Introduction

In this blog post, we will explore how to solve the LeetCode problem "1981" 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 an m x n integer matrix mat and an integer target. Choose one integer from each row in the matrix such that the absolute difference between target and the sum of the chosen elements is minimized. Return the minimum absolute difference. The absolute difference between two numbers a and b is the absolute value of a - b. Example 1: Input: mat = [[1,2,3],[4,5,6],[7,8,9]], target = 13 Output: 0 Explanation: One possible choice is to: - Choose 1 from the first row. - Choose 5 from the second row. - Choose 7 from the third row. The sum of the chosen elements is 13, which equals the target, so the absolute difference is 0. Example 2: Input: mat = [[1],[2],[3]], target = 100 Output: 94 Explanation: The best possible choice is to: - Choose 1 from the first row. - Choose 2 from the second row. - Choose 3 from the third row. The sum of the chosen elements is 6, and the absolute difference is 94. Example 3: Input: mat = [[1,2,9,8,7]], target = 6 Output: 1 Explanation: The best choice is to choose 7 from the first row. The absolute difference is 1. Constraints: m == mat.length n == mat[i].length 1 <= m, n <= 70 1 <= mat[i][j] <= 70 1 <= target <= 800

Explanation

Here's the solution to the problem, focusing on efficiency and optimality:

• Dynamic Programming with Set Optimization: The core idea is to build a set of possible sums achievable by considering rows one by one. A set is used to efficiently eliminate duplicate sums and keep the possible sums within a manageable range.
• Iterative Row Processing: Iterate through the rows of the matrix. For each row, update the set of possible sums by adding each element of the current row to the sums already present in the set.

• Minimize Absolute Difference: After processing all rows, iterate through the final set of possible sums and find the sum that minimizes the absolute difference with the target.

• Runtime Complexity: O(m * n * S), where 'm' is the number of rows, 'n' is the maximum number of elements in a row, and 'S' is the range of possible sums which is constrained. Storage Complexity: O(S), where 'S' is the range of possible sums.

Code

    def minimizeTheDifference(mat, target):
    m = len(mat)
    possible_sums = {0}

    for row in mat:
        new_sums = set()
        for s in possible_sums:
            for num in row:
                new_sums.add(s + num)
        possible_sums = new_sums

    min_diff = float('inf')
    for s in possible_sums:
        min_diff = min(min_diff, abs(target - s))

    return min_diff