Leetcode #2510: Check if There is a Path With Equal Number of 0's And 1's
In this guide, we solve Leetcode #2510 Check if There is a Path With Equal Number of 0's And 1's in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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).
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Array, Dynamic Programming, Matrix
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
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.
Python Solution
class Solution:
def isThereAPath(self, grid: List[List[int]]) -> bool:
def dfs(i, j, k):
if i >= m or j >= n:
return False
k += grid[i][j]
if k > s or i + j + 1 - k > s:
return False
if i == m - 1 and j == n - 1:
return k == s
return dfs(i + 1, j, k) or dfs(i, j + 1, k)
m, n = len(grid), len(grid[0])
s = m + n - 1
if s & 1:
return False
s >>= 1
return dfs(0, 0, 0)
Complexity
The time complexity is . The space complexity is O(n·m) or optimized.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.