2453. Destroy Sequential Targets

MediumArrayHash TableCounting
Leetcode Link

Problem Description

In this LeetCode problem, we are given an array nums, which represents targets on a number line, and a positive integer space. We also have a machine that can destroy targets when seeded with a number from nums. Once seeded with nums[i], the machine destroys all targets that can be expressed as nums[i] + c * space, where c is any non-negative integer. The goal is to destroy the maximum number of targets possible by seeding the machine with the minimum value from nums.

To put it simply, we want to find the smallest number in nums that, when used to seed the machine, results in the most targets being destroyed. Each time the machine is seeded, it destroys targets that fit the pattern based on the space value and the chosen seed from nums.

Intuition

The intuition behind the solution is to leverage modular arithmetic to find which seed enables the machine to destroy the most targets. Here's the thought process for arriving at the solution:

  1. The key insight is that targets are destroyed in intervals of space starting from nums[i]. So, if two targets a and b have the same remainder when divided by space, seeding the machine with either a or b will destroy the same set of targets aligned with that spacing.
  2. To understand which starting target affects the most other targets, we can count how many targets each possible starting value (based on the remainder when divided by space) would destroy.
  3. A Counter dictionary can be used to maintain the count of how many targets share the same value % space.
  4. We iterate through nums and update the maximum count of targets destroyed (mx) and the associated minimum seed value (ans), keeping track of the seed that destroys the most targets.
  5. If a new maximum count is found, we update ans to that new seed value. If we find an equal count, we choose the seed with the lower value to minimize the seed.

By leveraging the frequency of targets, modular arithmetic, and choosing the smallest value with the most targets destroyed, we efficiently solve the problem.

Solution Approach

The implementation of the solution utilizes Python's Counter class from the collections module along with a straightforward iteration over the list of nums. Here's a detailed walk-through of the implementation steps:

  1. cnt = Counter(v % space for v in nums): The first step involves creating a Counter object to count the occurrences of each unique remainder when dividing the targets in nums by space. This helps us understand the frequency distribution of how the targets line up with different offsets from 0 when considering the space intervals. Each unique remainder represents a potential starting point for the machine to destroy targets.

  2. ans = mx = 0: Two variables are initialized - ans for storing the minimum seed value needed to destroy the maximum number of targets, and mx for storing the maximum number of targets that can be destroyed from any given seed value (initially both are set to 0).

  3. The for loop - for v in nums: This loop iterates through each value v in nums and checks how many targets can be destroyed if the machine is seeded with v. This is found with reference to the remainder v % space.

  4. t = cnt[v % space]: For a given value v, the number of targets that can be destroyed from this seed v is retrieved from the Counter object we created. This gives us the count t.

  5. if t > mx or (t == mx and v < ans): This conditional block is the core logic that checks if the current count t of destroyable targets is greater than the maximum calculated so far mx or if it is equal to mx but the seed value v is smaller than the current answer ans.

  6. ans = v and mx = t: If the condition is true, then we update ans with v because we found a better (smaller) seed value that destroys an equal or greater number of targets compared to the previously stored answer. Likewise, mx is updated with t, reflecting the new maximum count of destroyable targets.

  7. return ans: Lastly, after going through all the values in nums, the value of ans now contains the minimum seed value that can destroy the maximum number of targets when the loop is complete. This value is returned as the final answer.

This approach smartly combines modular arithmetic with frequency counting and an iterative comparison to yield both the maximum destruction and the minimum seed value needed in a highly optimized way.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Suppose we are given nums = [3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5] and space = 2.

Step 1: Calculate remainders and count frequencies.

Using cnt = Counter(v % space for v in nums) will result in:

  • Remainder 0: {4, 6, 2} (3 targets, because 4, 6, and 2 are divisible by 2)
  • Remainder 1: {1, 3, 5, 9} (7 targets, because 1, 3, 5, 5, 3, 5, and 9 leave a remainder of 1 when divided by 2)

So our Counter object cnt would look like this: {1: 7, 0: 3}.

Step 2: Initialize variables for answer and maximum count.

set ans = 0 and mx = 0.

Step 3: Iterate through nums and determine the targets that can be destroyed using each number as a seed.

For v = 3, t = cnt[3 % 2] = cnt[1] = 7. Since t > mx, we update ans to 3 and mx to 7.

For the next value v = 1, t = cnt[1 % 2] = cnt[1] = 7. Now, t == mx, but since 1 < ans, we update ans to 1.

We continue this process for all nums. Throughout this process, ans will potentially be updated when we find a v such that either:

  • It destroys more targets, or
  • It destroys an equal number of targets and is a smaller number than the current ans.

Going through the rest of nums, we find that no other numbers will update ans, as they either have a remainder of 0 or are larger than 1 with a remainder of 1.

Finally, after checking all elements in nums, the final values of ans and mx will be 1 and 7, respectively.

Step 4: Return the answer.

The function would then return ans, which is 1, as the smallest number from nums that can be used to seed the machine to destroy the maximum number of targets, which, in this example, is 7 targets.

By following the implementation steps outlined in the description, we have found the minimum seed value that achieves the maximum destruction of targets according to the given space.

Python Solution

1from collections import Counter
2
3class Solution:
4    def destroyTargets(self, nums: List[int], space: int) -> int:
5        # Create a counter to keep track of the frequency of the modulus values
6        modulus_frequencies = Counter(value % space for value in nums)
7      
8        # Initialize variables to store the most frequent element and its frequency
9        most_frequent_element = 0
10        max_frequency = 0
11      
12        # Loop through the list of numbers
13        for value in nums:
14            # Get the frequency of the current element's modulus
15            current_frequency = modulus_frequencies[value % space]
16          
17            # Check if current element's modulus frequency is higher than the max frequency found
18            # Or if it's the same but the value is smaller (as per problem's requirement)
19            if current_frequency > max_frequency or (current_frequency == max_frequency and value < most_frequent_element):
20                # Update the most frequent element and the max frequency
21                most_frequent_element = value
22                max_frequency = current_frequency
23      
24        # Return the most frequent element after checking all elements
25        return most_frequent_element
26

Java Solution

1class Solution {
2    public int destroyTargets(int[] nums, int space) {
3        // Creating a map to store the frequency of each space-residual ('residue').
4        Map<Integer, Integer> residueFrequency = new HashMap<>();
5        for (int value : nums) {
6            // Compute the space-residual of the current number.
7            int residue = value % space;
8            // Increment the count of this residue in the map.
9            residueFrequency.put(residue, residueFrequency.getOrDefault(residue, 0) + 1);
10        }
11
12        // 'bestNumber' will hold the result, 'maxFrequency' the highest frequency found.
13        int bestNumber = 0, maxFrequency = 0;
14        for (int value : nums) {
15            // Get the frequency of the current number's space-residual.
16            int currentFrequency = residueFrequency.get(value % space);
17            // Update 'bestNumber' and 'maxFrequency' if a higher frequency is found,
18            // or if the frequency is equal and the value is less than the current 'bestNumber'.
19            if (currentFrequency > maxFrequency || (currentFrequency == maxFrequency && value < bestNumber)) {
20                bestNumber = value;
21                maxFrequency = currentFrequency;
22            }
23        }
24      
25        // Return the resultant number i.e., smallest number with the highest space-residual frequency.
26        return bestNumber;
27    }
28}
29

C++ Solution

1#include <vector>
2#include <unordered_map>
3
4class Solution {
5public:
6    // Function to determine the value to be destroyed based on given rules.
7    int destroyTargets(vector<int>& nums, int space) {
8        // Create a map to store the frequency of each number modulo space.
9        unordered_map<int, int> frequency;
10        // Fill the frequency map.
11        for (int value : nums) {
12            ++frequency[value % space];
13        }
14
15        int result = 0; // Store the number whose group to be destroyed.
16        int maxFrequency = 0; // Keep track of the max frequency found.
17      
18        // Iterate over the numbers to find the value with the
19        // highest frequency and still respect the rules for ties.
20        for (int value : nums) {
21            int currentFrequency = frequency[value % space];
22            // Check if the current value's frequency is greater than the max frequency so far
23            // or if the frequency is the same but the value is smaller (for tie-breaking).
24            if (currentFrequency > maxFrequency || (currentFrequency == maxFrequency && value < result)) {
25                result = value; // Update the result with the current value.
26                maxFrequency = currentFrequency; // Update the max frequency.
27            }
28        }
29
30        // Return the number whose group will be destroyed.
31        return result;
32    }
33};
34

Typescript Solution

1// Import statements for TypeScript (if needed in your environment)
2
3interface FrequencyMap {
4    [key: number]: number;
5}
6
7// Function to determine the value to be destroyed based on given rules.
8function destroyTargets(nums: number[], space: number): number {
9    // Create a map to store the frequency of each number modulo space.
10    const frequency: FrequencyMap = {};
11    // Fill the frequency map.
12    nums.forEach(value => {
13        const modValue = value % space;
14        frequency[modValue] = (frequency[modValue] || 0) + 1;
15    });
16
17    let result = 0; // Store the number whose group is to be destroyed.
18    let maxFrequency = 0; // Keep track of the max frequency found.
19
20    // Iterate over the numbers to find the value with the
21    // highest frequency and still adhere to the rules for ties.
22    nums.forEach(value => {
23        const modValue = value % space;
24        const currentFrequency = frequency[modValue];
25        // Check if the current value's frequency is greater than max frequency so far
26        // or if the frequency is the same but the value is smaller (for tie-breaking).
27        if (currentFrequency > maxFrequency || (currentFrequency === maxFrequency && value < result)) {
28            result = value; // Update the result with the current value.
29            maxFrequency = currentFrequency; // Update the max frequency.
30        }
31    });
32
33    // Return the number whose group will be destroyed.
34    return result;
35}
36
37// Example usage:
38// const nums = [2, 12, 4, 6, 8];
39// const space = 10;
40// console.log(destroyTargets(nums, space)); // Output would depend on the input array 'nums'
41

Time and Space Complexity

The given Python code defines a method destroyTargets, which takes a list of integers nums and an integer space, and returns the value of the integer from the list that has the highest number of occurrences mod space, with the smallest value chosen if there is a tie.

To analyze the time complexity:

  1. We start by creating a counter cnt for occurrences of each v % space, where v is each value in nums. The Counter operation has a time complexity of O(n) where n is the number of elements in the list nums, as it requires going through each element in the list once.

  2. Next, we run a for loop through each value in nums, which means the loop runs n times.

  3. Inside the loop, we perform a constant time operation checking if t (which is cnt[v % space]) is greater than mx or if it is equal to mx and v is less than ans, and perform at most two assignments if the condition is true.

Therefore, the loop has a time complexity of O(n), where each iteration is O(1) but we do it n times. When you combine the loop with the initial counter creation, the overall time complexity remains O(n) because both are linear operations with respect to the size of nums.

For space complexity:

  1. The Counter object cnt will at most have a unique count for each possible value of v % space, which is at most space different values. So the space complexity for cnt is O(space).

  2. No additional data structures grow with input size n are used; therefore, the space complexity is not directly dependent on n.

Considering both the Counter object and some auxiliary variables (ans, mx, t), the overall space complexity is O(space).

In summary:

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

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