## Description

https://leetcode.com/problems/kids-with-the-greatest-number-of-candies/

Given the array `candies`

and the integer `extraCandies`

, where `candies[i]`

represents the number of candies that the ** ith** kid has.

For each kid check if there is a way to distribute `extraCandies`

among the kids such that he or she can have the **greatest** number of candies among them. Notice that multiple kids can have the **greatest** number of candies.

**Example 1:**

Input:candies = [2,3,5,1,3], extraCandies = 3Output:[true,true,true,false,true]Explanation:Kid 1 has 2 candies and if he or she receives all extra candies (3) will have 5 candies --- the greatest number of candies among the kids. Kid 2 has 3 candies and if he or she receives at least 2 extra candies will have the greatest number of candies among the kids. Kid 3 has 5 candies and this is already the greatest number of candies among the kids. Kid 4 has 1 candy and even if he or she receives all extra candies will only have 4 candies. Kid 5 has 3 candies and if he or she receives at least 2 extra candies will have the greatest number of candies among the kids.

**Example 2:**

Input:candies = [4,2,1,1,2], extraCandies = 1Output:[true,false,false,false,false]Explanation:There is only 1 extra candy, therefore only kid 1 will have the greatest number of candies among the kids regardless of who takes the extra candy.

**Example 3:**

Input:candies = [12,1,12], extraCandies = 10Output:[true,false,true]

**Constraints:**

`2 <= candies.length <= 100`

`1 <= candies[i] <= 100`

`1 <= extraCandies <= 50`

## Explanation

Records the max candy number and then do compares with extra candy.

## Python Solution

```
class Solution:
def kidsWithCandies(self, candies: List[int], extraCandies: int) -> List[bool]:
max_candy = -1
for candy in candies:
if candy > max_candy:
max_candy = candy
results = []
for candy in candies:
if candy + extraCandies >= max_candy:
results.append(True)
else:
results.append(False)
return results
```

- Time complexity: O(N).
- Space complexity: O(N).