What is the time complexity of Destroy Sequential Targets?

The optimal solution has a time complexity of O(n).

What is the space complexity of Destroy Sequential Targets?

The space complexity is O(space).

What data structures are used to solve Destroy Sequential Targets?

This problem uses Array, Hash Table, Counting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Destroy Sequential Targets?

Insight 1: The key is to group numbers based on their remainder when divided by `space`. Numbers in the same group can be destroyed by the same seed value. Insight 2: We only need to track the count of numbers in each remainder group and the minimum value in each group.

Which companies ask LeetCode 2453. Destroy Sequential Targets in interviews?

This problem is frequently asked at Intuit.

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Destroy Sequential Targets - LeetCode 2453 Solution

Destroy Sequential Targets - Complete Solution Guide

Destroy Sequential Targets is LeetCode problem 2453, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 t

Detailed Explanation

The problem asks us to find the minimum starting value (`nums[i]`) that can destroy the maximum number of targets in the `nums` array. A target can be destroyed if its value is equal to `nums[i] + c * space`, where `c` is a non-negative integer. In simpler terms, we need to find a starting number such that a sequence of numbers starting from that number, incremented by `space` each time, has the largest possible overlap with the numbers in the `nums` array. We then need to return the smallest of such starting numbers.

Solution Approach

The solution uses a hash table (or an array in the C implementation) to group numbers based on their remainder when divided by `space`. For each number in `nums`, it calculates the remainder and increments the count for that remainder group. It also keeps track of the minimum number within that remainder group. After processing all numbers, the algorithm iterates through the groups, finding the group with the largest count. If multiple groups have the same largest count, the algorithm selects the group with the smallest minimum value. This minimum value is the desired seed value.

Step-by-Step Algorithm

Step 1: Create a hash table (or array) to store groups based on the remainder when each number in `nums` is divided by `space`.
Step 2: Iterate through the `nums` array. For each number:
Step 3: Calculate the remainder of the number when divided by `space`.
Step 4: Increment the count for that remainder group in the hash table.
Step 5: Update the minimum value for that remainder group in the hash table if the current number is smaller.
Step 6: After processing all numbers in `nums`, iterate through the hash table:
Step 7: Keep track of the maximum count seen so far and the corresponding minimum value (seed).
Step 8: If a group has a count larger than the current maximum count, update the maximum count and the seed.
Step 9: If a group has the same maximum count but a smaller minimum value, update the seed.
Step 10: Return the final seed value.

Key Insights

Insight 1: The key is to group numbers based on their remainder when divided by `space`. Numbers in the same group can be destroyed by the same seed value.
Insight 2: We only need to track the count of numbers in each remainder group and the minimum value in each group.
Insight 3: The minimum seed value is the smallest number among all groups that maximize the number of destroyed targets.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(space)

Topics

This problem involves: Array, Hash Table, Counting.

Companies

Asked at: Intuit.

Destroy Sequential Targets | LeetCode 2453

Medium

Array Hash Table Counting

Time: O(n)Space: O(space)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int destroyTargets(int[] nums, int space) {
        Map<Integer, int[]> groups = new HashMap<>();
        
        for (int num : nums) {
            int remainder = num % space;
            if (!groups.containsKey(remainder)) {
                groups.put(remainder, new int[]{0, Integer.MAX_VALUE});
            }
            groups.get(remainder)[0] += 1;
            groups.get(remainder)[1] = Math.min(groups.get(remainder)[1], num);
        }
        
        int max_destroyed = 0;
        int min_seed = Integer.MAX_VALUE;
        
        for (int[] group : groups.values()) {
            int count = group[0];
            int min_val = group[1];
            
            if (count > max_destroyed) {
                max_destroyed = count;
                min_seed = min_val;
            } else if (count == max_destroyed) {
                min_seed = Math.min(min_seed, min_val);
            }
        }
        
        return min_seed;
    }
}

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Problem Info

Description

Problem 2453: Destroy Sequential Targets - Detailed explanation and solution approaches.

Topics

Array Hash Table Counting

Companies

Intuit

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2453

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Create a hash table (or array) to store groups based on the remainder when each number in `nums` is divided by `space`.
2
Step 2: Iterate through the `nums` array. For each number:
3
Step 3: Calculate the remainder of the number when divided by `space`.
4
Step 4: Increment the count for that remainder group in the hash table.
5
Step 5: Update the minimum value for that remainder group in the hash table if the current number is smaller.
6
Step 6: After processing all numbers in `nums`, iterate through the hash table:
7
Step 7: Keep track of the maximum count seen so far and the corresponding minimum value (seed).
8
Step 8: If a group has a count larger than the current maximum count, update the maximum count and the seed.
9
Step 9: If a group has the same maximum count but a smaller minimum value, update the seed.
10
Step 10: Return the final seed value.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(space)

📝

Why?

The time complexity is dominated by the loop that iterates through the `nums` array. The space complexity depends on the number of distinct remainders when dividing by `space`, which can be at most `space`.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Sorting and Sliding Window

Sort the `nums` array. Then, for each element, use a sliding window to count the number of elements that can be destroyed, moving the window by `space`.

Time: O(n log n) for sorting + O(n * (max_num / space)) in worst case, so O(n log n + n * m) approximately, where m = max_num / space. This could be much larger than the O(n) approach.Space: O(1) if sorting is done in place, otherwise O(n) if a new sorted array is created.

Trade-offs: This approach can be less memory-intensive if in-place sorting is used, but it has a higher time complexity. It is generally not preferred.

✨

Key Insights

1Insight 1: The key is to group numbers based on their remainder when divided by `space`. Numbers in the same group can be destroyed by the same seed value.
2Insight 2: We only need to track the count of numbers in each remainder group and the minimum value in each group.
3Insight 3: The minimum seed value is the smallest number among all groups that maximize the number of destroyed targets.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: `nums` is empty. The provided code will likely not function correctly in this case (e.g., in C). Should probably add a check for numsSize == 0 and return appropriate value (e.g., -1 or something specified by the problem). The other language version will execute probably fine and return the correct default minimum value.
Edge case 2: All numbers in `nums` have the same remainder when divided by `space`. The solution correctly handles this case by finding the minimum number among them.
Edge case 3: `space` is 1. This means all numbers can be destroyed by any single number as they all belong to the same remainder group (0). The solution picks the smallest number in nums as a result.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the minimum value within each group. Only counting the number of targets destroyed is not enough; the minimum seed value must be returned.

✗Common mistake 2: Incorrectly calculating the remainder. It's crucial to use the modulo operator (`%`) correctly.

✗Common mistake 3: Not handling the case when multiple groups have the same maximum count. The solution needs to choose the group with the minimum starting value.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Scheduling tasks where tasks with similar deadlines (remainder when divided by a common period) can be grouped together for efficient processing. The minimum seed represents the earliest task to start.

🏦Real-world scenario 2: Distributing resources in a network where resources need to be allocated to clients based on a certain spacing requirement. The minimum seed represents the client who gets the first allocation within a certain spaced group.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 22, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Destroy Sequential Targets"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?