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LC240 - Search a 2D Matrix II

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Datastructure and Algorithm: DAY 74. Leetcode: 258/2736

Description

Write an efficient algorithm that searches for a value target in an m x n integer matrix matrix. This matrix has the following properties:
Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example 1:

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Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 5
Output: true

Example 2:

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Input: matrix = [[1,4,7,11,15],[2,5,8,12,19],[3,6,9,16,22],[10,13,14,17,24],[18,21,23,26,30]], target = 20
Output: false

Constraints:

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m == matrix.length
n == matrix[i].length
1 <= n, m <= 300
-109 <= matrix[i][j] <= 109
All the integers in each row are sorted in ascending order.
All the integers in each column are sorted in ascending order.
-109 <= target <= 109

Solution

BFS
Start the search from the upper right corner

  • search down if it is smaller than the target
  • search to the left if it is larger than the target)
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class Solution:
    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
        m, n = len(matrix), len(matrix[0])
        queue = deque([(0,n-1)])
        visited = set()
        while queue:
            x, y = queue.popleft()
            cur_val = matrix[x][y]
            if cur_val == target:
                return True
            
            if cur_val < target:
                nx, ny = x+1, y
                if 0<= nx < m and 0<= ny < n and (nx, ny) not in visited:
                    queue.append((nx, ny))
                    visited.add((nx, ny))
            else:
                nx, ny = x, y-1
                if 0<= nx < m and 0<= ny < n and (nx, ny) not in visited:
                    queue.append((nx, ny))
                    visited.add((nx, ny))
            
        return False