How do you solve Destroy Sequential Targets on LeetCode?

Destroy Sequential Targets (LeetCode #2453) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using remainders helps categorize targets efficiently.

What is the brute force solution for Destroy Sequential Targets?

The brute force approach for Destroy Sequential Targets is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying to seed the machine with each target in the array and counting how many targets can be destroyed for each seed. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Destroy Sequential Targets?

Destroy Sequential Targets uses the following concepts: Array, Hash Table, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Destroy Sequential Targets?

Always clarify the problem and constraints before jumping into coding. Discuss your thought process and potential approaches with the interviewer. Practice dry runs of your code with sample inputs to ensure correctness.

What are common mistakes when solving Destroy Sequential Targets?

Not considering all possible seeds when counting destroyed targets. Overlooking the importance of tracking both count and minimum seed.

#2453

Destroy Sequential Targets

Medium

ArrayHash TableCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

By using a mapping of remainders when divided by space, we can efficiently count how many targets can be destroyed for each unique remainder. This reduces the number of comparisons needed.

⚙️

Algorithm

3 steps

1Step 1: Create a frequency map to count how many numbers yield the same remainder when divided by space.
2Step 2: Iterate through the nums array to populate the frequency map.
3Step 3: Find the maximum count in the frequency map and track the corresponding minimum seed value.

solution.py19 lines

1# Full working Python code
2
3def max_targets(nums, space):
4    from collections import defaultdict
5    freq_map = defaultdict(int)
6    for num in nums:
7        remainder = num % space
8        freq_map[remainder] += 1
9    max_count = 0
10    min_seed = float('inf')
11    for num in nums:
12        remainder = num % space
13        if freq_map[remainder] > max_count or (freq_map[remainder] == max_count and num < min_seed):
14            max_count = freq_map[remainder]
15            min_seed = num
16    return min_seed
17
18# Example usage
19print(max_targets([3,7,8,1,1,5], 2))

ℹ

Complexity note: The time complexity is O(n) because we make two passes through the nums array: one to build the frequency map and another to determine the minimum seed. The space complexity is O(n) due to the storage of the frequency map.

1Using remainders helps categorize targets efficiently.
2The optimal solution reduces unnecessary comparisons.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.