Solving Leetcode Interviews in Seconds with AI: Equal Row and Column Pairs
Introduction
In this blog post, we will explore how to solve the LeetCode problem "2352" 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 a 0-indexed n x n integer matrix grid, return the number of pairs (ri, cj) such that row ri and column cj are equal. A row and column pair is considered equal if they contain the same elements in the same order (i.e., an equal array). Example 1: Input: grid = [[3,2,1],[1,7,6],[2,7,7]] Output: 1 Explanation: There is 1 equal row and column pair: - (Row 2, Column 1): [2,7,7] Example 2: Input: grid = [[3,1,2,2],[1,4,4,5],[2,4,2,2],[2,4,2,2]] Output: 3 Explanation: There are 3 equal row and column pairs: - (Row 0, Column 0): [3,1,2,2] - (Row 2, Column 2): [2,4,2,2] - (Row 3, Column 2): [2,4,2,2] Constraints: n == grid.length == grid[i].length 1 <= n <= 200 1 <= grid[i][j] <= 105
Explanation
- Hashing: Convert each row and column into a string or tuple representation and use a hash map (dictionary) to store the frequency of each row.
- Comparison: Iterate through the columns, convert each column to its string/tuple representation, and check if it exists in the row frequency map. If it does, increment the count by the frequency of that row.
- Return Count: The final count will be the number of equal row and column pairs.
- Runtime Complexity: O(n^2), Storage Complexity: O(n)
Code
def equalPairs(grid):
n = len(grid)
row_counts = {}
count = 0
# Convert rows to tuples and count their occurrences
for row in grid:
row_tuple = tuple(row)
row_counts[row_tuple] = row_counts.get(row_tuple, 0) + 1
# Iterate through columns and check for matches
for j in range(n):
column = []
for i in range(n):
column.append(grid[i][j])
col_tuple = tuple(column)
if col_tuple in row_counts:
count += row_counts[col_tuple]
return count