BS-26. Find Peak Element-II | Binary Search
HTML-код
- Опубликовано: 12 сен 2024
- Problem Link: bit.ly/3Ckb4Rb
Please watch part-1: • BS-9. Find Peak Element
Notes/C++/Java/Python codes:
We have solved the problem, and we have gone from brute force and ended with the most optimal solution. Every approach's code has been written in the video itself. Also, we have covered the algorithm with intuition.
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0:00 Introduction of Course
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..
solve by yourself - thats his intuition
Yes please🙏🙏
Yes Please
yes please sir
We love you Striverr!!!
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
I request you to please make videos in Bit Manipulation, Heaps, Strings, and Disjoint-set in the upcoming sessions.
He already has about 10 videos on disjoint set in the graph series. Maybe that can help.
Phle itna khatm krle
@@rekhabisht6709 Ab??
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.
Array and Strings are the most important for interviews, please make a playlist on strings.
nice solution i dont think i would come up with this solution on my own
clear cut explanation. Thank you Striver
This man is a genius!
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]
good
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
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 .
@@tusharyadav5874 I am sorry but I didn't understand what you wrote.
Please Start Linked Lists
sudo code is sufficient for any developer to write code in any language , understood thanks a lot !!
please make videos on string and other topic it is much needed no one explains like you!🙂
Striver you are a godsend to so many of us! Thank you 🙏
What are the upcoming topics?after this
Understood 🎉🎉
Thank You Much for this wonderful video..........🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻
ARTICLE column is empty for this problem. Please attach it.
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
high = n - 1 it is high = m- 1
What an approach mind blown 🙌🙌
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
Understood, even just looking at pseudo code, I was able to code it myself.
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]
love you striver ,clear explaination
your skin glowing in whole video
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?
Bhaiya please video continue to upload Kara Karo mera ye week topic haa🙏🙏
written article not there in description
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.
this works for it also, because even if it has multiple peaks, it will find ,watch the first video to understand why it works
Thanks for replying Stiver, Understood
Understood😊
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......
same elements wont be adjacent to each other in question
UNDERSTOOD;
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
bro plzz make a videos on string problems it was missed in play list bro strivers a to Z sheet plzz upload videos
Can there be a case where the part which have omitted peak element was present there and not where we are heading towards?
11:35 This is the part that answers your question
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
striver you are just too good
Understood, thank you.
bhaiya the java cpp notes's link is not given in the description. Please do share it.
Striver......The link to this problem on your page is not opening to view the code and intuition.
understood!
Bhaiya linkedlist start krdo iske badh
Understood :)
bhaiya please upload article of this problem
understood
long awaited
Problem seemed so intimidating at the begining, later on was simple variation of peak.
linked list striver bhaiya
striver please upload videos on strings it is difficult to go to others videos....
thanks bhaiya
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};
}
awesome
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};
}
};
didn't really understand but will try more
Very Nice!!!!
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
Understood!!
Understoood !!!!!
Thanks a lot Bhaiya
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};
}
};
Understood
Thank you Bhaiya
Ek month wait Kiya ek video ke liye
give its article also
Thanks a lot......!
sir khan thay aap
aap hi k lecture ka intizar tha
Done!!!
linked list series plss
tq
1st july mon 9.13
understoddddddddd
Undertsood
🙇♀🙇♀🙇♀🙇♀🙇♀🙇♀
You are calling col ko row ...
8:00
how many videos more to come
UnderStood
strings ke video upload krdo pls
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;
}
};
🎉❤
Jeetu bhaiya equivalent
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
100th comment of video
us
first view xD
Biltu kaka Zindabad
understood. thank you. you should sleep a bit more.
why am i getting a runtime error when i submit the solution on leetcode? @striver??
int left = mid-1>=0 ? mat[index][mid-1]:-1;
int right = mid+1
@@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]]
@@ajithc5505 thanks brother it works now
Understood, thank you.
Understood brother
understood
Understood
understood
Understood
Understood
understood
understood
understood