2910. Minimum Number of Groups to Create a Valid Assignment
Problem Description
In this problem, we're working with an array of integers called 'nums', and we've been tasked with organizing its indices into different groups. The challenge is to find the minimum number of groups needed to satisfy two specific criteria:
- Each group can only contain indices that point to the same value in 'nums'. That means if 'nums[i]' equals 'nums[j]', then 'i' and 'j' can be in the same group.
- Every group should be as close in size as possible. The size difference between any two groups must not be more than 1.
The goal is to achieve these conditions with the fewest number of groups possible.
Intuition
The solution strategy involves a clever use of counting and enumeration. Hereās how we arrive at the solution:
-
We first count how frequently each number occurs in our 'nums' array using a hash table called 'cnt'. This helps us know the distribution of values within 'nums'.
-
Understanding that the groups can have sizes that are either 'k' or 'k+1', where 'k' is the minimum occurrence count of any value, we can enumerate possible group sizes starting from 'k' down to 1.
-
For each potential group size 'k', we look at every number's occurrence 'v' and check if it's possible to divide 'v' into groups of size 'k' or 'k+1' without violating our conditions (where the size of the groups does not differ by more than 1).
-
If for any occurrence 'v', it is not possible to make such groups (that is, if 'v/k' is less than 'v % k'), we know that dividing into groups of this size would not satisfy the condition. So, we skip to the next group size.
-
If it is possible to form groups for a given 'k', we compute the number of groups by dividing 'v' by 'k+1' and rounding up to ensure we count partially filled groups as a whole one (this is done to minimize the number of groups).
-
Since we are trying group sizes from largest to smallest, as soon as we find a valid grouping that satisfies our conditions, we can be assured that itās the optimal one with the minimum groups required.
By tackling it step-by-step, we ensure that we're checking all possible group sizes and finding the most economical way to distribute the indices in accordance with both conditions.
Learn more about Greedy patterns.
Solution Approach
The chosen approach to solve this problem can be broken down into a few strategic steps, employing common algorithms, data structures, and patterns.
First, let's detail the use of Python's Counter
class for creating the frequency table, often referred to as a hash table, which is crucial for counting occurrences of each unique number from the array nums
.
-
Hash Table Creation: Using
Counter(nums)
, we count the occurrences of each number innums
. It allows us to easily access the frequencyv
of each unique value innums
. -
Enumeration of Group Sizes: We then attempt to find the minimum group sizes by starting from the minimum frequency
k
found in our hash table and decreasing towards 1. This takes advantage of the pattern that larger group sizes can potentially lead to fewer groups if such groupings are possible. -
Divisibility Check: For each proposed group size
k
, we iterate through all frequenciesv
and check whether each value can be divided into groups of sizek
ork+1
without exceeding the size difference constraint. This is done by checking iffloor(v/k) < v mod k
, wherefloor
is implicit in integer division andmod
is the modulo operation.- If this condition is true for any frequency
v
, it indicates that the sizek
cannot allow for a valid grouping, and we move to the next smaller size by breaking the current loop prematurely.
- If this condition is true for any frequency
-
Grouping Calculation: If all frequencies can be grouped by the current
k
, it means thatk
is a valid group size. To ensure we use as few groups as possible, we want to create groups ofk+1
first and only use groups of sizek
if necessary. The calculation(v + k) // (k + 1)
ensures we are creating as many full groups of sizek+1
as possible before resorting to any groups of sizek
.- This step is essential because we want the minimum number of groups, which means maximizing group size when possible while still respecting the rules.
-
Optimal Solution: Since we enumerate
k
from its maximum possible value down to 1, the firstk
for which a valid grouping exists will give us the minimum number of groups needed to satisfy the problem conditions.
This algorithm is both efficient and effective, employing a hash table for quick value access and enumeration to systematically check each group size. The first verified group size that holds the condition provides us with an optimal solution.
The formulae and concepts used:
- Hash table frequency count:
Counter(nums)
- Enumeration from max to min:
for k in range(min(cnt.values()), 0, -1)
- Divisibility check:
if v // k < v % k
- Group calculation:
ans += (v + k) // (k + 1)
- Optimal solution on first valid:
if ans: return ans
This approach elegantly combines these elements to guarantee the most efficient grouping under the given constraints.
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Start EvaluatorExample Walkthrough
Let's walk through a small example to illustrate the solution approach with the array nums = [3,3,3,3,1,1,1]
.
-
Hash Table Creation: First, we use
Counter(nums)
to create a frequency table.- We get
cnt = Counter({3: 4, 1: 3})
. This shows us that the number 3 occurs 4 times, and the number 1 occurs 3 times.
- We get
-
Enumeration of Group Sizes: Next, we look for the minimum group size starting from
min(cnt.values())
, which is 3 in this case, and decrement towards 1.- We iterate
k
from 3 down to 1.
- We iterate
-
Divisibility Check: For
k = 3
, we check each value incnt
to see if it can be divided into groups of size 3 or 4 (k+1
) while respecting the size difference condition.- For value
v = 4
(number 3 innums
), we check if4 // 3 < 4 % 3
.- This gives
1 < 1
, which is false, so a group size of 3 is valid for this occurrence.
- This gives
- For value
v = 3
(number 1 innums
), we check if3 // 3 < 3 % 3
.- This gives
1 < 0
, which is false, so a group size of 3 is valid for this occurrence as well.
- This gives
- Since both checks passed, we move to the Grouping Calculation step.
- For value
-
Grouping Calculation: Both numbers can create groups with size 3 or 4. We calculate the number of groups for each:
- For the number 3 in
nums
(v = 4
):ans += (4 + 3) // (4)
results inans += 2
. - For the number 1 in
nums
(v = 3
):ans += (3 + 3) // (4)
results inans += 1
. - Total
ans
now is2 + 1 = 3
.
- For the number 3 in
-
Optimal Solution: The first valid group size was 3, which required 3 groups. Since this is the first valid solution we've encountered as we decremented from
k = min(cnt.values())
, it is the optimal solution. Thus, we need a minimum of 3 groups to satisfy the problem conditions.
Putting it all together, we've found that the array nums = [3,3,3,3,1,1,1]
can be organized into a minimum of 3 groups such that each group contains indices that point to the same value, and the sizes of the groups differ by at most 1. The groups can be visualized as: [3, 3, 3]
, [3]
, and [1, 1, 1]
.
Solution Implementation
1from collections import Counter
2from typing import List
3
4class Solution:
5 def minGroupsForValidAssignment(self, nums: List[int]) -> int:
6 # Count the frequency of each number in the given list
7 frequency_count = Counter(nums)
8
9 # Iterate from the maximum frequency down to 1
10 for group_size in range(max(frequency_count.values()), 0, -1):
11 groups_needed = 0 # Initialize the number of groups needed for this group size
12
13 # Iterate through each frequency value
14 for frequency in frequency_count.values():
15 # If the distribution of frequency across groups is invalid, reset groups and break
16 if frequency // group_size < frequency % group_size:
17 groups_needed = 0
18 break
19
20 # Calculate the number of groups needed for each number with its frequency
21 groups_needed += -(-frequency // (group_size + 1)) # Same as ceil division
22
23 # If groups are successfully calculated, return the result
24 if groups_needed:
25 return groups_needed
26
27 # If unable to calculate number of groups, return 0
28 return 0 # As per the original code (although this line is unnecessary since the function implicitly returns None if no return is hit)
29
30# Example usage:
31# solution = Solution()
32# print(solution.minGroupsForValidAssignment([1,2,3,3,3,4,4])) # Replace with the actual numbers list
33
1class Solution {
2
3 /**
4 * Calculates the minimum number of groups for a valid assignment based on the input array.
5 * The method counts the frequency of each number in the array and determines the smallest
6 * number of groups such that the frequency of the numbers is proportionally distributed.
7 * @param nums The input array containing numbers.
8 * @return The minimum number of groups required.
9 */
10 public int minGroupsForValidAssignment(int[] nums) {
11 // Create a map to store the frequency count of each unique number in nums
12 Map<Integer, Integer> frequencyCount = new HashMap<>();
13 for (int number : nums) {
14 // Increment the frequency count for each number
15 frequencyCount.merge(number, 1, Integer::sum);
16 }
17
18 // Initialize k as the number of elements in nums, the maximum possible frequency
19 int k = nums.length;
20
21 // Find the smallest value among the frequencies to identify the initial group size
22 for (int frequency : frequencyCount.values()) {
23 k = Math.min(k, frequency);
24 }
25
26 // Continuously try smaller values of k to optimize the number of groups
27 while (k > 0) {
28 int groupsNeeded = 0;
29 for (int frequency : frequencyCount.values()) {
30 // If the frequency divided by k leaves a remainder larger than the quotient,
31 // the current value of k isn't a valid group size, break and try a smaller k
32 if (frequency % k > frequency / k) {
33 groupsNeeded = 0;
34 break;
35 }
36 // Calculate the number of groups needed for the current value of k
37 groupsNeeded += (frequency + k - 1) / k;
38 }
39 // If the number of needed groups is greater than zero, we've found a valid grouping
40 if (groupsNeeded > 0) {
41 return groupsNeeded;
42 }
43 // Decrement k and try again for a smaller group size
44 k--;
45 }
46
47 // The code should never reach this point
48 return -1; // This line is just for the sake of completeness; logically, it'll always return from the loop
49 }
50}
51
1#include <vector>
2#include <unordered_map>
3#include <algorithm>
4using namespace std;
5
6class Solution {
7public:
8 int minGroupsForValidAssignment(vector<int>& nums) {
9 // Create a map to store the frequency of each number in nums
10 unordered_map<int, int> frequency;
11 for (int num : nums) {
12 frequency[num]++;
13 }
14
15 // Initialize k with an arbitrarily large number
16 int k = 1e9;
17 // Variable to store the minimum number of groups
18 int minGroups = k;
19
20 // Find the smallest frequency to initialize the minimum number of groups
21 for (auto& element : frequency) {
22 minGroups = min(minGroups, element.second);
23 }
24
25 // Decrease k until a valid configuration is found
26 while (k > 0) {
27 // Temporary variable to store count of groups for the current k
28 int tempGroupCount = 0;
29
30 // Check if the configuration is valid for current k
31 for (auto& element : frequency) {
32 int freq = element.second;
33
34 // If at any point we cannot satisfy the condition, break out
35 if (freq / k < freq % k) {
36 tempGroupCount = 0;
37 break;
38 }
39
40 // Otherwise, add the number of groups needed for this frequency
41 tempGroupCount += (freq + k - 1) / k; // Use integer ceiling division
42 }
43
44 // If we found a valid group configuration, return it
45 if (tempGroupCount > 0) {
46 return tempGroupCount;
47 }
48
49 // Decrement k and try again
50 --k;
51 }
52
53 // In case no valid configuration is found (which won't happen for valid input),
54 // just return zero as a safe fallback.
55 return 0;
56 }
57};
58
1function minGroupsForValidAssignment(nums: number[]): number {
2 // A map to count the frequency of each number in the input array
3 const frequencyMap: Map<number, number> = new Map();
4
5 // Populating the frequency map
6 for (const num of nums) {
7 frequencyMap.set(num, (frequencyMap.get(num) || 0) + 1);
8 }
9
10 // Trying to find the minimum group size starting from the smallest frequency
11 for (let groupSize = Math.min(...frequencyMap.values()); ; --groupSize) {
12 let groupsNeeded = 0; // Variable to hold the number of groups needed
13
14 // Calculate the number of groups needed for each unique number
15 for (const [_, frequency] of frequencyMap) {
16
17 // Calculate how many full groups can be formed with the current frequency
18 const fullGroups = (frequency / groupSize) | 0;
19
20 // If the number of full groups is less than the remainder, we cannot form a valid group
21 if (fullGroups < frequency % groupSize) {
22 groupsNeeded = 0; // Reset groups needed, as the current group size is invalid
23 break;
24 }
25
26 // Increase the count of total groups needed
27 groupsNeeded += Math.ceil(frequency / (groupSize + 1));
28 }
29
30 // If we found a valid number of groups, return it
31 if (groupsNeeded) {
32 return groupsNeeded;
33 }
34
35 // Note: there is no explicit break in this loop for when group size reaches 0
36 // The loop relies on eventually finding a valid group size before that happens
37 }
38}
39
Time and Space Complexity
Time Complexity
The time complexity is actually not O(n)
in general, it depends on both n
, the length of nums
, and also the range of unique values as well as their frequencies. The time complexity can be analyzed as follows:
Counter(nums)
has a complexity ofO(n)
since it goes through each element ofnums
.- The outer loop runs at most
min(cnt.values())
times which depends on the minimum frequency of a number innums
. - The inner loop runs
O(u)
times whereu
is the number of unique elements innums
because it iterates through all values in the counter.
So, the complexity is O(n + min(min(cnt.values()), n/u) * u)
which is O(n + min_count * u)
if we let min_count
be min(cnt.values())
.
Giving a final verdict on time complexity without constraints of input can lead to a misleading statement since it can vary. If min_count
is small, it could be close to linear but could also go up to O(n^2)
in the worst scenario when all elements are unique.
Space Complexity
The space complexity is O(n)
for the counter dictionary that stores up to n
unique values from nums
where n
is the length of nums
. No other significant space is used.
Learn more about how to find time and space complexity quickly using problem constraints.
In a binary min heap, the maximum element can be found in:
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