Problem Description

In this problem, we are presented with n kids, each with a certain number of candies. The candies array represents the count of candies every kid has. We are also given extraCandies, which is a number of additional candies available to distribute to the kids. Our task is to determine, for each kid, whether giving them all the extraCandies would result in them having the greatest number of candies compared to all other kids. It's important to note that more than one kid can hold the title of having the most candies if they end up with the same number of candies after distributing extraCandies. We need to return a boolean array, where each element corresponds to a kid. The value is true if the kid can have the most candies after receiving extraCandies and false otherwise.

Intuition

To solve this problem, we look to determine the highest number of candies any kid currently has, which we can do by finding the maximum value in the candies array. Once we know this maximum value, the intuition is simple: for each kid, we add extraCandies to their current number of candies and compare it with the maximum number of candies. If the sum is greater than or equal to the maximum, it means that by receiving the extraCandies, this kid could potentially have the greatest number of candies. We can do this comparison for each kid, resulting in an array of boolean values that indicate whether each kid is enabled to reach or surpass the maximum candy count with the extra candies in hand.

Solution Approach

In the provided solution, we use a simple algorithm to accomplish the task. The implementation involves the following steps:

1. Find the maximum number of candies that any kid has by utilizing the built-in max() function, which is efficient for this purpose as it iterates through the list candies once and retrieves the highest value. This step is represented by the line of code:

   1mx = max(candies)

2. We then iterate through the list of candies using a list comprehension—a concise way to generate a new list by applying an expression to each item in an iterable. For each element (representing each kid's candy count) in the candies list, we add the number of extraCandies and check if the new total is greater than or equal to the maximum candy count found in the first step. This step effectively checks whether distributing the extraCandies to each kid in turn would make their total equal to or greater than the current maximum.

3. The comparison candy + extraCandies >= mx yields a boolean result (True or False) which indicates whether after receiving extraCandies, that particular kid will have candies not less than the kid with the most candies.

4. The list comprehension returns a new list of boolean values—a direct outcome of the comparisons—which answers the problem's question for each kid. The line of code for this step looks like this:

   1return [candy + extraCandies >= mx for candy in candies]

Using this approach, the algorithm achieves a time complexity of O(n), where n is the number of elements in the candies list, since the operations involve a single scan to find the maximum and another to evaluate the condition for each kid. As for space complexity, it is also O(n), as the output is a list of n boolean values.

The utilized data structure is a simple list, and the pattern applied is a straightforward linear scan for comparison. The beauty of this solution lies in its simplicity and direct approach to determining the outcome for each kid in relation to the others.

Example Walkthrough

Let's walk through a small example to demonstrate the solution approach. Assume we have five kids with the following number of candies: candies = [2, 3, 5, 1, 3], and we have extraCandies = 3 to distribute.

1. Find the maximum number of candies that any kid has:

   1mx = max(candies)  # mx = max([2, 3, 5, 1, 3]) = 5

   The maximum number of candies with any kid is 5.

2. With the extraCandies, we need to see if each kid can reach or surpass the maximum count of 5 candies. We do this by adding extraCandies to the current number of candies each kid has and checking if it's equal to or greater than 5.

3. Generate the new list of boolean values by comparing the current number of candies plus extraCandies to the maximum number:

   1result = [candy + extraCandies >= mx for candy in candies]
   2# result = [2+3 >= 5, 3+3 >= 5, 5+3 >= 5, 1+3 >= 5, 3+3 >= 5]
   3# result = [True, True, True, False, True]

   Kid 1: Starts with 2 candies; adding 3 extra candies gives them 5, which equals the maximum. Therefore, the first element is True.

   Kid 2: Starts with 3 candies; adding 3 extra candies gives them 6, which is more than the maximum. Therefore, the second element is True.

   Kid 3: Already has the maximum of 5 candies; adding 3 more makes it 8, hence the third element is True.

   Kid 4: Starts with 1 candy; adding 3 extra candies gives them only 4, which is less than the maximum. Therefore, the fourth element is False.

   Kid 5: Starts with 3 candies; adding 3 extra candies also gives them 6, resulting in the fifth element being True.

4. Returning this list gives us the final answer to the problem:

   1result = [True, True, True, False, True]

Using this example, each kid, except for the fourth one, can have the most candies if they receive all extraCandies. This is exactly what the boolean array indicates and is in alignment with the problem description and solution approach.

Solution Implementation

Time and Space Complexity

Time Complexity

The provided function involves iterating over the list candies twice - once to find the maximum value with the max function, and once in the list comprehension to check each child's candy count.

1. max(candies): This call takes O(n) time, where n is the number of elements in the candies list.
2. List comprehension: This also takes O(n) time as it iterates through the list candies once.

Adding both gives us a total time complexity of O(n) + O(n), which simplifies to O(n), because constant factors are ignored in big O notation.

Space Complexity

The space complexity of the function is determined by the additional space required for the output list:

1. List comprehension creates a new list with the same number of elements as the input list candies. This requires O(n) space.

There's no other significant additional space used in the function, so the total space complexity is O(n).

Learn more about how to find time and space complexity quickly using problem constraints.