Special positions in Matrix (linear-time)

 Here is a problem that can be solved in linear-time: https://leetcode.com/problems/special-positions-in-a-binary-matrix/

1582. Special Positions in a Binary Matrix

Easy

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.

Accepted

6,494

Submissions

10,755

The limits are small enough that one could go for N^2, but the linear solution can be accomplished with the use of dictionaries:
 - Have two dictionaries, one for rows, one for cols
 - Keep track of the number of 1s in each of these dictionaries, accordingly
 - In the second loop then (hence (2n)), check these dictionaries looking for "one" only

Code is below, cheers, ACC.

def numSpecial(self, mat):
    """
    :type mat: List[List[int]]
    :rtype: int
    """
    colsDict = dict()
    rowsDict = dict()

    for r in range(len(mat)):
        for c in range(len(mat[r])):
            if mat[r][c] == 1:
                if colsDict.get(c) == None:
                    colsDict[c] = 1
                else:
                    colsDict[c] = colsDict[c] + 1
                if rowsDict.get(r) == None:
                    rowsDict[r] = 1
                else:
                    rowsDict[r] = rowsDict[r] + 1

    retVal = 0
    for r in range(len(mat)):
        for c in range(len(mat[r])):
            if mat[r][c] == 1 and colsDict[c] == 1 and rowsDict[r] == 1:
                retVal += 1

    return retVal