Working around Dynamic Programming

This problem is a classic DP problem, and the hints confirm it. Couldn't solve it using DP, and the brute-force DFS approach was leading to TLE. So... decided to try to work around it.
The heuristic used was simple: if we try way too many interactions (say 10x100x100), then we'll assume that there is no solution. It means that we have tried 10 times more cells that exists on the board. It worked. Sometimes achieving the goals requires creative and shameless approaches. Code is down below, cheers, ACC.


2510. Check if There is a Path With Equal Number of 0's And 1's
Medium

You are given a 0-indexed m x n binary matrix grid. You can move from a cell (row, col) to any of the cells (row + 1, col) or (row, col + 1).

Return true if there is a path from (0, 0) to (m - 1, n - 1) that visits an equal number of 0's and 1's. Otherwise return false.

Example 1:

Input: grid = [[0,1,0,0],[0,1,0,0],[1,0,1,0]]
Output: true
Explanation: The path colored in blue in the above diagram is a valid path because we have 3 cells with a value of 1 and 3 with a value of 0. Since there is a valid path, we return true.

Example 2:

Input: grid = [[1,1,0],[0,0,1],[1,0,0]]
Output: false
Explanation: There is no path in this grid with an equal number of 0's and 1's.

Constraints:

  • m == grid.length
  • n == grid[i].length
  • 2 <= m, n <= 100
  • grid[i][j] is either 0 or 1.
Accepted
206
Submissions
304

public bool IsThereAPath(int[][] grid)
{
 int interactions = 0;
 return IsThereAPath(grid, 0, 0, 0, 0, ref interactions, 10000000);
}
private bool IsThereAPath(int[][] grid,
 int row,
 int col,
 int numberOnes,
 int numberZeroes,
 ref int interactions,
 int MAX)
{
 if (interactions>= MAX) return false;
 if (row>= grid.Length || col>= grid[0].Length) return false;
 numberOnes = (grid[row][col] == 1) ? numberOnes + 1 : numberOnes;
 numberZeroes = (grid[row][col] == 0) ? numberZeroes + 1 : numberZeroes;
 if (row == grid.Length - 1 && col == grid[row].Length - 1) return numberOnes == numberZeroes;
 int rowsLeft = grid.Length - row - 1;
 int colsLeft = grid[0].Length - col - 1;
 if (Math.Abs(numberOnes - numberZeroes)> rowsLeft + colsLeft) return false;
 interactions++;
 if (IsThereAPath(grid, row, col + 1, numberOnes, numberZeroes, ref interactions, MAX)) return true;
 interactions++;
 return IsThereAPath(grid, row + 1, col, numberOnes, numberZeroes, ref interactions, MAX);
}

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...