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Formatted question description: https://leetcode.ca/all/2510.html

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

## Description

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.

## Solutions

• class Solution {
private int s;
private int m;
private int n;
private int[][] grid;
private Boolean[][][] f;

public boolean isThereAPath(int[][] grid) {
m = grid.length;
n = grid[0].length;
this.grid = grid;
s = m + n - 1;
f = new Boolean[m][n][s];
if (s % 2 == 1) {
return false;
}
s >>= 1;
return dfs(0, 0, 0);
}

private boolean dfs(int i, int j, int k) {
if (i >= m || j >= n) {
return false;
}
k += grid[i][j];
if (f[i][j][k] != null) {
return f[i][j][k];
}
if (k > s || i + j + 1 - k > s) {
return false;
}
if (i == m - 1 && j == n - 1) {
return k == s;
}
f[i][j][k] = dfs(i + 1, j, k) || dfs(i, j + 1, k);
return f[i][j][k];
}
}

• class Solution {
public:
bool isThereAPath(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
int s = m + n - 1;
if (s & 1) return false;
int f[m][n][s];
s >>= 1;
memset(f, -1, sizeof f);
function<bool(int, int, int)> dfs = [&](int i, int j, int k) -> bool {
if (i >= m || j >= n) return false;
k += grid[i][j];
if (f[i][j][k] != -1) return f[i][j][k];
if (k > s || i + j + 1 - k > s) return false;
if (i == m - 1 && j == n - 1) return k == s;
f[i][j][k] = dfs(i + 1, j, k) || dfs(i, j + 1, k);
return f[i][j][k];
};
return dfs(0, 0, 0);
}
};

• class Solution:
def isThereAPath(self, grid: List[List[int]]) -> bool:
@cache
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)


• func isThereAPath(grid [][]int) bool {
m, n := len(grid), len(grid[0])
s := m + n - 1
if s%2 == 1 {
return false
}
s >>= 1
f := [100][100][200]int{}
var dfs func(i, j, k int) bool
dfs = func(i, j, k int) bool {
if i >= m || j >= n {
return false
}
k += grid[i][j]
if f[i][j][k] != 0 {
return f[i][j][k] == 1
}
f[i][j][k] = 2
if k > s || i+j+1-k > s {
return false
}
if i == m-1 && j == n-1 {
return k == s
}
res := dfs(i+1, j, k) || dfs(i, j+1, k)
if res {
f[i][j][k] = 1
}
return res
}
return dfs(0, 0, 0)
}