grid47 Exploring patterns and algorithms
Aug 20, 2024
5 min read
Solution to LeetCode 781: Rabbits in Forest Problem
In a forest, there are an unknown number of rabbits. We ask n rabbits, ‘How many rabbits of the same color as you are there?’ and collect their answers in an integer array answers, where answers[i] is the answer of the i-th rabbit. The task is to determine the minimum number of rabbits that could be in the forest.
Problem
Approach
Steps
Complexity
Input: The input consists of an array answers where each element answers[i] represents the number of rabbits that the i-th rabbit believes share the same color.
Example: Input: answers = [2, 2, 1]
Constraints:
• 1 <= answers.length <= 1000
• 0 <= answers[i] < 1000
Output: The output is a single integer representing the minimum number of rabbits that could be in the forest.
Example: Output: 6
Constraints:
• The minimum number of rabbits must satisfy the answers given by the rabbits.
Goal: The goal is to calculate the minimum number of rabbits in the forest that could explain the given answers.
Steps:
• Count the frequency of each answer.
• For each answer, calculate the minimum number of rabbits required to satisfy that answer. If there are X rabbits answering the same number Y, then the minimum number of rabbits for that answer group is ceil(X / (Y+1)) * (Y+1).
• Sum up the required rabbits for all answer groups to find the total minimum.
Goal: The solution must handle arrays with a length up to 1000, and handle answers in the range of 0 to 999.
Steps:
• The input array will have at least one element.
• The answers array will not have values greater than 999.
Assumptions:
• The array of answers is valid, and each answer is a non-negative integer.
• Input: Example 1: Input: answers = [2, 2, 1]
• Explanation: The two rabbits that answered '2' need to be in a group of at least 3 rabbits, and the one rabbit that answered '1' needs to be in a group of at least 2 rabbits. Therefore, the minimum number of rabbits in the forest is 6.
• Input: Example 2: Input: answers = [5, 5, 5]
• Explanation: The three rabbits that answered '5' need to be in a group of at least 6 rabbits, so the minimum number of rabbits in the forest is 6.
Approach: The approach is to count how many rabbits give the same answer, then compute the minimum number of rabbits required to form complete groups that match those answers.
Observations:
• The answers represent groups of rabbits with the same color, and the answers must form groups that fit within those constraints.
• To minimize the total number of rabbits, each group should be as small as possible while still matching the answers given by the rabbits.
Steps:
• Count the frequency of each answer.
• For each unique answer, calculate the minimum group size using the formula: ceil(X / (Y + 1)) * (Y + 1), where X is the frequency of the answer and Y is the answer value.
• Sum the minimum group sizes to get the total number of rabbits.
Empty Inputs:
• The input array will always have at least one element.
Large Inputs:
• Ensure the solution handles arrays with a length up to 1000.
Special Values:
• When all rabbits answer '0', only one rabbit is needed in the forest.
Constraints:
• The solution should handle all possible values of answers correctly, considering edge cases like answers with high values near 999.
intnumRabbits(vector<int>& answers) {
unordered_map<int, int> c;
for(inti: answers)
c[i]++;
int res =0;
for(autoi : c) res += (i.second + i.first) / (i.first +1) * (i.first +1);
return res;
}
1 : Function Definition
intnumRabbits(vector<int>& answers) {
Defines the function `numRabbits`, which takes a vector of integers `answers` representing the answers to the rabbit color question.
2 : Map Initialization
unordered_map<int, int> c;
Initializes an unordered map `c` to count the frequency of each answer (representing the number of rabbits with the same color).
3 : Loop Through Answers
for(inti: answers)
Loops through each element in the `answers` vector to count how many times each answer appears.
4 : Count Answers
c[i]++;
Increments the count for the current answer `i` in the map `c`.
5 : Result Initialization
int res =0;
Initializes a variable `res` to store the final result, which is the total number of rabbits.
6 : Calculate Total Rabbits
for(autoi : c) res += (i.second + i.first) / (i.first +1) * (i.first +1);
Iterates through the map `c` and calculates the total number of rabbits based on the frequency of each answer. The formula ensures that if there are `x` rabbits with the same answer, they will all be grouped together.
7 : Return Result
return res;
Returns the calculated total number of rabbits.
Best Case: O(n), where n is the number of rabbits, because we only need to count the frequency of answers.
Average Case: O(n), as the process of counting and grouping is linear with respect to the number of answers.
Worst Case: O(n), as the worst case also involves counting frequencies and calculating the group sizes.
Description: The time complexity is linear in the size of the input array.
Best Case: O(n), when all rabbits have unique answers.
Worst Case: O(n), where n is the number of unique answers, since we need to store the frequency of each answer.
Description: Space complexity is linear with respect to the number of unique answers in the input.
