Kids With the Greatest Number of Candies

Q: What is the time complexity of Kids With the Greatest Number of Candies?

O(n). The solution performs two passes over the candies array: one to find the maximum and one to build the result, each O(n).

Q: What pattern does Kids With the Greatest Number of Candies use?

The Pattern: Simulation (Direct State Evaluation) The Heuristic: When the problem asks to evaluate a condition for each element based on a global property (like max or min), first compute the global property in one pass, then simulate the condition check in a second pass to achieve linear time.

Easy Simulation

Problem

Given an integer array candies and an integer extraCandies, return a boolean array where each element indicates if the corresponding kid can have the greatest number of candies among all kids after receiving extraCandies.

2 ≤ candies.length ≤ 100
1 ≤ candies[i] ≤ 100
1 ≤ extraCandies ≤ 50

Example

Input: candies = [2,3,5,1,3], extraCandies = 3

Output: [true,true,true,false,true]

The maximum number of candies any kid currently has is 5. For each kid: - Kid 0: 2 + 3 = 5, which equals the max, so true. - Kid 1: 3 + 3 = 6, which is greater than max, so true. - Kid 2: 5 + 3 = 8, which is greater than max, so true. - Kid 3: 1 + 3 = 4, which is less than max, so false. - Kid 4: 3 + 3 = 6, which is greater than max, so true. The algorithm first finds the max candies (5), then iterates through each kid, checking if their candies plus extraCandies meet or exceed this max, appending the boolean result accordingly.

Approach

Straightforward Solution

A brute-force approach would check for each kid if adding extraCandies makes their total greater than or equal to every other kid's candies, resulting in O(n²) time complexity.

Core Observation

The problem reduces to comparing each kid's candies plus extraCandies against the maximum candies any kid currently has. This comparison determines if the kid can reach or exceed the current maximum.

Path to Optimal

Preview

Recognizing that the maximum candies among all kids is a fixed threshold allows a single pass to find this max in O(n) time. Then, a second pass compares each kid's candies plus extraCandies against this max, reducing the complexity to O(n)…

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Optimal Approach

Preview

First, find the maximum candies count in the array. Then, iterate through each kid, check if candies[i] + extraCandies >= max_candies, and append the boolean result to the output list…

Full step-by-step walkthrough on Pro →

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Time

O(n)

The solution performs two passes over the candies array: one to find the maximum and one to build the result, each O(n).

Space

O(n)

The output list stores one boolean per kid, resulting in O(n) auxiliary space proportional to the input size.

Pattern Spotlight

Simulation (Direct State Evaluation)

When the problem asks to evaluate a condition for each element based on a global property (like max or min), first compute the global property in one pass, then simulate the condition check in a second pass to achieve linear time.

Solution

Python

1class Solution:
2    def kidsWithCandies(self, candies: List[int], extraCandies: int) -> List[bool]:
3        max_c = max(candies)
4        res = []
5
6        for c in candies:
7            res.append(c + extraCandies >= max_c)
8
9        return res

Step-by-Step Solution

Determine Maximum Candies Among All Kids

3	`max_c = max(candies)`

Objective

To find the highest number of candies any kid currently has, establishing the threshold for comparison.

Key Insight

Identifying the maximum candies count upfront allows the algorithm to avoid repeated comparisons between kids. This single pass operation sets a fixed benchmark that simplifies subsequent checks.

Interview Quick-Check

Core Logic

Finding the maximum candies count is a prerequisite that enables efficient comparison for all kids in a single pass.

Complexity

This operation is O(n) time and O(1) auxiliary space, as it scans the array once without extra storage.

Evaluate Each Kid's Potential to Reach or Exceed Maximum

To iterate through each kid and determine if adding extraCandies allows them to have the greatest number of candies.

1 more step with full analysis available on Pro.

Line Analysis

This solution has 2 Critical lines interviewers watch for.

Line 7 Critical

res.append(c + extraCandies >= max_c)

Append True if the kid's candies plus extraCandies meet or exceed the maximum, else False.

This conditional check directly implements the problem's requirement, efficiently determining each kid's status in a single comparison.

Line 3 Critical

max_c = max(candies)

Find the maximum number of candies any kid currently has.

This line establishes the threshold against which all kids' potential candy counts will be compared, enabling a single-pass evaluation later.

Full line-by-line criticality + rationale for all 5 lines available on Pro.

Test Your Understanding

Why is it sufficient to compare each kid's candies plus extraCandies only against the maximum candies count, rather than against every other kid individually?

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1	`class Solution:`
2	`def kidsWithCandies(self, candies: List[int], extraCandies: int) -> List[bool]:`
3	`max_c = max(candies)`
4	`res = []`
5
6	`for c in candies:`
7	`res.append(c + extraCandies >= max_c)`
8
9	`return res`