HW Solution: Row with max 1s ✅✅ int rowWithMax1s(vector arr, int n, int m) { // code here int row = 0, col = m-1, max_count = 0, index = -1; while(row=0){ if(arr[row][col]==1){ max_count = m - col; index = row; col--; }else{ row++; } } return index; }
class Solution { public: bool search(vector& matrix, int n, int m, int x) { int i = 0, j = m - 1; // Start from the top-right corner
while (i < n && j >= 0) { if (matrix[i][j] == x) // Found the element return true; else if (matrix[i][j] > x) // Move left in the current row j--; else // Move down to the next row i++; }
no bro just think that the binary search is running for only 1 time when the condition is true it just no running for all rows it just ran for that row which matches the contdition . so TC will be n time the row loop run and 1 time binary search for it logn
Nhi, O(N + logM) hi hoga. Q ki humlog har ek row pe binary search nhi laga rhe hai. Bs 1 row pe laga rhe hai. So first vo row dhundne ke liye N time lagega, and then vo row me element dhundne ke liye logM time lagega. Agar har ek row me search karte karte ata then O(N*logM) hota complexity.
@@joydeep-halder am confused about how to find time complexity... Will you explain what are the parameters we must put in our mind to find time complexity??? Plz
@@Samatha5 Try to observe the number of iterations. Some common patterns: for loop, it is linear. But if the loop is dividing by k each time, then it will be logn base k, Similarly for nested loops it get multiplied. But it is not necessary all the times. You have to dry run and check. Always consider the worst case (if not asked best case)
Yeah bhai.. salute to sir🥺😢😭..i left babar sir after watching his video..im now his fan...i have completed 33 lec of him now...😍i will try to complete 2 lec daily from now onwards
class Solution { public: bool searchMatrix(vector& matrix, int target) { int m = matrix.size(); // Number of rows if (m == 0) return false; // Empty matrix int n = matrix[0].size(); // Number of columns
int i = 0, j = n - 1; // Start from the top-right corner
while (i < m && j >= 0) { if (matrix[i][j] == target) // Found the element return true; else if (matrix[i][j] > target) // Move left in the current row j--; else // Move down to the next row i++; }
solution for last problem . and here we start from the bottom. like from the 70. solution is bool search(vector matrix, int n, int m, int x) { // code here int i=n-1,j=0; while(i>=0 && jx){ i--; }else{ j++; } } return 0; }
Last wali question dekh kar yahi pta chalta hai, ki chor kahi bhi chhupa ho, rohit bhaiya ager aap jaise police ho to last last tak dhund hi lenge😅😅😅😅😅
last hW question //optimize o(n+log(m)) /*bool decOrderSearch(vector &matrix, int target) { int n = matrix.size(), m = matrix[0].size(); for (int i = 0; i < n; i++) { if (matrix[i][0] >= target && target >= matrix[i][m - 1]) { int start = 0, end = m - 1; while (start
Develop a plan == Discipline == consistency🚀
=Success❤
Correct
Bhaiyya please continue this never stop you're best teacher of dsa ❤
bhaiya aapne jaisa bataya hai aise to koi paid course me nahi padata itna achhha lecture to koi bhifree me nahi padayaga
Bhaiyya your way of teaching makes difficult concepts so easy and clear to understand, thank you so much!
❤nice explanation bhaiya maja aa gya padh ke 🙌🙌..... you are the great teacher
bhaiya chamak gaya achche se
Solution of the second question was just astonishing! Really enjoyed the video 😍😍
Unique content bhai ❤❤❤❤, zabardast explanation ❤❤❤❤
Very nice video bhaiya, mast concepts seekhe aaj
HW Solution: Row with max 1s ✅✅
int rowWithMax1s(vector arr, int n, int m) {
// code here
int row = 0, col = m-1, max_count = 0, index = -1;
while(row=0){
if(arr[row][col]==1){
max_count = m - col;
index = row;
col--;
}else{
row++;
}
}
return index;
}
18:53 ye approach mere dimmag m sabse pahle aaya tha sir ❤
thank you bhaiya your teaching is absoloutely incredible
Thanks a lot bhaiya ❤ for your this type of wonderful teaching ❤
RAM RAM BHAI Sarayiya ne ye hmara revision chal rha h rohit bhai ki jai ho
Bhai both your programming video & motivation video both are amazing. May Allah pak provide each type of success in your life.
happy for the reason that I tried myself and solved the binary search at 1st attempt with 2ms runtime
It's getting fun now. Thankyou bhaiya❤
17:09 chamak gaya🎉
Thank you sir for this quality content🙃
Chamk gaya bhaiya ye bhi pura thanku bhaiya
Wow wow wow just wow ❤❤❤❤❤
Love You Bhiya ❤🙏🏻
Ek baari mai hi chamak gya bhaiya🥳
Last problem krke mja aa gya bhaiya 🔥
sir i like your consistency❤
confidence aa geya bhai
Tq bhaiya, aaj maga aa gya, kuch seeka😄
Bhaiya lecture duration itna hi rakhiye, January se bada dijiye ga abhi itna hi rakhiye please kyuki semester exam bhi pass aa gaya hai🙏🙏
class Solution {
public:
bool search(vector& matrix, int n, int m, int x) {
int i = 0, j = m - 1; // Start from the top-right corner
while (i < n && j >= 0) {
if (matrix[i][j] == x) // Found the element
return true;
else if (matrix[i][j] > x) // Move left in the current row
j--;
else // Move down to the next row
i++;
}
// If element is not found
return false;
}
};
Done Bhaiya... Day 46/180 ✅
bhaiya apka teaching skill imrpove ho rhi h day by day hamari coding ki tarah😂
time complexity of binary search in 2d array is nLogn instead of n + logn
true
no bro just think that the binary search is running for only 1 time when the condition is true it just no running for all rows it just ran for that row which matches the contdition . so TC will be n time the row loop run and 1 time binary search for it logn
@@devprajapati5673 Thanks bro
Maja agay bhaiya❤
#day46/180
God morning bhaiya
maja aagya bhaiya 🤩🤩
Bhaiya ,@ 16:30 , time complexity will be O(n*logn) hogi ,na ki O(n+logn ).
Thanks ❤ for free teaching.
Nhi, O(N + logM) hi hoga. Q ki humlog har ek row pe binary search nhi laga rhe hai. Bs 1 row pe laga rhe hai. So first vo row dhundne ke liye N time lagega, and then vo row me element dhundne ke liye logM time lagega. Agar har ek row me search karte karte ata then O(N*logM) hota complexity.
gaddari kar be gaddari kar be
@@joydeep-halder am confused about how to find time complexity... Will you explain what are the parameters we must put in our mind to find time complexity??? Plz
@@joydeep-halder bhai for loop k andar hi binary search h ,to vo multiply hi hoga O(nlogn)
@@Samatha5 Try to observe the number of iterations. Some common patterns: for loop, it is linear. But if the loop is dividing by k each time, then it will be logn base k, Similarly for nested loops it get multiplied. But it is not necessary all the times. You have to dry run and check. Always consider the worst case (if not asked best case)
Awesome ❤ Chamak Gaya.
Chamak Gaya bhaiya ji❤❤
Good morning Bhaiya 😊
Good morning Rohit bhaiya ❤🙏
Good Morning bhaiya ❤❤
bhaiya appa ne kaal se pura maja la diya hai❤❤❤........
Present bhaiya, good morning bhaiya..💖😇
amazing lecture sir
chamak gya bhaiya
hah ji bhaiya😀😀😀
Bhaiya, Recursion ki videos/series aane mein kitna time hain ??
Good Morning Bhaiya
chamkaaaaa sirr op❤🔥🔥
Thankyou so much bhaiya bxz of u this soo easy to understand🤩💞
last problem is awesome ✌✌
Good morning boss ❤
Chamka bhaiya chamka 👍🏻
maja aa gya bhaiya apse dsa pdke
Good morning bhaiya
kasam se bhaiyya ab maja aan laga qns karne me dimag ki baat si jal ja jab aap samjhao ki yo qns is tro bhi solve ho sake
love U bhaiya
Thanks Bhaiya...❤🎉
Day 46 completed
Chamak gya ache se
Good morning 🌅
Chamak gta sir ji ❤❤
#46 out of 180
amazing class @RohitNegi Bhaiya
chamak gya bhaiya
chamak gaya bhaiya
Love babbar bhi bhaiya fail hai aapke concept ke samne😂
U R brilliant teacher
Yeah bhai.. salute to sir🥺😢😭..i left babar sir after watching his video..im now his fan...i have completed 33 lec of him now...😍i will try to complete 2 lec daily from now onwards
Day 46 ✅
Hw
int m,n;
int i=n-1, j=0 , x;
While (i>=0 && j
std::cout
Day 46/180 done ✅✅✅✅✅ # hard
Thank you❤
Thank you bhaiya
chamak gya!
Day 46 completed
#180dayscodingchallenges
#coderarmy😊
Day 46/180 Done
ram ram sir and everyone
Bhaia aapna kaha tha ke aap Time and space complexity pe ek aur lecture upload kroga ?? @Rohit Negi
Good morning bhiya
Ab mza aa raha hai bhaiya Ji
Jai Shree Ram🚩
Mast karke chamak Gaya
Thank u sir😊
class Solution {
public:
bool searchMatrix(vector& matrix, int target) {
int m = matrix.size(); // Number of rows
if (m == 0) return false; // Empty matrix
int n = matrix[0].size(); // Number of columns
int i = 0, j = n - 1; // Start from the top-right corner
while (i < m && j >= 0) {
if (matrix[i][j] == target) // Found the element
return true;
else if (matrix[i][j] > target) // Move left in the current row
j--;
else // Move down to the next row
i++;
}
// If element is not found
return false;
}
};
Sir, I am Biggner. Lower mila hai. To mujhe ye series karni chahiye ki nahi (C++,DSA ka full course hai)
solution for last problem . and here we start from the bottom. like from the 70.
solution is bool search(vector matrix, int n, int m, int x)
{
// code here
int i=n-1,j=0;
while(i>=0 && jx){
i--;
}else{
j++;
}
}
return 0;
}
❤
Day 46❤
Day 46/180☀☀🕕🕕...Surf Excel,Tide,Nirma,Wheel sab fail h bhaiya....Gajab chamak raha 😇😇
Day 46/180 Completed ✅✅✅
16:30 tc=n*logm not addition, please confirm bhaiya
Day 46/180 👍👍
❤❤❤❤
Consistency++
Last wali question dekh kar yahi pta chalta hai, ki chor kahi bhi chhupa ho, rohit bhaiya ager aap jaise police ho to last last tak dhund hi lenge😅😅😅😅😅
int rowWithMax1s(vector arr, int n, int m) {
int index=-1,maxi=0;
for(int i=0;i
Day 46
HW Solution: Binary search using logn + logm ✅✅
int n = matrix.size(), m = matrix[0].size();
int start = 0, end = n-1, mid, row=-1;
while(start
chamak gaya
Last hw logic start = m-1 and end = 0
Bro 1 question ki time complexity O(N*log(M)) honi chaiye na
16:19 i think time complexity is O(n *log(m)) correct me if i am.wrong
last hW question
//optimize o(n+log(m))
/*bool decOrderSearch(vector &matrix, int target)
{
int n = matrix.size(), m = matrix[0].size();
for (int i = 0; i < n; i++)
{
if (matrix[i][0] >= target && target >= matrix[i][m - 1])
{
int start = 0, end = m - 1;
while (start