LeetCode 202: Happy Number Solution

Problem Explanation

Explanation

To solve this problem, we can use a set to keep track of the numbers we have encountered while calculating the sum of the squares of digits. If we encounter a number that we have already seen, it means we are in a cycle and the number is not happy. If we reach 1, the number is happy. We keep iterating this process until we reach either 1 or a number we have seen before.

Algorithm:

Initialize an empty set to store seen numbers.
While the number is not 1 and the number is not in the set:
- Calculate the sum of the squares of digits.
- If the sum is 1, return true.
- Otherwise, add the sum to the set and update the number.
If the number is not 1, return false.

Time Complexity: O(log n) - Each step calculates the sum of squares of digits which involves log n digits in the number n. Space Complexity: O(log n) - The space used by the set to store seen numbers.

Solution Code

class Solution {
    public boolean isHappy(int n) {
        Set<Integer> seen = new HashSet<>();
        while (n != 1 && !seen.contains(n)) {
            seen.add(n);
            int sum = 0;
            while (n > 0) {
                int digit = n % 10;
                sum += digit * digit;
                n /= 10;
            }
            n = sum;
        }
        
        return n == 1;
    }
}

Frequently Asked Questions

How to solve LeetCode 202 (Happy Number)?

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What is the time complexity of LeetCode 202 (Happy Number)?

The time complexity for LeetCode 202 (Happy Number) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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