ShubhamBhatt-j7u]

Writing the code for this was very painful 😭😭

[@chiragsharma8905]

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

[@googleit2490]

Understood :) Too much to think... Jan'2, 2024 06:09 pm

[@shresthjain7557]

Wow ! what a solution

[@prerakunhale50]

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

[@KKKK-pl8yf]

Thanks Great Content!

[@thoughtsofkrishna8963]

Waiting for strings playlist

[@bbkivine3790]

ye dynamic programming hogya na

[@ShivamDangwal-n1o]

Understood 💯

[@deepanshu7295]

Bhaiya please start sde sheet challange 2024

[@sandeepxt99]

you are best❤

[@vkixgaming]

Simple Without DP , without recursion : solution class Solution { public: int minJumps(vector& arr) { int n = arr.size(); if (n

[@samitkumar18]

String please

[@Ybash2979]

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

[@Rahul_Mongia]

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

[@tanujaSangwan]

This is also a graph question. If you carefully see this problem of finding the minimum number of jumps in an array can be represented as a directed, unit-weight graph, BFS (Breadth-First Search) is an appropriate and efficient method to find the shortest path.

[@abhinay.k]

thank you sir

[@naviyas2305]

am not able to understand

[@StudyYuv]

got it bro!!!

[@atharv1002]

need strings lessonsssss

[@sujalsinha240]

Please bring heaps bro

[@DeadPoolx1712]

UNDERSTOOD;

[@NitinSharma-bk7dw]

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

[@parvahuja7618]

thankyou sir

[@subee128]

Thanks

[@ayushgaurabh8604]

awesome

[@SibiRanganathL]

Understood :)))

[@ManishKumar-dk8hl]

yehi too chahiye tha 😭

[@hautran7559]

love it

[@KartikeyTT]

ty sir

[@ANONYMOUS-xj1kd]

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; }

[@aarishfaiz7880]

Anyone in Dec.

[@Professor-du2pf]

Understood

[@worldfromhome4033]

Best!

[@AbhishekKumar-td5zu]

❤❤❤❤❤

[@surbhigupta5777]

US

[@084abhigna_y8]

Data Structures & Algorithm ❌ Data STRIVERS & Algorithm ✅

[@prateek4279]

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

[@firebout7675]

us

[@aggarwalsachin4854]

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

[@aryankumar3018]

Understood