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.

Examples

Example 1:

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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:

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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:

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

Solution

Method 1 - Find kid with max candies

Code

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public class Solution {
    public List<Boolean> kidsWithCandies(int[] candies, int extraCandies) {
        List<Boolean> result = new ArrayList<>();
        int maxCandies = Arrays.stream(candies).max().getAsInt();

        for (int candy : candies) {
            result.add(candy + extraCandies >= maxCandies);
        }

        return result;
    }

}
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def kids_with_candies(candies, extraCandies):
    max_candies = max(candies)
    result = []

    for candy in candies:
        result.append(candy + extraCandies >= max_candies)

    return result

Complexity

  • ⏰ Time complexity: O(n), for finding the maximum value and for generating the result array.
  • 🧺 Space complexity: O(n), for the result array.