# 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](https://www.chatmagic.app), 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 n<sup>2</sup>. 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) = a<sup>2</sup> - b<sup>2</sup>. Dividing the second equation by the first gives a + b. Now we can solve for a and b.

*   **Runtime Complexity:** O(n<sup>2</sup>), **Storage Complexity:** O(1)

	
	# Code
	```python
	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]
	```