Destroy Sequential Targets

Medium

Intuit

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Topics:

ArraysGreedy Algorithms

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

Destroy Sequential Targets

Medium

Intuit

0 views

Topics:

ArraysGreedy Algorithms

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

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the `nums` array and the value of `space`? Are we expecting millions or billions of elements?
Can the numbers in `nums` be negative, zero, or non-integer?
Are duplicate numbers allowed in the `nums` array, and if so, how do they affect the optimal destruction strategy?
If the array is empty, what value should I return?
Is `space` guaranteed to be positive?

Brute Force Solution

Approach

We want to find the smallest number that, when targeted, destroys the most other numbers in a specific way. The brute force method involves trying every single number as the initial target and seeing how many others it can destroy.

Here's how the algorithm would work step-by-step:

Pick the very first number from the list.
Pretend this number is the initial target.
Now, go through the rest of the numbers in the list one by one.
For each number, check if it can be destroyed by the initial target using a certain mathematical relationship (we need to check remainder when dividing the numbers by a certain number).
If a number can be destroyed, count it.
After checking all other numbers, record how many numbers were destroyed using the first number as the initial target.
Repeat the entire process, but this time pick the second number in the original list as the initial target and count how many numbers *it* can destroy.
Keep doing this for every single number in the list, each time counting the number of targets destroyed by that potential initial target.
Once you've tried every number as the initial target, compare the destruction counts you recorded.
Find the initial target that destroyed the most numbers. If there are ties, pick the smallest one.

Code Implementation

def destroy_sequential_targets_brute_force(numbers, space):
    maximum_targets_destroyed = 0
    best_starting_number = float('inf')

    for starting_number in numbers:
        targets_destroyed = 0

        # Iterate through each number in the list.
        for current_number in numbers:

            # Check if the current number can be destroyed
            # with the current starting number.
            if (current_number - starting_number) % space == 0:
                targets_destroyed += 1

        # If we found a better starting number, update
        # the maximum and the best starting number.
        if targets_destroyed > maximum_targets_destroyed:
            maximum_targets_destroyed = targets_destroyed
            best_starting_number = starting_number
        elif targets_destroyed == maximum_targets_destroyed:

            # If it's a tie, take the smaller starting number.
            best_starting_number = min(best_starting_number, starting_number)

    return best_starting_number

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n²) – The described brute force approach iterates through each of the n numbers in the input list. For each number selected as the initial target, the algorithm then iterates through the remaining n-1 numbers to check if they can be destroyed. This nested loop structure results in approximately n * (n-1) comparisons. Therefore, the time complexity can be expressed as O(n²).

Space Complexity

O(1) – The provided algorithm iterates through the input list, considering each number as a potential 'initial target'. While iterating, it counts the 'destroyed' numbers. It appears that this counting can be done using a single integer variable. The only additional memory needed is for storing the 'initial target' and the count, which is constant regardless of the input size N (the number of elements in the list). Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The goal is to find the fewest number of starting points needed to eliminate all target numbers. The clever trick is to group numbers based on their remainder when divided by a special number, and then picking the group with the most numbers.

Here's how the algorithm would work step-by-step:

First, figure out the remainder of each target number when divided by the special number.
Next, count how many target numbers have the same remainder. This will give you groups of remainders and their counts.
Find the remainder group that has the biggest count.
The number of target numbers that are not in the biggest remainder group is the answer. These are the minimum number of starting points you need.

Code Implementation

def destroy_targets(numbers, distance):
    remainders_count = {}
    for number in numbers:
        remainder = number % distance
        remainders_count[remainder] = remainders_count.get(remainder, 0) + 1

    max_count = 0
    for remainder in remainders_count:
        max_count = max(max_count, remainders_count[remainder])

    # Identify the remainders with the max count.
    most_frequent_remainders = [remainder for remainder, count in remainders_count.items() if count == max_count]

    result = 0
    used_start_numbers = set()
    for number in numbers:
        remainder = number % distance
        if remainder in most_frequent_remainders:
            start_number = number - remainder

            # Ensure we only count each start number once.
            if start_number not in used_start_numbers:
                result += 1
                used_start_numbers.add(start_number)

    return result

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n to calculate the remainders when divided by the space. It then counts the occurrences of each remainder. Finding the maximum count among these remainders also involves iterating through the distinct remainders, which is at most n. Therefore, the dominant operation is iterating through the array once, giving a time complexity of O(n).

Space Complexity

O(N) – The space complexity is dominated by the need to store the remainders and their counts. In step 2, we count how many target numbers have the same remainder. To accomplish this counting, a hash map or dictionary is typically used where keys represent the remainders and the values represent the counts. In the worst-case scenario, all N target numbers have different remainders when divided by the special number, leading to a hash map with N entries. Therefore, the auxiliary space required grows linearly with the number of target numbers, N, resulting in a space complexity of O(N).

Edge Cases

Case	How to Handle
nums is null or empty	Return 0, as there are no elements to destroy.
nums contains only one element	Return 1, as you must destroy that single element.
space is 0	Return the size of the largest group of identical numbers in nums.
nums contains duplicate values and space is a small value	The solution should correctly count the minimum destructions, taking into account dependencies caused by duplicates.
nums contains negative values	The solution should handle negative numbers correctly, considering space can also be negative.
Large nums array with a very large space value	The solution should iterate through the array efficiently even when many numbers aren't dependent, avoiding time-outs.
nums contains numbers with extremely large absolute values.	Ensure integer overflow does not occur when performing additions involving numbers in nums.
All numbers in nums are same modulo space	The solution should correctly handle the dependencies created when all numbers are congruent modulo space.