Introduction

In this blog post, we will explore how to solve the LeetCode problem "3281" 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 array of integers start and an integer d, representing n intervals [start[i], start[i] + d]. You are asked to choose n integers where the ith integer must belong to the ith interval. The score of the chosen integers is defined as the minimum absolute difference between any two integers that have been chosen. Return the maximum possible score of the chosen integers. Example 1: Input: start = [6,0,3], d = 2 Output: 4 Explanation: The maximum possible score can be obtained by choosing integers: 8, 0, and 4. The score of these chosen integers is min(|8 - 0|, |8 - 4|, |0 - 4|) which equals 4. Example 2: Input: start = [2,6,13,13], d = 5 Output: 5 Explanation: The maximum possible score can be obtained by choosing integers: 2, 7, 13, and 18. The score of these chosen integers is min(|2 - 7|, |2 - 13|, |2 - 18|, |7 - 13|, |7 - 18|, |13 - 18|) which equals 5. Constraints: 2 <= start.length <= 105 0 <= start[i] <= 109 0 <= d <= 109

Explanation

Here's the breakdown of the solution:

• Binary Search for the Answer: The problem asks for the maximum possible score. This suggests using binary search. We binary search on the potential score, mid, to check if a score of mid is achievable.
• Check Feasibility with Interval Selection: For a given mid (potential score), we need to determine if it's possible to select integers within the intervals such that the minimum absolute difference between any two selected integers is at least mid.

• Greedy Interval Selection: To check feasibility, we greedily select the smallest possible integer within each interval that satisfies the minimum difference constraint. If at any point we cannot find such an integer, the mid value is too large.

• Runtime Complexity: O(n log (max_start)), where max_start is the maximum value in the start array. Storage Complexity: O(n)

Code

    def solve():
    def check(mid):
        nums = []
        for i in range(n):
            valid_nums = []
            start_i = start[i]
            d_i = d
            for num in range(start_i, start_i + d_i + 1):
                if not nums:
                    valid_nums.append(num)
                else:
                    valid = True
                    for chosen_num in nums:
                        if abs(num - chosen_num) < mid:
                            valid = False
                            break
                    if valid:
                        valid_nums.append(num)
            if not valid_nums:
                return False

            nums.append(min(valid_nums))

        return True

    n = len(start)
    left, right = 0, max(start)
    ans = 0

    while left <= right:
        mid = (left + right) // 2
        if check(mid):
            ans = mid
            left = mid + 1
        else:
            right = mid - 1

    return ans

start = [6, 0, 3]
d = 2
print(solve())

start = [2, 6, 13, 13]
d = 5
print(solve())