Reference: LeetCode
Difficulty: Easy

Problem

Write an algorithm to determine if a number is “happy”.

A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers.

Example: 

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Input: 19
Output: true
Explanation: 
12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1

Follow up: Use $O(1)$ space.

Analysis

Hash Set

Since you will get repeat numbers again and again, we can put all visited numbers in a hash set to know whether we can stop trying.

I list two ways of writing squareSum().

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public boolean isHappy(int n) {
  Set<Integer> seen = new HashSet<>();
  while (n != 1 && !seen.contains(n)) {
    seen.add(n);
    n = squareSum(n);
  }
  return n == 1;
}
// 1
private int squareSum(int n) {
  int sum = 0;
  do {
    int d = n % 10;
    sum += d * d;
    n /= 10;
  } while (n != 0);
}
// 2
private int squareSum(int n) {
  String s = "" + n;
  for (char ch : s.toCharArray()) {
    int digit = ch - '0';
    sum += digit * digit;
  }
  return sum;
}

Time: N/A
Space: N/A It depends on the input.

Cycle Detection

We can solve it without extra space.

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public boolean isHappy(int n) {
  int slow = n;
  int fast = n;
  do {
    slow = squareSum(slow);
    fast = squareSum(squareSum(fast));
  } while (slow != fast); // stops at slow == fast
  return slow == 1;
}

Time: N/A
Space: $O(1)$