1431. Kids With the Greatest Number of Candies

Problem

There are n kids with candies. You are given an integer array candies, where each candies[i] represents the number of candies the ith kid has, and an integer extraCandies, denoting the number of extra candies that you have.

Return **a boolean array *result* of length n, where result[i] is true if, after giving the ith kid all the extraCandies, they will have the greatest number of candies among all the kids****, or *false* otherwise**.

Note that multiple kids can have the greatest number of candies.

  Example 1:

Input: candies = [2,3,5,1,3], extraCandies = 3
Output: [true,true,true,false,true] 
Explanation: If you give all 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.

Example 2:

Input: candies = [4,2,1,1,2], extraCandies = 1
Output: [true,false,false,false,false] 
Explanation: There is only 1 extra candy.
Kid 1 will always have the greatest number of candies, even if a different kid is given the extra candy.

Example 3:

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

  Constraints:

• n == candies.length

• 2 <= n <= 100

• 1 <= candies[i] <= 100

• 1 <= extraCandies <= 50

Solution (Java)

class Solution {
    public List<Boolean> kidsWithCandies(int[] candies, int extraCandies) {
        int max = 0;
        for (int i : candies) {
            max = Math.max(max, i);
        }
        List<Boolean> result = new ArrayList<>();
        for (int candy : candies) {
            result.add(candy + extraCandies >= max);
        }
        return result;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).