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The idea is to use a HashSet to track all numbers we've seen so far. If we encounter a number that we've seen before, it means we're in a cycle and the number is not happy. We keep replacing the number with the sum of the squares of its digits until we either reach 1 or the number starts repeating.
Time Complexity: O(log n), where n is the input number. Space Complexity: O(log n), as we store seen numbers.
1#include <unordered_set>
2
3bool isHappy(int n) {
4 auto getNext = [](int n) {
5 int totalSum = 0;
6 while (n > 0) {
7 int d = n % 10;
8 n /= 10;
9 totalSum += d * d;
10 }
11 return totalSum;
12 };
13 std::unordered_set<int> seen;
14 while (n != 1 && seen.find(n) == seen.end()) {
15 seen.insert(n);
16 n = getNext(n);
17 }
18 return n == 1;
19}The unordered_set is used to keep track of all previously seen numbers. If we encounter a number that we have seen before, it means we are in a cycle. We repeatedly replace the number with the sum of the squares of its digits until we reach 1 or a cycle is detected.
This approach utilizes Floyd's Cycle-Finding Algorithm (also known as Tortoise and Hare). Instead of using a hash set, we can use two pointers: a slow pointer and a fast pointer. The slow pointer moves one step at a time, whereas the fast pointer moves two steps at a time. If they meet, it means the sequence is cyclic.
Time Complexity: O(log n). Space Complexity: O(1), since no extra space is used apart from variables.
1 public bool IsHappy(int n) {
int slow = n, fast = GetNext(n);
while (fast != 1 && slow != fast) {
slow = GetNext(slow);
fast = GetNext(GetNext(fast));
}
return fast == 1;
}
private int GetNext(int n) {
int totalSum = 0;
while (n > 0) {
int d = n % 10;
n /= 10;
totalSum += d * d;
}
return totalSum;
}
}C# solution implements the cycle detection technique using two pointers. Slow pointer moves gradually, fast pointer leaps two steps. Their alignment indicates a loop unless 1 interrupts.