How do you solve Kids With the Greatest Number of Candies on LeetCode?

Kids With the Greatest Number of Candies (LeetCode #1431) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Finding the maximum number of candies only once significantly reduces the time complexity.

What is the time complexity of Kids With the Greatest Number of Candies?

The optimal solution for Kids With the Greatest Number of Candies runs in O(n) time and uses O(n) space. The time complexity is O(n) because we find the maximum once (O(n)) and then check each kid (O(n)). This is efficient for larger inputs.

What is the brute force solution for Kids With the Greatest Number of Candies?

The brute force approach for Kids With the Greatest Number of Candies is Brute Force, with O(n²) time complexity and O(1) space complexity. For each kid, we will check if giving them all the extra candies makes their total greater than or equal to the maximum candies any kid has. This is straightforward but inefficient because we repeatedly find the maximum for each kid. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Kids With the Greatest Number of Candies?

Kids With the Greatest Number of Candies uses the following concepts: Array. The recommended patterns to study are: Array, Max/Min Search.

What are the interview tips for Kids With the Greatest Number of Candies?

Always start with a brute-force solution to understand the problem better. Look for opportunities to optimize your solution by reducing repeated calculations. Explain your thought process clearly during the interview; it shows your understanding of the problem.

What are common mistakes when solving Kids With the Greatest Number of Candies?

Not finding the maximum candies before checking each kid, leading to unnecessary repeated calculations. Overcomplicating the solution by trying to implement more complex data structures when a simple array suffices.

#1431

Kids With the Greatest Number of Candies

Easy

ArrayArrayMax/Min Search

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Instead of finding the maximum for each kid, we can find the maximum once and then use it to check each kid's total candies. This reduces the number of times we search for the maximum, making it much faster.

⚙️

Algorithm

3 steps

1Step 1: Find the maximum number of candies any kid currently has.
2Step 2: Create a result array initialized to false.
3Step 3: For each kid, check if their candies plus extraCandies is greater than or equal to the maximum candies. If true, set the corresponding index in the result array to true.

solution.py3 lines

1def kidsWithCandies(candies, extraCandies):
2    max_candies = max(candies)
3    return [c + extraCandies >= max_candies for c in candies]

ℹ

Complexity note: The time complexity is O(n) because we find the maximum once (O(n)) and then check each kid (O(n)). This is efficient for larger inputs.

1Finding the maximum number of candies only once significantly reduces the time complexity.
2Understanding the problem allows for a more efficient solution rather than brute-forcing through each possibility.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.