Introduction

In this blog post, we will explore how to solve the LeetCode problem "1329" 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

A matrix diagonal is a diagonal line of cells starting from some cell in either the topmost row or leftmost column and going in the bottom-right direction until reaching the matrix's end. For example, the matrix diagonal starting from mat[2][0], where mat is a 6 x 3 matrix, includes cells mat[2][0], mat[3][1], and mat[4][2]. Given an m x n matrix mat of integers, sort each matrix diagonal in ascending order and return the resulting matrix. Example 1: Input: mat = [[3,3,1,1],[2,2,1,2],[1,1,1,2]] Output: [[1,1,1,1],[1,2,2,2],[1,2,3,3]] Example 2: Input: mat = [[11,25,66,1,69,7],[23,55,17,45,15,52],[75,31,36,44,58,8],[22,27,33,25,68,4],[84,28,14,11,5,50]] Output: [[5,17,4,1,52,7],[11,11,25,45,8,69],[14,23,25,44,58,15],[22,27,31,36,50,66],[84,28,75,33,55,68]] Constraints: m == mat.length n == mat[i].length 1 <= m, n <= 100 1 <= mat[i][j] <= 100

Explanation

Here's a solution to the problem, along with an explanation of the approach:

• Identify Diagonals: Each diagonal can be uniquely identified by the difference between its row and column indices (row - col). We iterate through the matrix and extract the elements belonging to each diagonal.
• Sort Diagonals: Store the diagonals in a dictionary (or hashmap) using the row - col difference as the key. Sort the elements within each diagonal.

• Reconstruct Matrix: Iterate through the matrix again, placing the sorted elements back into their corresponding positions based on the diagonal to which they belong.

• Runtime Complexity: O(m*n*log(min(m,n))) where m and n are the dimensions of the matrix. Storage Complexity: O(m*n)

Code

    def diagonalSort(mat):
    m, n = len(mat), len(mat[0])
    diagonals = {}

    # Extract and store diagonals
    for i in range(m):
        for j in range(n):
            diff = i - j
            if diff not in diagonals:
                diagonals[diff] = []
            diagonals[diff].append(mat[i][j])

    # Sort diagonals
    for diff in diagonals:
        diagonals[diff].sort()

    # Reconstruct matrix
    for i in range(m):
        for j in range(n):
            diff = i - j
            mat[i][j] = diagonals[diff].pop(0)

    return mat