Kids With the Greatest Number of Candies
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
candies = [2,3,5,1,3], extraCandies = 3[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
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). This avoids redundant comparisons and leverages the problem's structure.
Optimal Approach
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. This approach is efficient, simple, and uses O(n) time and O(n) space.
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
| 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 |
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
| 4 | res = [] |
| 6 | for c in candies: |
| 7 | res.append(c + extraCandies >= max_c) |
| 9 | return res |
Objective
To iterate through each kid and determine if adding extraCandies allows them to have the greatest number of candies.
Key Insight
By comparing candies[i] + extraCandies against the precomputed maximum, the algorithm efficiently decides the boolean outcome for each kid without nested loops or redundant checks.
Interview Quick-Check
Core Logic
The condition c + extraCandies >= max_c directly tests if the kid can reach or surpass the current maximum candies count.
State & Boundaries
Appending the boolean result for each kid preserves the order and ensures the output aligns with the input array.
Common Pitfalls & Bugs
A common mistake is to compare each kid against all others individually, leading to O(n²) complexity instead of the optimal O(n).
Line Analysis
This solution has 2 Critical lines interviewers watch for.
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.
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.
return resReturn the list of boolean results indicating each kid's potential.
Returning the accumulated results completes the function, providing the required output in the correct format.
for c in candies:Iterate through each kid's candy count.
This loop processes each kid individually to evaluate their potential after receiving extraCandies.
res = []Initialize the result list to store boolean outcomes for each kid.
This list accumulates the final answers, preserving the input order and enabling efficient appends.
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?
Because the maximum candies count represents the highest threshold any kid currently has, so if a kid's candies plus extraCandies meet or exceed this value, they are guaranteed to have the greatest number of candies.
Related Problems
Simulation pattern
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