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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.
1using System.Collections.Generic;
2
3public class Solution {
4 public bool IsHappy(int n) {
5 HashSet<int> seen = new HashSet<int>();
6 while (n != 1 && !seen.Contains(n)) {
7 seen.Add(n);
8 n = GetNext(n);
9 }
10 return n == 1;
11 }
12
13 private int GetNext(int n) {
14 int totalSum = 0;
15 while (n > 0) {
16 int d = n % 10;
17 n /= 10;
18 totalSum += d * d;
19 }
20 return totalSum;
21 }
22}This C# solution uses a HashSet to memorize numbers as they are checked. If a number is revisited, a cycle is detected. The GetNext function calculates the sum of the squares of the digits.
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
int getNext(int n) {
int totalSum = 0;
while (n > 0) {
int d = n % 10;
n /= 10;
totalSum += d * d;
}
return totalSum;
}
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;
}In this C++ solution, two pointers are used. Slow moves one step, and fast moves two. If they meet, a cycle is confirmed unless one of them reaches 1 first.