Destroy Sequential Targets solution and explanation
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The problem
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 <= 1051 <= nums[i] <= 1091 <= space <= 109
Approach overview
Problem Overview: You are given an array of integers representing targets and an integer space. Choosing a number destroys every target that differs from it by multiples of space. The task is to pick the starting value that destroys the maximum number of targets, returning the smallest such number.
Approach 1: Modulo Grouping Approach (O(n) time, O(n) space)
The key observation is that numbers that differ by multiples of space share the same remainder when divided by space. If two values have num % space equal, they belong to the same destroyable chain. Iterate through the array once and maintain a hash map counting how many numbers appear for each remainder. While counting, track the smallest value belonging to the remainder group with the highest frequency. This uses constant-time hash lookups and avoids sorting, giving linear runtime. This approach relies on a hash table for grouping and works well for large inputs where sorting would be unnecessary overhead.
Approach 2: Frequency Counting with Sorting (O(n log n) time, O(n) space)
Another way to reason about the same grouping is to sort the array first. Sorting ensures you encounter smaller values earlier, which simplifies picking the smallest valid answer when multiple groups have equal size. After sorting, iterate through the array and compute num % space, storing frequencies in a map. For each remainder group, update the best answer if its count exceeds the previous maximum. Sorting adds O(n log n) overhead but keeps the implementation straightforward, especially in languages where ordered traversal helps with tie-breaking. This approach combines array traversal with counting using a map.
Recommended for interviews: The modulo grouping approach is the expected optimal solution. It demonstrates the key mathematical insight that sequential targets share the same remainder when divided by space. Showing the sorted counting version first can help explain the grouping idea, but the linear-time hash map solution proves you recognize the optimization and can implement it efficiently.
Approach 1: Modulo Grouping Approach
By considering each number modulo space, we can group numbers into equivalence classes based on their remainder when divided by space. For a given number, all other numbers it can destroy have the same remainder modulo space. We can use this grouping to determine the largest group and select the smallest number from it to satisfy the conditions of the problem.
def destroy_targets(nums, space):
from collections import defaultdict
modulo_count = defaultdict(int)
for num in nums:
remainder = num % space
modulo_count[(remainder, num)] += 1
max_count = 0
min_num = float('inf')
for (remainder, num), count in modulo_count.items():
if count > max_count or (count == max_count and num < min_num):
max_count = count
min_num = num
return min_num
# Example usage:
nums = [3, 7, 8, 1, 1, 5]
space = 2
print(destroy_targets(nums, space)) # Output: 1#include <iostream>
#include <vector>
#include <unordered_map>
#include <climits>
using namespace std;
int destroyTargets(vector<int>& nums, int space) {
unordered_map<int, pair<int, int>> modulo_count;
for (int num : nums) {
int remainder = num % space;
if (modulo_count.find(remainder) == modulo_count.end())
modulo_count[remainder] = {0, INT_MAX};
modulo_count[remainder].first++;
modulo_count[remainder].second = min(modulo_count[remainder].second, num);
}
int max_count = 0;
int min_num = INT_MAX;
for (auto &[remainder, pair] : modulo_count) {
if (pair.first > max_count || (pair.first == max_count && pair.second < min_num)) {
max_count = pair.first;
min_num = pair.second;
}
}
return min_num;
}
int main() {
vector<int> nums = {3, 7, 8, 1, 1, 5};
int space = 2;
cout << destroyTargets(nums, space) << endl; // Output: 1
return 0;
}This solution groups each number in nums by its remainder when divided by space. It uses a dictionary to count how many numbers fall into each group. Then, it finds the group with the most numbers and selects the smallest number from that group. This approach ensures that we can quickly find the best number to start the destruction sequence.
Time Complexity: O(n), where n is the length of nums, since we iterate over the array once.
Space Complexity: O(n), for storing the counts of each group based on remainders.
Approach 2: Frequency Counting with Sorting
Instead of directly using modulo and counting, this approach sorts the array first. The idea is to sort the numbers and then count the potential sequence each number can form with others if seeded from it. Start from each distinct number and count how many numbers can be destroyed if that number is the seed, given the space.
import java.util.Arrays;
public class DestroyTargets {
public static int destroyTargets(int[] nums, int space) {
Arrays.sort(nums);
int maxCount = 0;
int minNum = nums[0];
for (int i = 0; i < nums.length; ++i) {
int start = nums[i];
int count = 1;
for (int j = i + 1; j < nums.length; ++j) {
if ((nums[j] - start) % space == 0) count++;
}
if (count > maxCount) {
maxCount = count;
minNum = start;
}
}
return minNum;
}
public static void main(String[] args) {
int[] nums = {3, 7, 8, 1, 1, 5};
int space = 2;
System.out.println(destroyTargets(nums, space)); // Output: 1
}
}function destroyTargets(nums, space) {
nums.sort((a, b) => a - b);
let maxCount = 0;
let minNum = nums[0];
for (let i = 0; i < nums.length; i++) {
let count = 1;
for (let j = i + 1; j < nums.length; j++) {
if ((nums[j] - nums[i]) % space === 0) count++;
}
if (count > maxCount) {
maxCount = count;
minNum = nums[i];
}
}
return minNum;
}
// Example usage:
const nums = [3, 7, 8, 1, 1, 5];
const space = 2;
console.log(destroyTargets(nums, space)); // Output: 1This Java solution first sorts the numbers. Then, for every number in the array, it calculates how many other numbers can be sequentially destroyed based on the space constraint. It keeps track of the maximum destruction possible and updates the smallest starting number accordingly. Sorting helps to manage the sequence and compare numbers efficiently.
Time Complexity: O(n^2) due to the nested loop over the sorted array.
Space Complexity: O(1) since we only use additional variables for counting, not extra storage.
Approach 3: Modulo + Enumeration
We traverse the array nums and use a hash table cnt to count the frequency of each number modulo space. The higher the frequency, the more targets can be destroyed. We find the group with the highest frequency and take the minimum value in the group.
The time complexity is O(n), and the space complexity is O(n). Here, n is the length of the array nums.
func destroyTargets(nums []int, space int) int {
cnt := map[int]int{}
for _, v := range nums {
cnt[v%space]++
}
ans, mx := 0, 0
for _, v := range nums {
t := cnt[v%space]
if t > mx || (t == mx && v < ans) {
ans = v
mx = t
}
}
return ans
}Complexity comparison
| Approach | Complexity |
|---|---|
| Modulo Grouping Approach | Time Complexity: O(n), where n is the length of |
| Frequency Counting with Sorting | Time Complexity: O(n^2) due to the nested loop over the sorted array. |
| Modulo + Enumeration | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Modulo Grouping with Hash Map | O(n) | O(n) | Best general solution. Avoids sorting and scales well for large arrays. |
| Frequency Counting with Sorting | O(n log n) | O(n) | Useful when sorted order simplifies tie-breaking or debugging. |
Frequently asked questions
Is Destroy Sequential Targets easy or hard?
Destroy Sequential Targets is generally considered a medium-level problem. The difficulty comes from recognizing the modulo grouping insight. Once you see that numbers differing by multiples of space share the same remainder, the implementation becomes straightforward.
Destroy Sequential Targets Python/Java solution
Python and C++ implementations typically use a dictionary or unordered_map to track remainder frequencies in O(n) time. Java and JavaScript solutions often follow the same logic, sometimes sorting first and then using a HashMap or object for counting.
How to solve Destroy Sequential Targets in O(n)?
Compute num % space for every number and store counts in a hash map. All numbers with the same remainder form a valid destruction sequence. Track the remainder with the highest frequency and return the smallest number belonging to that group. The single pass through the array keeps the complexity linear.
What is the best approach for Destroy Sequential Targets?
The best approach groups numbers by their remainder when divided by space. Targets with the same remainder can be destroyed together because their differences are multiples of space. Using a hash map to count frequencies gives an O(n) time and O(n) space solution while tracking the smallest value in the largest group.
Is Destroy Sequential Targets asked at Google/Amazon/Meta?
Problems involving modulo grouping, hash maps, and counting patterns are common in interviews at companies like Amazon, Google, and Meta. While this exact problem may not appear frequently, the underlying technique of grouping by remainder is a common interview pattern.
What data structure is used in Destroy Sequential Targets?
The main data structure is a hash map that stores the frequency of each remainder (num % space). This allows constant-time updates and quick identification of the remainder group that produces the most destroyed targets.
What is the time complexity of Destroy Sequential Targets?
The optimal modulo grouping solution runs in O(n) time because the array is scanned once and each remainder count is updated in constant time using a hash map. Space complexity is O(n) for storing remainder frequencies. A sorting-based variant takes O(n log n) time.