Introduction

In this blog post, we will explore how to solve the LeetCode problem "1582" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given an m x n binary matrix mat, 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. Constraints: m == mat.length n == mat[i].length 1 <= m, n <= 100 mat[i][j] is either 0 or 1.

Explanation

Here's an efficient solution to the problem, focusing on minimizing redundant checks:

• Precompute Row and Column Sums: Calculate the sum of elements in each row and each column. This allows for O(1) checking if a row or column has more than one '1'.
• Iterate and Check: Iterate through the matrix. If an element is '1', check its corresponding row and column sums. If both sums are '1', it's a special position.

• Count Special Positions: Increment a counter for each special position found.

• Runtime Complexity: O(m*n), where m is the number of rows and n is the number of columns. Storage Complexity: O(m + n) due to the row and column sum arrays.

Code

    def numSpecial(mat: list[list[int]]) -> int:
    """
    Finds the number of special positions in a binary matrix.

    Args:
        mat: The input binary matrix.

    Returns:
        The number of special positions in the matrix.
    """

    m = len(mat)
    n = len(mat[0])

    row_sums = [sum(row) for row in mat]
    col_sums = [sum(mat[i][j] for i in range(m)) for j in range(n)]

    special_count = 0
    for i in range(m):
        for j in range(n):
            if mat[i][j] == 1 and row_sums[i] == 1 and col_sums[j] == 1:
                special_count += 1

    return special_count