# 1431. Kids With the Greatest Number of Candies

 Given an integer array, where `candies[i]` represents the number of candies the $$i\_{th}$$ kid has. 

Return a boolean array `result` of length `n`, where `result[i]` is `true` if post giving the  $$i\_{th}$$  kid all the `extraCandies`, they will have the **greatest** number of candies among all the kids, or `false` otherwise.

```
Input: candies = [2,3,5,1,3], extraCandies = 3
Output: [true,true,true,false,true] 
Explanation: If give extraCandies to:
- Kid 1, they will have 2 + 3 = 5 candies, which is the greatest among the kids.
- Kid 2, they will have 3 + 3 = 6 candies, which is the greatest among the kids.
- Kid 3, they will have 5 + 3 = 8 candies, which is the greatest among the kids.
- Kid 4, they will have 1 + 3 = 4 candies, which is not the greatest among the kids.
- Kid 5, they will have 3 + 3 = 6 candies, which is the greatest among the kids.
```

***

## Intuition

* Find the kid with maximum candies & compare with each kid if given extra candies he will have the most or not.

## Approach

* Iterate over the array to find the kid with maximum candies.
* Iterate over the array again to check if `candies[i] + extraCandies` is greater or equal to the kid with maximum candies.

## Complexity

| Space Complexity | Time Complexity |
| ---------------- | --------------- |
| $$\text{O}(1)$$  | $$\text{O}(n)$$ |

* Only extra space is used to store the result
* We itreate over the array only once

## Code

```java
class Solution {
    public List<Boolean> kidsWithCandies(int[] candies, int extraCandies) {
        int maxCandiesOnAKid = 0;
        for(int i = 0; i < candies.length; ++i) {
            maxCandiesOnAKid = Math.max(maxCandiesOnAKid,candies[i]);
        }

        List<Boolean> result = new ArrayList<>();
        for(int i = 0; i < candies.length; ++i) {
            result.add(candies[i] + extraCandies>= maxCandiesOnAKid ); 
        }        

        return result;
    }
}
```


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