Is Check if There is a Path With Equal Number of 0's And 1's easy or hard?

The problem is rated Medium on LeetCode. The challenge is recognizing the parity constraint and converting the path search into a dynamic programming problem with memoization and pruning.

Check if There is a Path With Equal Number of 0's And 1's Python/Java solution

Implement a recursive DFS that tracks position and balance, treating 1 as +1 and 0 as -1. Use a memoization set or DP table to skip repeated states. The same logic translates directly across Python, Java, C++, and Go.

How to solve Check if There is a Path With Equal Number of 0's And 1's in O(n)?

An O(n) solution is not possible because the algorithm must consider multiple states across the grid. The practical optimal solution uses DFS with memoization over the matrix, leading to about O(m * n * (m+n)) complexity with pruning based on remaining steps.

What is the best approach for Check if There is a Path With Equal Number of 0's And 1's?

DFS with memoization (dynamic programming) is the most effective approach. Track the grid position and the current balance between ones and zeros, and cache visited states to avoid recomputation. This reduces the search from exponential paths to roughly O(m * n * (m+n)).

Is Check if There is a Path With Equal Number of 0's And 1's asked at Google/Amazon/Meta?

Grid dynamic programming and path constraint problems like this appear frequently in interviews at companies such as Google, Amazon, and Meta. The problem tests DFS state exploration, memoization, and pruning techniques.

What data structure is used in Check if There is a Path With Equal Number of 0's And 1's?

The main structures are a 2D matrix representing the grid and a memoization cache (often a hash set or 3D DP structure) that stores visited states defined by row, column, and balance of zeros vs ones.

What is the time complexity of Check if There is a Path With Equal Number of 0's And 1's?

The optimized solution runs in O(m * n * (m+n)) time. Each grid cell can be visited with different balance values representing the difference between counts of 1s and 0s along the path.

Description Hints 2 Solution Video 10:39

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

Medium51.9% acceptance1 companyPremium elsewhere, free here

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.

Asked at 1 company

Google

Topics

Hints open one at a time

Hint 1

Can you use dynamic programming to solve the problem?

Hint 2

Let dp[i][j][diff] be true if there is a path from the cell (i, j) to (m - 1, n - 1) such that the difference between the number of 0’s and the number of 1’s that we visited so far is diff, or false otherwise. The answer to the problem will be dp[0][0][0]. How do you compute this dp?

10:39

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