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

Problem Description

The problem provides us with an m x n binary matrix, grid, indexed at 0. We can navigate this grid, moving either right to the next column or down to the next row from our current position. The goal is to determine whether there exists a path from the top-left cell (0, 0) to the bottom-right cell (m - 1, n - 1) where the number of 0s encountered is exactly equal to the number of 1s encountered along the path.

Intuition

To solve this problem, we can employ a depth-first search (DFS) that keeps track of our position in the grid and the count of 1s that we've seen so far. We define a criterion for a successful path: it has to reach the bottom-right corner of the grid and, upon reaching it, the count of 1s must be equal to the count of 0s. Since any path from (0, 0) to (m - 1, n - 1) is m + n - 1 steps long and we want equal numbers of 1s and 0s, the total number of each must be half of m + n - 1. This is only possible if m + n - 1 is even, hence we check for that early on and return False if it's not.

If the condition is met, we perform the DFS. The key to making the DFS manageable is to cache the results of previous paths using the @cache decorator to prevent re-computation and to cut off paths early if they become impossible. This happens if too many 1s or 0s are encountered before reaching the end, indicated by our counts exceeding half of m + n - 1.

The recursive DFS is defined as such:

• We increase our count of 1s (k) as we encounter them along the path.
• If our current position is out of bounds, we return False.
• If the count of 1s or 0s exceeds half of the length of a potential correct path (s), we also return False.
• If we reach the bottom-right corner, we confirm if our count of 1s is equal to s (as 0s will automatically be equal due to path length constraints) and return True or False accordingly.
• At each cell, we explore both possible next cells (right and down), if either returns True, we have found a valid path.

The solution is initialized by setting m and n to the grid dimensions and computing s, which is half of the path length. DFS starts at (0, 0) with an initial count of 0. If a path that satisfies the conditions is found, dfs will eventually return True, otherwise, False.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution approach uses a recursive depth-first search (DFS) technique to navigate through the matrix. The DFS is augmented with memoization through the @cache decorator, which is a way to store the results of expensive function calls and return the cached result when the same inputs occur again, ensuring that each state is only computed once. This is critical for efficiency, especially in a matrix where there could be overlapping subproblems.

Here's the breakdown of the implementation:

• We initialize two variables m and n to store the number of rows and columns of the grid, respectively. Then we calculate s as half the length of a balanced path from (0, 0) to (m - 1, n - 1), which is (m + n - 1) / 2. This division by 2 is represented by the right shift operator s >>= 1, which is an efficient way to divide by 2.
• We use the dfs(i, j, k) function where i and j are the current row and column in the grid, and k is the current count of 1s encountered on the path.
• Inside the dfs function:
  • We first check if the current position is out of bounds (i >= m or j >= n), returning False since it is not a valid path.
  • We update the count of 1s found (k += grid[i][j]).
  • If the number of 1s (k) exceeds s or the number of 0s (i + j + 1 - k) exceeds s, the path cannot be balanced, so we return False.
  • When the bottom-right corner (i == m - 1 and j == n - 1) is reached, we check if the number of 1s is exactly s. If it is, we've found a valid path, otherwise, the path is invalid.
  • For all other cases, we recurse to the right dfs(i, j + 1, k) and down dfs(i + 1, j, k). If either of these paths return True, we return True, indicating a valid path has been found.
• Before starting the DFS, we check if s & 1 is truthy, meaning m + n - 1 is odd and thus cannot be evenly split between 0s and 1s. In such a case, we return False.
• Finally, we call the dfs(0, 0, 0) function to start the path search from the top-left corner with 0 1s counted so far. If the function eventually returns True, then a balanced path exists, and we return True; otherwise, we return False.

The use of recursion and memoization (through @cache) is the key algorithmic pattern. This combination ensures that once a certain state (in terms of location and current count of 1s) has been computed, it won't be recomputed unnecessarily, thus reducing the overall time complexity from exponential to polynomial.

Example Walkthrough

To illustrate the described solution approach, let's consider a small 3x3 binary matrix example:

1grid = [
2  [0, 1, 0],
3  [0, 0, 1],
4  [1, 0, 1]
5]

• First, we determine m and n. In this case, m = 3 and n = 3.
• We calculate s as half the length of a balanced path. Since the path from (0, 0) to (m - 1, n - 1) has m + n - 1 = 5 steps, and 5 is odd, we can conclude right away that it's not possible to have a balanced path with an equal number of 0s and 1s. Therefore, in this situation, our algorithm would return False.

Suppose we had a scenario where m + n - 1 is even. Let's change the grid to make that possible:

1grid = [
2  [0, 1],
3  [1, 0]
4]

• m = 2, n = 2, and s = (2 + 2 - 1) / 2 = 1.5. Since we deal with whole numbers, we can only have an equal number of 0s and 1s if s is an integer, so 3 would be rounded down to 1.

Now, let's walk through a successful step-by-step DFS:

1. Start DFS with dfs(0, 0, 0).
2. At [0][0], our count k remains 0. We call dfs(0, 1, 0) to move right and dfs(1, 0, 0) to move down.
3. The right move leads to [0][1], which is a 1, so k is incremented. The count k is now 1, so we cannot move right anymore as it would leave the grid. We move down with dfs(1, 1, 1).
4. The down move from step 2 leads to [1][0], which is a 1, so we increment k. We move right to [1][1] with dfs(1, 1, 1).
5. Now at [1][1], we are in the bottom-right corner. Here k is 1, and since s is 1, we have an equal number of 0s and 1s, which satisfies the conditions of the path.

Therefore, a valid path exists, and the function dfs(0, 0, 0) eventually returns True.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the isThereAPath function is O(mns), where m is the number of rows, n is the number of columns, and s is half the perimeter of the grid (sum of rows m and columns n, minus 1, all divided by 2).

This is because in the worst case, the dfs function is called for every possible position (i, j) and for every possible sum k up to s. Since there are m*n positions and up to s possible values for k, the result is m*n*s.

The recursive function might visit each cell multiple times with different values of k, but the memoization using @cache ensures that it will only recompute if it finds a cell with a distinct k value that it has not yet explored.

Space Complexity

The space complexity of the solution is O(mns), which comes from the call stack used in recursion and the cache used for memoization.

The recursive depth of the stack can go as deep as m + n in the worst case (if a path involves all rows and columns). Additionally, the memoization employed by @cache will store results for each (i, j, k) tuple which has been computed and not yet been visited.

Since there are m*n positions and up to s distinct values for each position's sum k, the memoization cache can take up as much space as there are combinations of positions and sums, leading to a space complexity of O(mns).

Learn more about how to find time and space complexity quickly using problem constraints.