Solving Leetcode Interviews in Seconds with AI: Find Missing and Repeated Values
Introduction
In this blog post, we will explore how to solve the LeetCode problem "2965" 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
You are given a 0-indexed 2D integer matrix grid of size n n with values in the range [1, n2]. Each integer appears exactly once except a which appears twice and b which is missing. The task is to find the repeating and missing numbers a and b. Return a 0-indexed integer array ans of size 2 where ans[0] equals to a and ans[1] equals to b. Example 1: Input: grid = [[1,3],[2,2]] Output: [2,4] Explanation: Number 2 is repeated and number 4 is missing so the answer is [2,4]. Example 2: Input: grid = [[9,1,7],[8,9,2],[3,4,6]] Output: [9,5] Explanation: Number 9 is repeated and number 5 is missing so the answer is [9,5]. Constraints: 2 <= n == grid.length == grid[i].length <= 50 1 <= grid[i][j] <= n n For all x that 1 <= x <= n n there is exactly one x that is not equal to any of the grid members. For all x that 1 <= x <= n n there is exactly one x that is equal to exactly two of the grid members. For all x that 1 <= x <= n * n except two of them there is exactly one pair of i, j that 0 <= i, j <= n - 1 and grid[i][j] == x.
Explanation
Here's the breakdown of the solution:
- Sum and Sum of Squares: Calculate the actual sum and sum of squares of all elements in the grid. Compare these with the expected sum and sum of squares of numbers from 1 to n2. The differences will give us equations involving the repeating number (a) and the missing number (b).
Solving Equations: Solve the two equations derived in the previous step to find the values of 'a' and 'b'. We have two equations: (Sum of grid elements - Expected sum) = a - b and (Sum of squares of grid elements - Expected sum of squares) = a2 - b2. Dividing the second equation by the first gives a + b. Now we can solve for a and b.
Runtime Complexity: O(n2), Storage Complexity: O(1)
Code
def find_repeating_and_missing(grid):
"""
Finds the repeating and missing numbers in a 2D grid.
Args:
grid: A 0-indexed 2D integer matrix of size n * n.
Returns:
A 0-indexed integer array ans of size 2 where ans[0] equals the repeating number
and ans[1] equals the missing number.
"""
n = len(grid)
total_elements = n * n
actual_sum = 0
actual_sum_sq = 0
for i in range(n):
for j in range(n):
actual_sum += grid[i][j]
actual_sum_sq += grid[i][j] * grid[i][j]
expected_sum = total_elements * (total_elements + 1) // 2
expected_sum_sq = total_elements * (total_elements + 1) * (2 * total_elements + 1) // 6
diff_sum = actual_sum - expected_sum
diff_sum_sq = actual_sum_sq - expected_sum_sq
sum_ab = diff_sum_sq // diff_sum # a + b
diff_ab = diff_sum # a - b
a = (sum_ab + diff_ab) // 2
b = sum_ab - a
return [a, b]