#1431 Kids With the Greatest Number of Candies

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the problem

In plain words

candies[i] is how many candies kid i has. For each kid, imagine giving them all extraCandies. Would they then have at least as many as every other kid? Return one true/false per kid as a List<Boolean>.

in candies = [2,3,5,1,3], extraCandies = 3

out [true,true,true,false,true]

Only kid 3 (1 + 3 = 4) stays below the current max of 5.

in candies = [4,2,1,1,2], extraCandies = 1

out [true,false,false,false,false]

in candies = [12,1,12], extraCandies = 10

out [true,false,true]

2 ≤ n ≤ 100
1 ≤ candies[i] ≤ 100
1 ≤ extraCandies ≤ 50

your turn

Try it before you peek

Read the examples, say the brute force out loud, then write code: here or on LeetCode ↗. Stuck for 20 minutes? Open one hint at a time.

Solve it in the browser

LeetCode-style: complete the class. Compiled by the real javac (Java 17) and judged right here — nothing is sent anywhere.

Solutions are hidden so you can try first. Even 20 minutes of real effort makes the answer stick far better than reading it.

Marking the problem solved, or opening every hint, also shows it.

the aha

Key insight

patternSimulation

"At least as many as everyone" only depends on the current maximum. Find it once, then each answer is one comparison.

solutions

From brute force to optimal

For each kid, check that no other kid has more than candies[i] + extra.

For each i, loop over every j.
If some candies[j] > candies[i] + extra → false.
Otherwise true.

timeO(n²)

n kids, each compared with n kids.

spaceO(1) extra

Only the output list.

javajava.util.*

1class Solution {2    public List<Boolean> kidsWithCandies(int[] candies, int extraCandies) {3        List<Boolean> result = new ArrayList<>(candies.length);4        for (int i = 0; i < candies.length; i++) {5            boolean greatest = true;6            for (int j = 0; j < candies.length; j++) {7                if (candies[j] > candies[i] + extraCandies) {   // someone still has more8                    greatest = false;9                    break;10                }11            }12            result.add(greatest);                    // autoboxed to Boolean13        }14        return result;15    }16}

The only kid you really have to match is the current leader.

Pass 1: max = largest candies[i].
Pass 2: add candies[i] + extra >= max for each kid.

timeO(n)

Two passes.

spaceO(1) extra

One int besides the output list.

javajava.util.*

1class Solution {2    public List<Boolean> kidsWithCandies(int[] candies, int extraCandies) {3        int max = 0;4        for (int c : candies) max = Math.max(max, c); // pass 1: the current record5        List<Boolean> result = new ArrayList<>(candies.length); // capacity known: no resizing6        for (int c : candies) result.add(c + extraCandies >= max); // pass 2: ties count7        return result;8    }9}

in the room

How to say it in the interview

talk track

Clarify that ties count as "greatest".
Say why one max is enough: the gift is imaginary, so the max never changes between kids.

traps

Using > instead of >= — ties must be true.
Returning boolean[]: the method must return List<Boolean> (boxed values).

what they ask next

Follow-ups

What if each gift were real and the candies stayed?

Then the max can change after every gift, so update it as you go — still one pass, but the order of kids now matters.

before you submit

Edge cases to test

Ties count: equal to the max is true.
The kid who already has the max is always true.

practise next