# Solving Leetcode Interviews in Seconds with AI: Destroy Sequential Targets


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2453" 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 array nums consisting of positive integers, representing targets on a number line. You are also given an integer space. You have a machine which can destroy targets. Seeding the machine with some nums[i] allows it to destroy all targets with values that can be represented as nums[i] + c * space, where c is any non-negative integer. You want to destroy the maximum number of targets in nums. Return the minimum value of nums[i] you can seed the machine with to destroy the maximum number of targets.   Example 1:  Input: nums = [3,7,8,1,1,5], space = 2 Output: 1 Explanation: If we seed the machine with nums[3], then we destroy all targets equal to 1,3,5,7,9,...  In this case, we would destroy 5 total targets (all except for nums[2]).  It is impossible to destroy more than 5 targets, so we return nums[3].  Example 2:  Input: nums = [1,3,5,2,4,6], space = 2 Output: 1 Explanation: Seeding the machine with nums[0], or nums[3] destroys 3 targets.  It is not possible to destroy more than 3 targets. Since nums[0] is the minimal integer that can destroy 3 targets, we return 1.  Example 3:  Input: nums = [6,2,5], space = 100 Output: 2 Explanation: Whatever initial seed we select, we can only destroy 1 target. The minimal seed is nums[1].    Constraints:  1 <= nums.length <= 105 1 <= nums[i] <= 109 1 <= space <= 109  

	# Explanation
	Here's the breakdown of the approach and the Python code:

*   **High-Level Approach:**
    *   Iterate through the `nums` array. For each number, calculate how many other numbers in `nums` it can destroy based on the given `space`.
    *   Keep track of the maximum number of targets destroyed so far and the minimum seed that achieves this maximum.
    *   If a new seed destroys more targets than the current maximum, update the maximum count and the minimum seed. If a new seed destroys the same number of targets but is smaller than the current minimum seed, update the minimum seed.

*   **Complexity:**
    *   Runtime Complexity: O(N^2) in the worst case due to nested loops. However, on average, if the values in `nums` are somewhat spread out and `space` is relatively large, the inner loop might not iterate through all elements for each outer loop iteration, making it closer to O(N*K) where K is the average number of elements within the space distance.
    *   Storage Complexity: O(1) (excluding the input array).

	
	# Code
	```python
	def destroyTargets(nums, space):
    max_destroyed = 0
    min_seed = float('inf')

    for i in range(len(nums)):
        destroyed_count = 0
        for j in range(len(nums)):
            if (nums[j] - nums[i]) >= 0 and (nums[j] - nums[i]) % space == 0:
                destroyed_count += 1

        if destroyed_count > max_destroyed:
            max_destroyed = destroyed_count
            min_seed = nums[i]
        elif destroyed_count == max_destroyed and nums[i] < min_seed:
            min_seed = nums[i]

    return min_seed
	```