LeetCode is an excellent platform for enhancing coding skills and preparing for technical interviews. One problem that often captures the attention of beginners is the “Kids With the Greatest Number of Candies” problem. This problem is perfect for those who are just starting with Python or are new to algorithmic thinking.

In this article, we’ll go over:

- The Problem Statement
- A Basic Approach to the Problem
- Optimizing the Solution
- Python Code Implementations
- Edge Cases and Testing
- Time and Space Complexity Analysis
- Conclusion

## 1. Problem Statement

Given an array `candies`

, representing the number of candies that each kid has, and an integer `extraCandies`

, denoting the number of extra candies that you have, return a Boolean list result of length `n`

, where `result[i]`

is true if, after giving the `i-th`

kid all the `extraCandies`

, they will have the **greatest** number of candies among all the kids, or false otherwise.

#### Example 1:

Input: `candies = [2,3,5,1,3]`

, `extraCandies = 3`

Output: `[True, True, True, False, True]`

#### Example 2:

Input: `candies = [4,2,1,1,2]`

, `extraCandies = 1`

Output: `[True, False, False, False, False]`

## 2. A Basic Approach to the Problem

A simple approach is to:

- Identify the kid who has the maximum number of candies initially.
- Loop through each kid, add the extra candies, and compare their total candies with the maximum.

## 3. Optimizing the Solution

The problem can be easily optimized by finding the maximum number of candies in a single loop, thereby making the solution more efficient.

## 4. Python Code Implementations

### Basic Implementation

```
def kidsWithCandies(candies, extraCandies):
max_candies = max(candies)
result = []
for candy in candies:
if candy + extraCandies >= max_candies:
result.append(True)
else:
result.append(False)
return result
```

### Optimized Implementation

```
def kidsWithCandies(candies, extraCandies):
max_candies = max(candies)
return [candy + extraCandies >= max_candies for candy in candies]
```

## 5. Edge Cases and Testing

To validate both implementations, we can check the function with different test cases, including edge cases.

```
# Test 1
assert kidsWithCandies([2,3,5,1,3], 3) == [True, True, True, False, True]
# Test 2
assert kidsWithCandies([4,2,1,1,2], 1) == [True, False, False, False, False]
# Test 3 (Edge case)
assert kidsWithCandies([1], 1) == [True]
```

## 6. Time and Space Complexity Analysis

### Time Complexity

Both approaches have a time complexity of O(n), where n is the number of kids.

### Space Complexity

The space complexity for both solutions is O(n) as we’re creating a list of the same length as the input list.

## 7. Conclusion

The “Kids With the Greatest Number of Candies” problem is a great entry-level problem for beginners in Python and algorithmic problem-solving. It offers a good way to practice array manipulation and conditional logic.

In this article, we explored both a basic approach and an optimized approach to solve the problem, including their Python implementations. We also looked at test cases to validate the solution and performed a time and space complexity analysis.