Congrats josh !!!
a cartoon character with a smile on his face is standing on a white background with a star tamil chat logo .
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media.tenor.com
September 24, 2025 at 3:39 PM
Congrats josh !!!
Can't wait for this course josh!
a cartoon of spongebob saying ready in front of spongebob squarepants
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media.tenor.com
July 16, 2025 at 7:39 AM
Can't wait for this course josh!
Same here
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ALT: a man in a suit and tie is standing in front of a door with blinds .
media.tenor.com
May 22, 2025 at 8:46 AM
Same here
Solution with optimal solution:
- Search vertically for rows that have the potential to hold the target value using binary search
- check if the target is not in the matrix vertically then return false
- Look for the target in the line that has been found
time complexity:
O(log(m*n))
- Search vertically for rows that have the potential to hold the target value using binary search
- check if the target is not in the matrix vertically then return false
- Look for the target in the line that has been found
time complexity:
O(log(m*n))
May 1, 2025 at 5:55 AM
Solution with optimal solution:
- Search vertically for rows that have the potential to hold the target value using binary search
- check if the target is not in the matrix vertically then return false
- Look for the target in the line that has been found
time complexity:
O(log(m*n))
- Search vertically for rows that have the potential to hold the target value using binary search
- check if the target is not in the matrix vertically then return false
- Look for the target in the line that has been found
time complexity:
O(log(m*n))
Solution with brute force:
We can solve this problem by checking the values one by one in the 2-dimensional matrix
time complexity:
O(n^2)
We can solve this problem by checking the values one by one in the 2-dimensional matrix
time complexity:
O(n^2)
May 1, 2025 at 5:54 AM
Solution with brute force:
We can solve this problem by checking the values one by one in the 2-dimensional matrix
time complexity:
O(n^2)
We can solve this problem by checking the values one by one in the 2-dimensional matrix
time complexity:
O(n^2)
Leetcode #74 - Search a 2D Matrix
Problem:
Search num target in matrix 2 dimension, return true if target is in matrix or false otherwise.
Write a solution with time complexity:
O(log(m * n))
link: leetcode.com/problems/sea...
Problem:
Search num target in matrix 2 dimension, return true if target is in matrix or false otherwise.
Write a solution with time complexity:
O(log(m * n))
link: leetcode.com/problems/sea...
Search a 2D Matrix - LeetCode
Can you solve this real interview question? Search a 2D Matrix - You are given an m x n integer matrix matrix with the following two properties:
* Each row is sorted in non-decreasing order.
* The ...
leetcode.com
May 1, 2025 at 5:52 AM
Leetcode #74 - Search a 2D Matrix
Problem:
Search num target in matrix 2 dimension, return true if target is in matrix or false otherwise.
Write a solution with time complexity:
O(log(m * n))
link: leetcode.com/problems/sea...
Problem:
Search num target in matrix 2 dimension, return true if target is in matrix or false otherwise.
Write a solution with time complexity:
O(log(m * n))
link: leetcode.com/problems/sea...
Is there any way, we can check if its refurbished or not?
beside chec the first number serial, cycle counts, or check directly apple cooverage?
beside chec the first number serial, cycle counts, or check directly apple cooverage?
March 26, 2025 at 8:29 AM
Is there any way, we can check if its refurbished or not?
beside chec the first number serial, cycle counts, or check directly apple cooverage?
beside chec the first number serial, cycle counts, or check directly apple cooverage?
Decided to buy a MBP M3 Pro for my future patner
a man in a suit and tie is holding a wooden gavel and says " decided "
ALT: a man in a suit and tie is holding a wooden gavel and says " decided "
media.tenor.com
March 19, 2025 at 3:04 AM
Decided to buy a MBP M3 Pro for my future patner
Wow its really cool, we can do that only use css
March 14, 2025 at 7:01 PM
Wow its really cool, we can do that only use css