Kids With the Greatest Number of Candies

The problem is about distributing extra candies to kids and checking if each kid could potentially have the greatest number of candies.

Here’s a solution that works as follows:

Find the Maximum Number of Candies: Determine the maximum number of candies that any kid has currently.
Check for Each Kid: Iterate through the candies, and for each kid, check if the number of candies they currently have plus the extra candies is greater than or equal to the maximum number of candies. If it is, the kid could have the greatest number of candies, so we add true to the result; otherwise, we add false.

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class Solution:
    def kidsWithCandies(self, candies: List[int], extraCandies: int) -> List[bool]:
        # Find the maximum number of candies any kid currently has
        max_candies = max(candies)

        # Initialize an empty result list
        result = []

        # Iterate through the candies
        for candy in candies:
            # Check if the current kid could have the greatest number of candies
            could_have_greatest = candy + extraCandies >= max_candies

            # Append the result to the result list
            result.append(could_have_greatest)

        return result

Explanation

The line max_candies = max(candies) calculates the maximum number of candies among all kids. This helps us to compare whether each kid could potentially reach or exceed this number.
The line result = [candy + extraCandies >= max_candies for candy in candies] creates a boolean list, where each value corresponds to whether the ith kid could have the greatest number of candies. It uses a list comprehension to create this list in a concise manner.

This code will return the required boolean array, showing whether each kid could potentially have the greatest number of candies if given all the extra candies. The time complexity of this code is ( O(n) ), where ( n ) is the number of kids, and it meets the provided constraints.

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# @param {Integer[]} candies
# @param {Integer} extra_candies
# @return {Boolean[]}
def kids_with_candies(candies, extra_candies)
  max = candies.max
  
  result = []
  
  for i in (0..candies.size-1)
    result << (candies[i] + extra_candies >= max)
  end
  
  result
end

Abstract Representation of the Problem

An abstract representation emphasizes the underlying structure and key elements of a problem. Here’s an abstract formulation of the given problem:

Elements: A set of entities, each having an associated numerical value.
Operations: There’s a specific quantity that can be added to each individual numerical value.
Constraints: A comparison is made between the sum of each entity’s numerical value and the specific quantity, against a certain benchmark (maximum among all entities’ numerical values).
Goal: Determine for each entity whether the sum of its numerical value and the specific quantity is greater than or equal to the benchmark.

You have a collection of n entities, each possessing a numerical attribute.
You have an additional quantity that can be added to the numerical attribute of each entity.
You want to find if, by adding this quantity to the numerical attribute of each entity, it meets or exceeds a certain benchmark.
The benchmark is defined as the maximum value of the numerical attribute among all entities.
The output is a collection of Boolean values, corresponding to whether each entity meets the condition.

This abstraction removes specific details related to candies, kids, and the real-world context, focusing only on the logical structure of the problem. It can be applied to various other scenarios with different context but similar underlying logic.

Greedy Approach

The solution to the “Kids With Candies” problem can be considered to utilize a Greedy approach. Here’s why:

The problem is about determining if each kid can have the greatest number of candies by adding the extra candies.
In the context of this problem, a Greedy approach refers to making a local decision (i.e., deciding for each kid independently whether they can have the greatest number of candies) without considering the overall system’s global state.
By checking whether the number of candies each kid has, plus the extra candies, is greater than or equal to the maximum number of candies any kid has, you are making the best local decision for each kid.
This decision is Greedy because it doesn’t take into account any other kids or a larger overall strategy. The algorithm makes the best decision for each kid individually, based on the current data.

The Greedy approach works here because the problem’s structure allows for making independent decisions for each kid based on the current state, leading to the correct overall result.

Practice Session

Reading the explanation of the output was a bit scary. After looking at the hint, it was easy to code the solution because the condition to check was expressed directly in the code.

The input and output are:

 @param {Integer[]} candies
 @param {Integer} extra_candies
 @return {Boolean[]}

The base case is when the input is nil. This can happen when the input to the program is nil or when we reach the leaf node (it has no children, nothing to compute). We return without doing any computation.

Greedy Approach

Solution:

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def kids_with_candies(candies, extra_candies)
  result = []
  max = candies.max

  candies.each do |candy|
    result << (candy + extra_candies >= max) 
  end

  result
end