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This approach involves counting the frequency of each card number. Once we have the frequencies, the problem reduces to checking if there exists a number x greater than 1 that divides all these frequencies. This is efficiently achieved by calculating the GCD of the list of frequencies.
Time Complexity: O(n + m log m), where n is the length of the deck and m is the number of unique numbers.
Space Complexity: O(m), where m is the number of unique numbers due to the frequency array.
1import java.util.HashMap;
2
3public class Solution {
4 private int gcd(int a, int b) {
5 while (b != 0) {
6 int temp = b;
7 b = a % b;
8 a = temp;
9 }
10 return a;
11 }
12
13 public boolean hasGroupsSizeX(int[] deck) {
14 HashMap<Integer, Integer> freq = new HashMap<>();
15 for (int num : deck) {
16 freq.put(num, freq.getOrDefault(num, 0) + 1);
17 }
18
19 int X = 0;
20 for (int count : freq.values()) {
21 X = gcd(X, count);
22 }
23
24 return X > 1;
25 }
26
27 public static void main(String[] args) {
28 Solution sol = new Solution();
29 int[] deck = {1,2,3,4,4,3,2,1};
30 System.out.println(sol.hasGroupsSizeX(deck) ? "true" : "false");
31 }
32}This Java solution applies a similar strategy: it uses a HashMap for counting, then checks the GCD of the counts. The necessary calculations follow after considering all unique counts.
This approach also involves counting the frequency of each card in the deck. However, rather than immediately calculating the GCD, it sorts the frequency list and iteratively checks for the smallest divisor of these frequencies greater than 1.
Time Complexity: O(n + m√n), where n is the number of cards and m is the number of unique cards.
Space Complexity: O(m) for the frequency array.
1#include <vector>
#include <unordered_map>
#include <algorithm>
bool hasGroupsSizeX(std::vector<int>& deck) {
std::unordered_map<int, int> freq;
for (int num : deck) {
freq[num]++;
}
int min_freq = INT_MAX;
for (auto& entry : freq) {
if (entry.second < min_freq) {
min_freq = entry.second;
}
}
for (int x = 2; x <= min_freq; x++) {
bool canDivide = true;
for (auto& entry : freq) {
if (entry.second % x != 0) {
canDivide = false;
break;
}
}
if (canDivide) {
return true;
}
}
return false;
}
int main() {
std::vector<int> deck = {1,1,1,2,2,2,3,3};
std::cout << (hasGroupsSizeX(deck) ? "true" : "false") << std::endl;
return 0;
}Similarly to the previous implementation, this C++ function calculates frequency and attempts division by numbers ranging from 2 up to the smallest frequency, returning true if feasible.