LeetCode #1431 — EASY

Kids With the Greatest Number of Candies

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

There are n kids with candies. You are given an integer array candies, where each candies[i] represents the number of candies the i^th 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 i^th 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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[2,3,5,1,3]
3

Example 2

[4,2,1,1,2]
1

Example 3

[12,1,12]
10

Step 02

Core Insight

What unlocks the optimal approach

For each kid check if candies[i] + extraCandies ≥ maximum in Candies[i].

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1431: Kids With the Greatest Number of Candies
class Solution {
    public List<Boolean> kidsWithCandies(int[] candies, int extraCandies) {
        int mx = 0;
        for (int candy : candies) {
            mx = Math.max(mx, candy);
        }
        List<Boolean> res = new ArrayList<>();
        for (int candy : candies) {
            res.add(candy + extraCandies >= mx);
        }
        return res;
    }
}

// Accepted solution for LeetCode #1431: Kids With the Greatest Number of Candies
func kidsWithCandies(candies []int, extraCandies int) (ans []bool) {
	mx := slices.Max(candies)
	for _, candy := range candies {
		ans = append(ans, candy+extraCandies >= mx)
	}
	return
}

# Accepted solution for LeetCode #1431: Kids With the Greatest Number of Candies
class Solution:
    def kidsWithCandies(self, candies: List[int], extraCandies: int) -> List[bool]:
        mx = max(candies)
        return [candy + extraCandies >= mx for candy in candies]

// Accepted solution for LeetCode #1431: Kids With the Greatest Number of Candies
impl Solution {
    pub fn kids_with_candies(candies: Vec<i32>, extra_candies: i32) -> Vec<bool> {
        let max = *candies.iter().max().unwrap();
        candies.iter().map(|v| v + extra_candies >= max).collect()
    }
}

// Accepted solution for LeetCode #1431: Kids With the Greatest Number of Candies
function kidsWithCandies(candies: number[], extraCandies: number): boolean[] {
    const max = candies.reduce((r, v) => Math.max(r, v));
    return candies.map(v => v + extraCandies >= max);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.