LeetCode 2453 - Destroy Sequential Targets

The problem presents a scenario where we have a list of positive integers nums representing targets placed on a number line. We also have an integer space representing the step interval of a machine.

leetcodemediumarrayhash-tablecounting

LeetCode Problem 2453

Difficulty: 🟡 Medium
Topics: Array, Hash Table, Counting

Solution

Problem Understanding

The problem presents a scenario where we have a list of positive integers nums representing targets placed on a number line. We also have an integer space representing the step interval of a machine. The machine can destroy all targets starting from an initial seed value nums[i] in steps of space. In other words, if we seed the machine with nums[i], it destroys all numbers in the form nums[i] + c * space where c is any non-negative integer.

The task is to determine the initial seed value nums[i] that allows us to destroy the maximum number of targets, and in case of ties, return the smallest such seed. The problem is constrained by potentially large arrays (nums.length <= 10^5) and large numbers (nums[i] <= 10^9), meaning a naive approach that checks every possible target sequence individually would be too slow.

Important edge cases include arrays where all numbers are unique or very large space values that prevent any sequence from destroying more than one target. The input guarantees at least one number, so empty array handling is not needed.

Approaches

The brute-force approach would involve, for each element in nums, simulating the destruction process and counting how many numbers it destroys by checking every other number for congruence with the starting number modulo space. While correct, this approach would take O(n²) time, which is unacceptable for n up to 100,000.

The optimal approach relies on a key observation: numbers that are destroyed in one sequence all have the same remainder modulo space. Therefore, we can group numbers by their num % space value and count how many numbers are in each group. The group with the largest count represents the sequence that can destroy the maximum number of targets. Among numbers in this group, we pick the smallest number as the seed.

Approach Time Complexity Space Complexity Notes
Brute Force O(n²) O(1) Simulates each sequence separately by checking all numbers
Optimal O(n) O(n) Groups numbers by num % space and selects the smallest seed from the largest group

Algorithm Walkthrough

  1. Initialize a hash map (dictionary) to store counts for each remainder when numbers in nums are divided by space. The key is the remainder, and the value is a list containing the count of numbers and the minimal number observed for that remainder.
  2. Iterate over each number num in nums. Compute the remainder r = num % space. If this remainder is not in the map, initialize it with the current number as the minimum and count 1. If it exists, increment the count and update the minimum if the current number is smaller than the stored minimum.
  3. After processing all numbers, iterate over the hash map to find the remainder with the largest count. In case of ties, select the remainder whose stored minimum number is the smallest.
  4. Return the minimum number corresponding to that remainder.

Why it works: All numbers in the same modulo group can be destroyed starting from any number in the group because adding multiples of space keeps numbers in that group. By tracking counts and minima, we efficiently find the optimal starting seed.

Python Solution

from typing import List
from collections import defaultdict

class Solution:
    def destroyTargets(self, nums: List[int], space: int) -> int:
        remainder_map = defaultdict(lambda: [0, float('inf')])
        
        for num in nums:
            r = num % space
            remainder_map[r][0] += 1
            remainder_map[r][1] = min(remainder_map[r][1], num)
        
        max_count = 0
        result = float('inf')
        
        for count, minimum in remainder_map.values():
            if count > max_count or (count == max_count and minimum < result):
                max_count = count
                result = minimum
        
        return result

Explanation: We use defaultdict to store [count, minimum] for each remainder. As we process each number, we update the count and minimal number efficiently. Finally, we select the group with the maximum count and return the minimal seed.

Go Solution

func destroyTargets(nums []int, space int) int {
    type Pair struct {
        count int
        minNum int
    }
    
    remainderMap := make(map[int]Pair)
    
    for _, num := range nums {
        r := num % space
        if p, exists := remainderMap[r]; exists {
            p.count++
            if num < p.minNum {
                p.minNum = num
            }
            remainderMap[r] = p
        } else {
            remainderMap[r] = Pair{1, num}
        }
    }
    
    maxCount := 0
    result := int(1e9 + 1)
    
    for _, p := range remainderMap {
        if p.count > maxCount || (p.count == maxCount && p.minNum < result) {
            maxCount = p.count
            result = p.minNum
        }
    }
    
    return result
}

Go-specific notes: We define a struct Pair to hold count and minimum number, since Go does not have tuple types. We update the map in place by reassigning the struct because map values are copied on access.

Worked Examples

Example 1: nums = [3,7,8,1,1,5], space = 2

num remainder (num % 2) remainder_map after step
3 1 {1: [1,3]}
7 1 {1: [2,3]}
8 0 {1: [2,3], 0: [1,8]}
1 1 {1: [3,1], 0: [1,8]}
1 1 {1: [4,1], 0: [1,8]}
5 1 {1: [5,1], 0: [1,8]}

Maximum count is 5 for remainder 1. Minimum number in this group is 1, so output is 1.

Example 2: nums = [1,3,5,2,4,6], space = 2

Remainders:

Remainder 0 Remainder 1
2,4,6 1,3,5

Both groups have count 3. Minimum numbers are 2 and 1 respectively. Return 1.

Example 3: nums = [6,2,5], space = 100

Remainders: 6 % 100 = 6, 2 % 100 = 2, 5 % 100 = 5

All counts are 1. Minimum number is 2, so return 2.

Complexity Analysis

Measure Complexity Explanation
Time O(n) Single pass over nums to populate the map, then linear pass over map values
Space O(n) Storing counts and minima for up to n distinct remainders

Because the number of distinct remainders is at most space but bounded by n, space is O(n) in the worst case.

Test Cases

# Provided examples
assert Solution().destroyTargets([3,7,8,1,1,5], 2) == 1  # maximum destruction is 5 targets
assert Solution().destroyTargets([1,3,5,2,4,6], 2) == 1  # tie between groups, minimal seed is 1
assert Solution().destroyTargets([6,2,5], 100) == 2      # only single targets destroyed

# Edge cases
assert Solution().destroyTargets([1], 1) == 1            # single element
assert Solution().destroyTargets([10,20,30,40], 10) == 10 # all divisible by 10, pick smallest
assert Solution().destroyTargets([5,11,17], 6) == 5      # no repeats, pick minimal
assert Solution().destroyTargets([7,7,7,7], 3) == 7      # all identical
Test Why
[3,7,8,1,1,5], 2 Typical scenario with multiple groups and duplicates
[1,3,5,2,4,6], 2 Tie between groups, ensure minimal seed chosen
[6,2,5], 100 Large space, minimal destruction
[1], 1 Single element array
[10,20,30,40], 10 All divisible by space, minimal selection
[5,11,17], 6 No sequence has more than one target
[7,7,7,7], 3 All identical, ensure duplicates handled

Edge Cases

1. Single-element array: If nums has only one number, the machine can destroy only that target. The implementation correctly returns the only number.

2. Large space exceeding nums differences: When space is much larger than the difference between numbers, each sequence can destroy at most one target. The algorithm