Don't be ashamed of N^4 (well, in some cases)

Polynomial solutions are of course exponentially better than non-polynomial solutions (pun-intended). Comparing an N^2 with a 2^N is a no brainer, same for LogN with a SQRT(N) (the former is way better than the latter). The value of N clearly matters, though. The problem below is solved with an N^4 solution, where N=100. Now it is pushing the limits of LC, which won't ever accept a 10^9 solution (billion), but in some cases 100M is tractable. That's the case here. Best way to solve it IMO is to have a function to check whether the boundaries of the sub-matrix form a valid square. That one is tricky, requires some careful calculations. It runs in O(N). Once you have it, then you still need an N^3 nested-loops to check each sub-matrix. If N is relatively small, don't be ashamed of N^4. There must be a way to do it faster, but this is the best that your friend here could come up with. Cheers, ACC.


1139. Largest 1-Bordered Square
Medium

Given a 2D grid of 0s and 1s, return the number of elements in the largest square subgrid that has all 1s on its border, or 0 if such a subgrid doesn't exist in the grid.

Example 1:

Input: grid = [[1,1,1],[1,0,1],[1,1,1]]
Output: 9

Example 2:

Input: grid = [[1,1,0,0]]
Output: 1

Constraints:

  • 1 <= grid.length <= 100
  • 1 <= grid[0].length <= 100
  • grid[i][j] is 0 or 1
Accepted
24,869
Submissions
49,582



public int Largest1BorderedSquare(int[][] grid)
{
 int retVal = 0;
 for (int r = 0; r < grid.Length; r++)
 {
 for (int c = 0; c < grid[r].Length; c++)
 {
 int cDelta = 0;
 int len = 0;
 while (c + cDelta < grid[r].Length && grid[r][c + cDelta] == 1)
 {
 len++;
 if (IsSquare(grid, r, c + cDelta, len))
 retVal = Math.Max(retVal, len * len);
 cDelta++;
 }
 }
 }
 return retVal;
}
private bool IsSquare(int[][] grid, int initRow, int endCol, int len)
{
 //Down
 bool isSquare = true;
 for (int r = 0; r < len; r++)
 {
 if (initRow + r >= grid.Length || 
 grid[initRow + r][endCol] != 1)
 {
 isSquare = false;
 break;
 }
 }
 if (!isSquare) return false;
 //Left
 for (int c = 0; c < len; c++)
 {
 if (initRow + len - 1 >= grid.Length ||
 endCol - c < 0 ||
 grid[initRow + len - 1][endCol - c] != 1)
 {
 isSquare = false;
 break;
 }
 }
 if (!isSquare) return false;
 //Up
 for (int r = 0; r < len; r++)
 {
 if (initRow + r >= grid.Length ||
 endCol - len + 1 < 0 ||
 grid[initRow + r][endCol - len + 1] != 1)
 {
 isSquare = false;
 break;
 }
 }
 return isSquare;
}

Comments

Post a Comment

[フレーム]

Popular posts from this blog

Quasi FSM (Finite State Machine) problem + Vibe

Not really an FSM problem since the state isn't changing, it is just defined by the current input. Simply following the instructions should do it. Using VSCode IDE you can also engage the help of Cline or Copilot for a combo of coding and vibe coding, see below screenshot. Cheers, ACC. Process String with Special Operations I - LeetCode You are given a string  s  consisting of lowercase English letters and the special characters:  * ,  # , and  % . Build a new string  result  by processing  s  according to the following rules from left to right: If the letter is a  lowercase  English letter append it to  result . A  '*'   removes  the last character from  result , if it exists. A  '#'   duplicates  the current  result  and  appends  it to itself. A  '%'   reverses  the current  result . Return the final string  result  after processing all char...

Shortest Bridge – A BFS Story (with a Twist)

Here's another one from the Google 30 Days challenge on LeetCode — 934. Shortest Bridge . The goal? Given a 2D binary grid where two islands (groups of 1s) are separated by water (0s), flip the fewest number of 0s to 1s to connect them. Easy to describe. Sneaky to implement well. 🧭 My Approach My solution follows a two-phase Breadth-First Search (BFS) strategy: Find and mark one island : I start by scanning the grid until I find the first 1 , then use BFS to mark all connected land cells as 2 . I store their positions for later use. Bridge-building BFS : For each cell in the marked island, I run a BFS looking for the second island. Each BFS stops as soon as it hits a cell with value 1 . The minimum distance across all these searches gives the shortest bridge. πŸ” Code Snippet Here's the core logic simplified: public int ShortestBridge(int[][] grid) { // 1. Mark one island as '2' and gather its coordinates List<int> island = FindAndMark...

Classic Dynamic Programming IX

A bit of vibe code together with OpenAI O3. I asked O3 to just generate the sieve due to laziness. Sieve is used to calculate the first M primes (when I was using Miller-Rabin, was giving me TLE). The DP follows from that in a straightforward way: calculate the numbers from i..n-1, then n follows by calculating the min over all M primes. Notice that I made use of Goldbach's Conjecture as a way to optimize the code too. Goldbach's Conjecture estates that any even number greater than 2 is the sum of 2 primes. The conjecture is applied in the highlighted line. Cheers, ACC. PS: the prompt for the sieve was the following, again using Open AI O3 Advanced Reasoning: " give me a sieve to find the first M prime numbers in C#. The code should produce a List<int> with the first M primes " Minimum Number of Primes to Sum to Target - LeetCode You are given two integers  n  and  m . You have to select a multiset of  prime numbers  from the  first   m  pri...