BS-26. Find Peak Element-II | Binary Search

[@iammysterious-ud9cf]

Striver..... PLEASE CONTINUE A TO Z DSA Course....From Step 5(strings) , it is incomplete(no videos for the questions)... Your videos are very useful in intuition building and finding approach... PLEASE DONT LEAVE US IN THE MIDDLE😢😢😢...I hope you respond to the requests for the guys like me..

[@bhubeshsr6281]

solve by yourself - thats his intuition

[@sachinchoudhary4142]

Yes please🙏🙏

[@Dipanshutripathi2407]

Yes Please

[@krishnayadav5601]

yes please sir

[@sachitkumbhat2004]

We love you Striverr!!!

[@kathakalisaha9735]

Those who watches your videos completely will never say you use cpp , they are the ones who dont watch the intuition part, directly jump into the code

[@saibingi6980]

I request you to please make videos in Bit Manipulation, Heaps, Strings, and Disjoint-set in the upcoming sessions.

[@godson200]

He already has about 10 videos on disjoint set in the graph series. Maybe that can help.

[@rekhabisht6709]

Phle itna khatm krle

[@dumpster-jackson]

@@rekhabisht6709 Ab??

[@LemesaElias]

Even if you didn't provide the pseudo code and just went to implement the C++ code, I don't think it matters because it's actually your explanation of the solutions that actually helps anyone that considers him/herself a problem solver or anyone that's learning DSA. And as someone who uses/used multiple languages, I think given the explanation , any code from any programming language is pretty much understandable for someone that knows at least one programming language.

[@divyanshusingh6008]

Yup , thats why i dont even watch his pseuodo code , because i also consider this as a cheating

[@ritikadhangar2979]

Sir I can't express my job while taking your lecture. before that I am scared of DSA but now I enjoy due to you thankyou so muchhh

[@sundaramkumarsingh8448]

clear cut explanation. Thank you Striver

[@zerobear-xf7qh]

nice solution i dont think i would come up with this solution on my own

[@cenacr007]

Array and Strings are the most important for interviews, please make a playlist on strings.

[@ayushagarwal5271]

This man is a genius!

[@tanvirkekan6111]

This is an interesting soln: x,y = 0, len(mat[0]) - 1 while x < len(mat) or y >= 0: if x + 1 < len(mat) and mat[x + 1][y] > mat[x][y]: x += 1 elif y - 1 >= 0 and mat[x][y - 1] > mat[x][y]: y -= 1 else: break return [x, y]

[@TarunGautam-r7g]

good

[@abdulrehmanamer4252]

[ [13,10], [12,11] ] Your code fails on this testcase

[@shwetachoudhary9003]

Sir plz complete this series🥺🥺 i have learned a lot from this series so plz don't leave me in middle🙏🏻🥺

[@utkarshtrivedi9866]

Understood, even just looking at pseudo code, I was able to code it myself.

[@LearnwithEase20]

please make videos on string and other topic it is much needed no one explains like you!🙂

[@stith_pragya]

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

[@JothiprakashThangaraj]

sudo code is sufficient for any developer to write code in any language , understood thanks a lot !!

[@iammysterious-ud9cf]

Please Start Linked Lists

[@securelyinsaycure]

Striver you are a godsend to so many of us! Thank you 🙏

[@anonymous-ms2en]

love you striver ,clear explaination

[@jeehub041]

What an approach mind blown 🙌🙌

[@harshit.53]

i don't understand why people complain about only using cpp...first of all he discusses the idea behind the code and then writes a simple pseudo code...secondly, even if he uses cpp, any good coder can easily convert that code into any other language, maybe in other languages the function name and syntax is different, so u just need to use the functions provided by ur language

[@JagadeeswarN-ur7of]

Great Explanation!! Thanks!!

[@GeetainSaar]

your skin glowing in whole video

[@adilkevin6220]

ARTICLE column is empty for this problem. Please attach it.

[@rahulgoel7652]

I am unsure if first method would have O(mn *4) TC. You are traversing the matrix once, i.e. O(mn). For each element, checking the neighbors is constant time, so it wont be 4 times the number of elements

[@tusharyadav5874]

We will say it as O(m*n*4). So thats why he said we can we can slightly improve this by just finding the largest element in matrix . Then will be not using that 4 .

[@rahulgoel7652]

@@tusharyadav5874 I am sorry but I didn't understand what you wrote.

[@AshishSingh-he2qo]

Understood 🎉🎉

[@shivamdubey3798]

Bhaiya please video continue to upload Kara Karo mera ye week topic haa🙏🙏

[@sidharthsharma341]

What are the upcoming topics?after this

[@swagcoder]

For this problem, in leetcode , it’ll fail one test case where there are multiple peaks in the same row, that one is different scenario though, in this problem it is assumed that there will be only one peak per row , because you can’t eliminate directly in case of multiple peaks.

[@takeUforward]

this works for it also, because even if it has multiple peaks, it will find ,watch the first video to understand why it works

[@swagcoder]

Thanks for replying Stiver, Understood

[@vikassinghbisht7305]

striver you are just too good

[@t3ch_r4id]

class Solution { public: int maxElement(vector& arr,int n,int m, int col){ int index = -1; int maxele = INT_MIN; for (int i = 0; i < m; i++){ if(arr[i][col] > maxele){ maxele = arr[i][col]; index = i; } } return index; } vector findPeakGrid(vector& mat){ //n no of cloumns are there int n = mat.size();//size of each row //m no of rows are there int m = mat[0].size();//size of each column int low = 0, high = n - 1; while(low = 0)? mat[row][mid - 1] : -1; int right = (mid+1 < n)? mat[row][mid + 1] : -1; if(left < mat[row][mid] && mat[row][mid] > right){ return {row, mid}; } else if(left > mat[row][mid]){//cut out right part high = mid - 1; } else{//cut out the left part low = mid + 1; } } return {-1, -1}; } }; can anyone find the mistake please🙏 it's showing runtime error (something overflow) for a very large input matrix

[@roshankoshy6141]

high = n - 1 it is high = m- 1

[@AYUSHNIGAM97]

Can't we find the vertical peak with Binary search as well and reduce complexity to O(logn * logm)?

[@ashish7604]

don't you think if we keep udpating ele with bigger ele found then it will take O(m+n) time better than current time comp

[@anusmitahait2329]

Problem seemed so intimidating at the begining, later on was simple variation of peak.

[@lokeshnegi5051]

even if you are writing the code in cpp what's wrong in that it's so easy nowadays to convert it to any language of your choice with so many AI tools, the main part is the intuition

[@TheAI-Tutor]

But instead of traversing vertically, can't we just Traverse horizontally and find the maximum element to do STL and then check for the upper and lower element? and discard the upper half/lower half then?

[@NazeerBashaShaik]

Understood, thank you.

[@Fe-ironman]

finally didn't understood one of your videos lol....understood all of the rest in series one remaining tho

[@chandandasari6351]

bro plzz make a videos on string problems it was missed in play list bro strivers a to Z sheet plzz upload videos

[@MJBZG]

didn't really understand but will try more

[@vcfirefox]

superb

[@gugulothsrilakshmi8729]

written article not there in description

[@VishalKumar-uk5uc]

Ek month wait Kiya ek video ke liye

[@learn_in_shorts]

Bhaiya mey aapka binary tree ka playlist se tree padh rha hu but jab aap iss video mey jaise padhete hai to sab samjh aata hai but tree wala mey aisa kuch samjh nahi aa rha hai jab jab aap board pr padhye jab tak to sab samjh aaya but uske baad kuch jyada nahi aaya 😊 plz bhaiya aap iss video ki format mey videos banaya kariye plz and do something trees also

[@DeadPoolx1712]

UNDERSTOOD;

[@RitikaMajumdar__F]

bhaiya the java cpp notes's link is not given in the description. Please do share it.

[@ddevarapaga5134]

Understood brother

[@SurajYadav-bt9jb]

I have one doubt regarding BS that it can apply on sorted ones, but here 2D Matrix is not sorted so how can we apply BS on it , please resolve my confusion

[@ranjeetkumaryadav2967]

bhaiya please upload article of this problem

[@GeetainSaar]

give its article also

[@AbdulRehman-ee8zc]

sir khan thay aap aap hi k lecture ka intizar tha

[@VarshaSingh-hi2sb]

Can there be a case where the part which have omitted peak element was present there and not where we are heading towards?

[@brp3522]

11:35 This is the part that answers your question

[@guttulavybhav1030]

striver please upload videos on strings it is difficult to go to others videos....

[@sanketatmaram]

understood!

[@prikshit_sharma4071]

Bhaiya linkedlist start krdo iske badh

[@Learnprogramming-q7f]

Thank you Bhaiya

[@mahimagautam1810]

class Solution { public int findMaxIndex(int[][] mat,int mid,int n) { int maxi = -1; int ind = -1; for (int i = 0; i < n; i++) { if (mat[i][mid] > maxi) { maxi = mat[i][mid]; ind = i; } } return ind; } public int[] findPeakGrid(int[][] mat) { int ans[] = {-1,-1}; int n = mat.length; int m = mat[0].length; int low =0; int high = m-1; while(low=0 ? mat[index][mid-1]:-1; int right = mid+1left && mat[index][mid]>right){ return new int[]{index, mid}; } else if (mat[index][mid]

[@Harshitbejsbsksjnw]

Just a little doubt, when you say we can find the maximum element in the matrix and it will surely be the peak as well. What about the fact that there can be two maximums in the matrix and both lie adjacent to each other? And the question says that the element should be strictly greater than its adjacents......

[@ShanmugaSundaramS-o6i]

same elements wont be adjacent to each other in question

[@hashcodez757]

Understood!!

[@vishious14]

Understoood !!!!!

[@saketjaiswalSJ]

linked list striver bhaiya

[@DeepakPatel-d5v]

Thanks a lot Bhaiya

[@amanpatel6203]

long awaited

[@oyeesharme]

thanks bhaiya

[@daniyalhussain5231]

Solution in C++: class Solution { public: int findMaxIndex(vector&mat,int col,int rows) { int maxi=INT_MIN; int maxIndex; for(int i=0;imaxi) { maxi=mat[i][col]; maxIndex=i; } } return maxIndex; } vector findPeakGrid(vector& mat) { vectorans(2); int rows=mat.size(); int cols=mat[0].size(); int low=0,high=cols-1; while(low=0)left=mat[maxElementRowIndex][mid-1];//edge case int right=-1; if(mid+1left&&curr>right) { ans[0]=maxElementRowIndex; ans[1]=mid; return ans; } else if(curr

[@viveksoni3269]

Very Nice!!!!

[@nayankhuman1043]

Understood :)

[@soumiyamuthuraj3516]

awesome

[@anand_yv]

Understood

[@mithileshkumbhar7968]

Striver......The link to this problem on your page is not opening to view the code and intuition.

[@lakshsinghania]

bhaiya i did convert from 2d array to 1d array but only 45/55 test cases is passing pls help class Solution { public: vector findPeakGrid(vector& matrix) { // binary search on answer int m = matrix.size(); int n = matrix[0].size(); // tc = o(log(m*n)) int start = 0; // created an imaginery 1D array of size from 0 to n*m-1 int end = (n * m) - 1; while (start 0) ? matrix[i - 1][j] : -1; int bottom = (i < m - 1) ? matrix[i + 1][j] : -1; int left = (j > 0) ? matrix[i][j - 1] : -1; int right = (j < n - 1) ? matrix[i][j + 1] : -1; if (mid > start && mid < n * m - 1) { if (matrix[i][j] > top && matrix[i][j] > bottom && matrix[i][j] > left && matrix[i][j] > right) { return {i, j}; // Peak element found } // If left or top neighbor is greater, move to the left or top // half if (left > matrix[i][j] || top > matrix[i][j]) { end = mid - 1; } else { // Else move to the right or bottom half start = mid + 1; } } else if (mid == start) { // Only move right if there's no left neighbor if (matrix[i][j] > right && matrix[i][j] > bottom) { return {i, j}; } else { start = mid + 1; } } else if(mid == end){ // Only move left if there's no right neighbor if (matrix[i][j] > left && matrix[i][j] > top) { return {i, j}; } else { end = mid - 1; } } } return {-1, -1}; } };

[@vijeshsshetty]

linked list series plss

[@surbhirathore._]

Done!!!

[@sanjayp.m7008]

why cant i use the logic of matching the row and col to mid and considering this matrix as a 2d array. i tried this but only 46 testcases passed in the leetcode qn. class Solution { public: vector findPeakGrid(vector& mat) { int n = mat.size(); int m = mat[0].size(); int low = 0; int high = n * m - 1; while (low = 0) ? mat[row][col - 1] : INT_MIN; int bottom = (row + 1 < n) ? mat[row + 1][col] : INT_MIN; int top = (row - 1 >= 0) ? mat[row - 1][col] : INT_MIN; if (el > right && el > left && el > top && el > bottom) { return {row, col}; } if (right > el) { low = mid + 1; } else if (left > el) { high = mid - 1; } else if (bottom > el) { low = mid + 1; } else { high = mid - 1; } } return {-1, -1}; } };

[@binarybardic4667]

Thanks a lot......!

[@viratlover6206]

understood

[@nishantsatere9350]

understoddddddddd

[@mohitsinghnitjsr2652]

how many videos more to come

[@codeman3828]

Undertsood

[@rahulhembram4519]

UnderStood

[@shivamkumar-hz7jt]

strings ke video upload krdo pls

[@AkashBisht-cq3ys]

1st july mon 9.13

[@PriyashiChoudhary-l5k]

Jeetu bhaiya equivalent

[@shaiksoofi3741]

tq

[@charuprabha9696]

🙇‍♀🙇‍♀🙇‍♀🙇‍♀🙇‍♀🙇‍♀

[@infinity2creation551]

You are calling col ko row ...

[@PrinceKumar-ef1xf]

vector findPeakGrid(vector& mat) { int startCol = 0, endCol = mat[0].size()-1; while(startCol = startCol+1 && mat[maxRow][midCol-1] > mat[maxRow][midCol]; bool rightIsBig = midCol mat[maxRow][midCol]; if(!leftIsBig && !rightIsBig) return vector{ maxRow, midCol}; else if(rightIsBig) startCol = midCol+1; else endCol = midCol-1; } return vector{-1,-1}; }

[@dhakadkumman4239]

100th comment of video

[@crazybro4383]

8:00

[@dayashankarlakhotia4943]

public class solution { public static int []find Peek Grid(int [][]grid){ int n=grid. length,m=grid [0].length; int l=0,r=m-1; while (lgrid [idx][mid +1]){ r=mid; }else { l=mid-1; } return new int []{l,find max(grid [l])}; } public static int find max(int []col){ int index =0,n=col length; for(int i=0;icol[index]) index =i; } return index; } }; 🎉❤

[@Lucifer0872]

Biltu kaka Zindabad

[@AvaneeshSrivastava-lm5vj]

understood. thank you. you should sleep a bit more.

[@RAJPATEL-ir7ly]

O(n+m) BFS class Solution { public: vector findPeakGrid(vector& mat) { int i = 0 ; int j = 0 ; int n = mat.size() ; int m = mat[0].size() ; while(true){ int flag = 0 ; vector coordinates = {{1,0} , {0,1} , {-1,0} , {0,-1}} ; int maxi= -1 ; int newRow =0 ; int newCol = 0 ; for(auto coordinate : coordinates){ int x = i+coordinate.first ; int y = j+coordinate.second ; if(x>=0 && y>=0 && x

[@cenacr007]

us

[@adityauditsingh5986]

first view xD

[@ishangujarathi10]

why am i getting a runtime error when i submit the solution on leetcode? @striver??

[@ajithc5505]

int left = mid-1>=0 ? mat[index][mid-1]:-1; int right = mid+1

[@tejaswaniguttula5961]

@@ajithc5505 Hi vector findPeakGrid(vector &g){ // Write your code here. int rows = g.size(); int cols = g[0].size(); int low = 0, high = cols-1; while(low max_ele){ max_ele = g[i][mid]; index = i; } } int left = mid-1 >= 0 ? g[index][mid-1] : -1; int right = mid +1 < cols ? g[index][mid+1] : -1; if(max_ele > left && max_ele > right){ return {index, mid}; } else if(max_ele < g[index][mid-1]){ high = mid-1; } else{ low = mid+1; } } return {-1, -1}; } }; Though I applied edge cases why I am getting runtime error for this testcase?? Test case: [[10,50,40,30,20],[1,500,2,3,4]]

[@ParshvaDani]

@@ajithc5505 thanks brother it works now

[@NazeerBashaShaik]

Understood, thank you.

[@music-loverFam]

Understood😊

[@_sf_editz1870]

Understood

[@culeforever5408]

understood

[@abhishekprasad010]

Understood