LeetCode 1582. Special Positions in a Binary Matrix

Description

https://leetcode.com/problems/special-positions-in-a-binary-matrix/

Given a rows x cols matrix mat, where mat[i][j] is either 0 or 1, return the number of special positions in mat.

A position (i,j) is called special if mat[i][j] == 1 and all other elements in row i and column j are 0 (rows and columns are 0-indexed).

Example 1:

Input: mat = [[1,0,0],
              [0,0,1],
              [1,0,0]]
Output: 1
Explanation: (1,2) is a special position because mat[1][2] == 1 and all other elements in row 1 and column 2 are 0.

Example 2:

Input: mat = [[1,0,0],
              [0,1,0],
              [0,0,1]]
Output: 3
Explanation: (0,0), (1,1) and (2,2) are special positions. 

Example 3:

Input: mat = [[0,0,0,1],
              [1,0,0,0],
              [0,1,1,0],
              [0,0,0,0]]
Output: 2

Example 4:

Input: mat = [[0,0,0,0,0],
              [1,0,0,0,0],
              [0,1,0,0,0],
              [0,0,1,0,0],
              [0,0,0,1,1]]
Output: 3

Constraints:

  • rows == mat.length
  • cols == mat[i].length
  • 1 <= rows, cols <= 100
  • mat[i][j] is 0 or 1.

Explanation

Find all the positions which are the only ‘1’ in their rows and check if these position are also the only ‘1’ in their columns.

Python Solution

class Solution:
    def numSpecial(self, mat: List[List[int]]) -> int:
        result = 0
        
        special_rows = []
        
        for i in range(len(mat)):
            row = mat[i]
            
            count = row.count(1)
            
            if count == 1:
                j = row.index(1)
                    
                special_rows.append([i, j])
        
        for x, y in special_rows:
            
            is_special = True
            for i in range(0, len(mat)):
                if i != x and mat[i][y] == 1:

                    is_special = False
                    break
            
            if is_special:
                result += 1
        
        return result
  • Time Complexity: O(N).
  • Space Complexity: O(1).

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