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:
- Kid 1: 2 + 3 = 5 candies, which is the greatest among the kids.
- Kid 2: 3 + 3 = 6 candies, which is the greatest among the kids.
- Kid 3: 5 + 3 = 8 candies, which is the greatest among the kids.
- Kid 4: 1 + 3 = 4 candies, which is not the greatest among the kids.
- Kid 5: 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:
- 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
Initial Thought Process:
The basic approach is to:
- Find the maximum number of candies that any kid currently has.
- Iterate through each kid, and check if giving them all the
extraCandies
makes their total candies greater than or equal to the current maximum number of candies. - Return a boolean array where each element indicates whether that kid can have the greatest number of candies.
Basic Solution:
Code:
function kidsWithCandiesBasic(candies: number[], extraCandies: number): boolean[] {
let maxCandies = Math.max(...candies);
let result: boolean[] = [];
for (let i = 0; i < candies.length; i++) {
if (candies[i] + extraCandies >= maxCandies) {
result.push(true);
} else {
result.push(false);
}
}
return result;
}
Time Complexity Analysis:
- Time Complexity: O(n), where n is the number of kids. Finding the maximum candies takes O(n), and iterating through the candies array also takes O(n).
- Space Complexity: O(n), for the result array of boolean values.
Limitations:
This solution is efficient given the constraints. It works within the allowed time and space complexities.
Optimized Solution:
The basic solution is already optimal in terms of time complexity. However, we can focus on making the code more concise and clean.
Code:
function kidsWithCandiesOptimized(candies: number[], extraCandies: number): boolean[] {
const maxCandies = Math.max(...candies);
return candies.map(candy => candy + extraCandies >= maxCandies);
}
Time Complexity Analysis:
- Time Complexity: O(n), where n is the number of kids. Finding the maximum candies takes O(n), and mapping through the candies array also takes O(n).
- Space Complexity: O(n), for the result array of boolean values.
Improvements Over Basic Solution:
- The optimized solution uses
Array.prototype.map
, which makes the code more concise and readable.
Edge Cases and Testing:
Edge Cases:
-
candies
array has minimum and maximum values. -
extraCandies
is equal to the number of candies the kid with the most candies has. -
extraCandies
is much smaller than the number of candies the kid with the most candies has.
Test Cases:
console.log(kidsWithCandiesBasic([2,3,5,1,3], 3)); // [true, true, true, false, true]
console.log(kidsWithCandiesBasic([4,2,1,1,2], 1)); // [true, false, false, false, false]
console.log(kidsWithCandiesBasic([12,1,12], 10)); // [true, false, true]
console.log(kidsWithCandiesOptimized([2,3,5,1,3], 3)); // [true, true, true, false, true]
console.log(kidsWithCandiesOptimized([4,2,1,1,2], 1)); // [true, false, false, false, false]
console.log(kidsWithCandiesOptimized([12,1,12], 10)); // [true, false, true]
General Problem-Solving Strategies:
- Understand the Problem: Carefully read the problem statement and constraints to understand what is required.
- Identify Key Operations: Determine the key operations needed, such as finding the maximum value and iterating through the array.
-
Optimize for Readability: Use built-in functions like
Math.max
andArray.prototype.map
to make the code concise and readable. - Test Thoroughly: Test the solution with various cases, including edge cases, to ensure correctness.
Identifying Similar Problems:
-
Finding the Maximum Element:
- Problems where you need to determine the maximum element in an array.
- Example: Finding the maximum score in a game leaderboard.
-
Conditional Array Mapping:
- Problems where you need to create a new array based on a condition applied to each element of the original array.
- Example: Creating an array of booleans indicating whether students passed based on their scores.
-
Comparison with Extra Values:
- Problems where you need to compare elements of an array with an additional value to determine a condition.
- Example: Checking if adding a bonus to employees' scores makes them eligible for a reward.
Conclusion:
- The problem of determining if kids can have the greatest number of candies after adding extra candies can be efficiently solved using a straightforward approach.
- Understanding the problem and breaking it down into manageable parts is crucial.
- Using built-in functions can make the code more concise and readable.
- Testing with various edge cases ensures robustness.
- Recognizing patterns in problems can help apply similar solutions to other challenges.
By practicing such problems and strategies, you can improve your problem-solving skills and be better prepared for various coding challenges.
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