LeetCode #2453 — MEDIUM

Destroy Sequential Targets

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[3,7,8,1,1,5]
2

Example 2

[1,3,5,2,4,6]
2

Example 3

[6,2,5]
100

Core Insight

What unlocks the optimal approach

Keep track of nums[i] modulo k.
Iterate over nums in sorted order.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2453: Destroy Sequential Targets
class Solution {
    public int destroyTargets(int[] nums, int space) {
        Map<Integer, Integer> cnt = new HashMap<>();
        for (int v : nums) {
            v %= space;
            cnt.put(v, cnt.getOrDefault(v, 0) + 1);
        }
        int ans = 0, mx = 0;
        for (int v : nums) {
            int t = cnt.get(v % space);
            if (t > mx || (t == mx && v < ans)) {
                ans = v;
                mx = t;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2453: Destroy Sequential Targets
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
}

# Accepted solution for LeetCode #2453: Destroy Sequential Targets
class Solution:
    def destroyTargets(self, nums: List[int], space: int) -> int:
        cnt = Counter(v % space for v in nums)
        ans = mx = 0
        for v in nums:
            t = cnt[v % space]
            if t > mx or (t == mx and v < ans):
                ans = v
                mx = t
        return ans

// Accepted solution for LeetCode #2453: Destroy Sequential Targets
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2453: Destroy Sequential Targets
// class Solution {
//     public int destroyTargets(int[] nums, int space) {
//         Map<Integer, Integer> cnt = new HashMap<>();
//         for (int v : nums) {
//             v %= space;
//             cnt.put(v, cnt.getOrDefault(v, 0) + 1);
//         }
//         int ans = 0, mx = 0;
//         for (int v : nums) {
//             int t = cnt.get(v % space);
//             if (t > mx || (t == mx && v < ans)) {
//                 ans = v;
//                 mx = t;
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2453: Destroy Sequential Targets
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2453: Destroy Sequential Targets
// class Solution {
//     public int destroyTargets(int[] nums, int space) {
//         Map<Integer, Integer> cnt = new HashMap<>();
//         for (int v : nums) {
//             v %= space;
//             cnt.put(v, cnt.getOrDefault(v, 0) + 1);
//         }
//         int ans = 0, mx = 0;
//         for (int v : nums) {
//             int t = cnt.get(v % space);
//             if (t > mx || (t == mx && v < ans)) {
//                 ans = v;
//                 mx = t;
//             }
//         }
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.