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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

• Java
• C++
• Python
• Go

• 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)
  }
  

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