LeetCode 1431. Kids With the Greatest Number of Candies Java Solution

Problem

Kids With the Greatest Number of Candies - LeetCode

Approach

• Algorithm
  • First, find the maximum value in the given array.
  • Traverse each element in the array, and for each kid, check if adding the extra candies makes their candy count greater than or equal to the maximum count.
  • If it is greater, add true to the result list, otherwise add false.
• There is no need to put candy + extraCandies inside parentheses in the if statement (if(candy + extraCandies < max)). Java operator precedence will handle it correctly without the need for explicit grouping.
• The if statement can be replaced with the ternary operator:

result.add(candy + extraCandies >= max ? true : false);

References

Java Operator Precedence

What is Java Ternary Operator(?:)

Github Link

https://github.com/eunhanlee/LeetCode_1431_KidsWiththeGreatestNumberofCandies_Solution.git

Time Complexity: O(n), Space Complexity: O(n)

/**
 * Determines whether each kid can have the maximum number of candies considering the extra candies.
 *
 * @param candies       An integer array representing the number of candies each kid has.
 * @param extraCandies  The number of extra candies each kid can have.
 * @return              A list of booleans representing whether each kid can have the maximum number of candies.
 */
public List<Boolean> kidsWithCandies(int[] candies, int extraCandies) {
    // Find the maximum number of candies in the given array.
    int max = 0;
    for (int candy : candies) {
        max = Math.max(max, candy);
    }

    // Check if each kid can have the maximum number of candies.
    List<Boolean> result = new ArrayList<>(candies.length);
    for (int candy : candies) {
        // If adding the extra candies makes the current kid's candy count greater than or equal to the maximum count,
        // consider them able to have the maximum number of candies.
        result.add(candy + extraCandies >= max ? true : false);
    }

    return result;
}