Leetcode 2358: Maximum Number of Groups Entering a Competition
You are given a positive integer array grades which represents the grades of students. Your goal is to group these students into ordered non-empty groups where:
- The sum of the grades in the i-th group is less than the sum of the grades in the (i + 1)-th group.
- The number of students in the i-th group is less than the number of students in the (i + 1)-th group. Return the maximum number of groups that can be formed.
Problem
Approach
Steps
Complexity
Input: The input consists of an array grades representing the grades of students.
Example: Input: grades = [4, 1, 8, 3, 6, 7]
Constraints:
• 1 <= grades.length <= 100
• 1 <= grades[i] <= 100
Output: Return the maximum number of groups that can be formed.
Example: Output: 3
Constraints:
• The output will always be a positive integer.
Goal: Group students such that the total number of students and the sum of their grades satisfy the given conditions for each group.
Steps:
• Sort the grades array.
• Iterate through the grades, keeping track of the number of students and the sum of their grades for each group.
• Check whether the current group satisfies the condition for being formed. If it does, increment the total number of groups.
Goal: The input constraints are small enough (grades.length <= 100) to use a sorting-based approach.
Steps:
• The number of grades will always be between 1 and 100.
• Each grade is between 1 and 100.
Assumptions:
• All grades are positive integers.
• There are enough students to form at least one group.
• Input: Example 1: grades = [4, 1, 8, 3, 6, 7]
• Explanation: In this case, the array is sorted, and the students can be grouped as follows: Group 1: [8] (Sum: 8, Students: 1), Group 2: [6, 7] (Sum: 13, Students: 2), Group 3: [4, 1, 3] (Sum: 8, Students: 3).
• Input: Example 2: grades = [3, 3]
• Explanation: In this case, only one group can be formed, as dividing it into two would violate the condition of having a different number of students in each group.
Approach: Sort the grades array and try to form the maximum number of groups by checking if each group satisfies the conditions.
Observations:
• Sorting the array helps us easily form groups with increasing sums and sizes.
• We need to iterate through the array and ensure that each new group meets the sum and size conditions.
Steps:
• Sort the grades array.
• Iterate through the array and try to form groups by adding students to the current group.
• Check if the current group satisfies the sum and size conditions compared to the previous group.
Empty Inputs:
• If there are no grades provided, no groups can be formed.
Large Inputs:
• Even with the largest input (100 students), sorting and iterating through the array will be efficient.
Special Values:
• If all grades are equal, only one group can be formed.
Constraints:
• Ensure that the groups formed have increasing sizes and sums.
int maximumGroups(vector<int>& nums) {
sort(nums.begin(), nums.end());
int cur_people = 1;
int cur_grade = nums[0];
int tmp_people = 0;
int tmp_grade = 0;
int total = 1;
int n = nums.size();
for(int i = 1; i < n; i++) {
if(tmp_people <= cur_people || tmp_grade <= cur_grade) {
tmp_grade += nums[i];
tmp_people += 1;
} else {
total += 1;
cur_people = tmp_people;
cur_grade = tmp_grade;
tmp_people = 1;
tmp_grade = nums[i];
}
}
return (tmp_people > cur_people && tmp_grade > cur_grade)? total + 1: total;
}
1 : Function Definition
int maximumGroups(vector<int>& nums) {
Defines the function `maximumGroups`, which takes a vector of integers `nums` and returns an integer representing the maximum number of groups that can be formed.
2 : Sorting
sort(nums.begin(), nums.end());
Sorts the vector `nums` in ascending order to allow for the grouping of values based on increasing order of grades.
3 : Variable Initialization
int cur_people = 1;
Initializes `cur_people` to track the number of people in the current group.
4 : Variable Initialization
int cur_grade = nums[0];
Initializes `cur_grade` to the first grade in the sorted array to start building the first group.
5 : Variable Initialization
int tmp_people = 0;
Initializes `tmp_people` to track the number of people in the temporary group during iteration.
6 : Variable Initialization
int tmp_grade = 0;
Initializes `tmp_grade` to track the sum of grades in the temporary group during iteration.
7 : Variable Initialization
int total = 1;
Initializes `total` to 1, since the first group is always formed before the loop starts.
8 : Size Calculation
int n = nums.size();
Calculates the size of the `nums` vector to be used in the loop iteration.
9 : Loop
for(int i = 1; i < n; i++) {
Starts a loop from the second element to the last in the sorted `nums` array to attempt grouping people.
10 : Condition Check
if(tmp_people <= cur_people || tmp_grade <= cur_grade) {
Checks if the number of people and the sum of grades in the current group are still less than or equal to the previous group's values.
11 : Group Update
tmp_grade += nums[i];
Adds the current grade to the `tmp_grade` for the current temporary group.
12 : Group Update
tmp_people += 1;
Increments the number of people in the current temporary group.
13 : Else Block
} else {
If the condition is not satisfied, the current temporary group is complete, and a new group is formed.
14 : Group Transition
total += 1;
Increments the `total` count of groups when a new group is formed.
15 : Group Transition
cur_people = tmp_people;
Sets `cur_people` to the number of people in the temporary group to start tracking the new group.
16 : Group Transition
cur_grade = tmp_grade;
Sets `cur_grade` to the sum of grades in the temporary group to start tracking the new group.
17 : Group Reset
tmp_people = 1;
Resets `tmp_people` to 1 for the new group starting with the current person.
18 : Group Reset
tmp_grade = nums[i];
Resets `tmp_grade` to the current person's grade for the new group.
19 : Return Result
return (tmp_people > cur_people && tmp_grade > cur_grade)? total + 1: total;
Returns the total number of groups formed, adding 1 if the last group formed is valid based on the condition.
Best Case: O(n log n)
Average Case: O(n log n)
Worst Case: O(n log n)
Description: The time complexity is dominated by the sorting step, which is O(n log n).
Best Case: O(1)
Worst Case: O(n)
Description: The space complexity is O(n) for storing the sorted array.
| LeetCode Solutions Library / DSA Sheets / Course Catalog |
|---|
comments powered by Disqus