Leetcode 2928: Distribute Candies Among Children I
You are given two positive integers, ’n’ (number of candies) and ’limit’ (maximum number of candies each child can receive). Your task is to calculate how many different ways you can distribute ’n’ candies among 3 children such that no child receives more than ’limit’ candies.
Problem
Approach
Steps
Complexity
Input: You are given two integers: 'n' representing the total number of candies and 'limit' representing the maximum number of candies any child can receive.
Example: For n = 4 and limit = 2, the result will be 10.
Constraints:
• 1 <= n <= 50
• 1 <= limit <= 50
Output: Return the total number of valid ways to distribute the candies among the children such that no child receives more than 'limit' candies.
Example: For n = 3 and limit = 3, the output will be 10.
Constraints:
• The result will be a positive integer or zero, depending on the number of valid distributions.
Goal: The goal is to find all valid distributions of 'n' candies among 3 children such that no child gets more than 'limit' candies.
Steps:
• Iterate through all possible distributions of candies among the 3 children (i, j, k) where i + j + k = n.
• For each combination, check if i, j, and k are within the limit specified.
• Count all valid combinations and return the total.
Goal: The values for 'n' and 'limit' are relatively small, so brute force checking all combinations is feasible.
Steps:
• n and limit are both between 1 and 50.
• The total number of valid combinations can be computed by checking every possible distribution of candies.
Assumptions:
• It is assumed that n and limit are both positive integers within the specified ranges.
• Input: Example 1: n = 5, limit = 2
• Explanation: In this case, we are distributing 5 candies among 3 children, and no child can receive more than 2 candies. The valid distributions are: (1, 2, 2), (2, 1, 2), and (2, 2, 1), making the total number of ways = 3.
• Input: Example 2: n = 4, limit = 2
• Explanation: Here, we are distributing 4 candies among 3 children, where no child can receive more than 2 candies. The valid distributions are: (2, 1, 1), (1, 2, 1), (1, 1, 2), and permutations, which give a total of 10 ways.
Approach: The task is to calculate how many ways we can distribute the candies among 3 children while respecting the limit condition.
Observations:
• This is a problem that can be solved by brute force by checking all possible combinations of candies for the three children.
• Since the total number of candies 'n' and the limit are both small, checking each combination is computationally feasible.
Steps:
• Start by iterating over all possible values of the first child's candies (i) from 0 to limit.
• For each i, iterate over the second child's candies (j) from 0 to limit.
• For each pair (i, j), calculate the third child's candies (k = n - i - j).
• If k is between 0 and limit, then this combination is valid, so increase the counter.
• Return the total valid combinations.
Empty Inputs:
• The problem guarantees that n and limit are both positive integers, so no need to handle empty inputs.
Large Inputs:
• The problem constraints ensure that 'n' and 'limit' are small, so large inputs do not need to be considered.
Special Values:
• When n = 1, there is only one way to distribute 1 candy among the three children: (1, 0, 0), and permutations.
Constraints:
• The approach should handle edge cases like when n equals limit or when there is only one possible distribution.
int distributeCandies(int n, int limit) {
int res = 0;
for(int i = 0; i <= limit; ++i){
for(int j = 0; j <= limit; ++j){
for(int k = 0; k <= limit; ++k){
if(i + j + k == n) { res++; }
}
}
}
return res;
}
1 : Function Definition
int distributeCandies(int n, int limit) {
Defines the function 'distributeCandies' which takes in two parameters: the total number of candies 'n' and the maximum limit 'limit' for each person.
2 : Variable Initialization
int res = 0;
Initializes a variable 'res' to store the number of valid distributions where the sum of candies given to three people equals 'n'. Initially, it's set to 0.
3 : First Loop
for(int i = 0; i <= limit; ++i){
Starts the first loop over 'i', representing the number of candies given to the first person. The loop iterates from 0 to 'limit' candies.
4 : Second Loop
for(int j = 0; j <= limit; ++j){
Starts the second loop over 'j', representing the number of candies given to the second person. This loop also iterates from 0 to 'limit'.
5 : Third Loop
for(int k = 0; k <= limit; ++k){
Starts the third loop over 'k', representing the number of candies given to the third person. Like the other loops, it iterates from 0 to 'limit'.
6 : Condition Check
if(i + j + k == n) { res++; }
Checks if the total number of candies given to the three people (i + j + k) equals 'n'. If so, increments 'res' to count this valid distribution.
7 : Return Result
return res;
Returns the total count of valid distributions stored in 'res'.
Best Case: O(n^2)
Average Case: O(n^2)
Worst Case: O(n^2)
Description: The time complexity is quadratic in nature due to the nested loops for iterating through possible candy distributions.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is constant because we only need a few variables to keep track of the counts and limits.
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