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..
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.
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]
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
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 .
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.
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]
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
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 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
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
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
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......
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
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}; } };
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; } }; 🎉❤
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
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}; } };
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
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!!!
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??
nice solution i dont think i would come up with this solution on my own
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.
Yup , thats why i dont even watch his pseuodo code , because i also consider this as a cheating
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
This man is a genius!
clear cut explanation. Thank you Striver
Array and Strings are the most important for interviews, please make a playlist on strings.
Understood, even just looking at pseudo code, I was able to code it myself.
Please Start Linked Lists
Thank You Much for this wonderful video..........🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻
sudo code is sufficient for any developer to write code in any language , understood thanks a lot !!
JUST WOW! SO GREAT INTUITION BUILDING, MUCH LOVE TO YOUR WORK SIR :)
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 🙏
love you striver ,clear explaination
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
[
[13,10],
[12,11]
]
Your code fails on this testcase
your skin glowing in whole video
What an approach mind blown 🙌🙌
Great Explanation!! Thanks!!
Sir plz complete this series🥺🥺 i have learned a lot from this series so plz don't leave me in middle🙏🏻🥺
Understood 🎉🎉
striver you are just too good
Bhaiya please video continue to upload Kara Karo mera ye week topic haa🙏🙏
What are the upcoming topics?after this
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
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.
Understood 😀
Understood, thank you.
Ek month wait Kiya ek video ke liye
Problem seemed so intimidating at the begining, later on was simple variation of peak.
bro plzz make a videos on string problems it was missed in play list bro strivers a to Z sheet plzz upload videos
didn't really understand but will try more
superb
UNDERSTOOD;
Understood brother
Thank you Bhaiya
ARTICLE column is empty for this problem. Please attach it.
Bhaiya linkedlist start krdo iske badh
give its article also
Thanks a lot Bhaiya
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
finally didn't understood one of your videos lol....understood all of the rest in series one remaining tho
understoddddddddd
linked list striver bhaiya
thanks bhaiya
long awaited
sir khan thay aap
aap hi k lecture ka intizar tha
Understood!!
understood!
Understoood !!!!!
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]
Very Nice!!!!
striver please upload videos on strings it is difficult to go to others videos....
bhaiya please upload article of this problem
awesome
Understood
Understood :)
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
Undertsood
understood
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
written article not there in description
Thanks a lot......!
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 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
linked list series plss
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
Can't we find the vertical peak with Binary search as well and reduce complexity to O(logn * logm)?
8:00
Done!!!
UnderStood
tq
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
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
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
1st july mon 9.13
bhaiya the java cpp notes's link is not given in the description. Please do share it.
strings ke video upload krdo pls
is it a sorted matrix?
nope
🙇♀🙇♀🙇♀🙇♀🙇♀🙇♀
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
how many videos more to come
Striver......The link to this problem on your page is not opening to view the code and intuition.
Jeetu bhaiya equivalent
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};
}
You are calling col ko row ...
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};
}
};
100th comment of video
us
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;
}
};
🎉❤
Biltu kaka Zindabad
first view xD
understood. thank you. you should sleep a bit more.
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
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};
}
};
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😊
Understood, thank you.