# Solution: Happy Number - Grokking Coding Interview Patterns in Java

Need some help in understanding the time complexity for this

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Hello @Sai_Teja,

The space complexity of the given code is `O(1)`, which means it uses a constant amount of extra space that does not depend on the size of the input.

Letâ€™s break down the space complexity analysis:

1. The code uses a constant number of variables regardless of the input size.
2. The loop in the `sumOfSquaredDigits` function is not recursive, and it doesnâ€™t use additional space that scales with the input.
3. The code does not use any data structures (e.g., arrays, lists, or other dynamic data structures) that would grow with the input size.

I hope this clarifies your confusion regarding the space complexity of the provided solution. Feel free to share more feedbacks and suggestions, weâ€™d be happy to help. Thanks!

Happy learning!

@Dian_Us_Suqlain he asked for clarification on the Time Complexity not Space Complexity. I am also confused on this point and need clarification on the Time Complexity for this problem.

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Hello @Jesse_Martinez and @Sai_Teja,

Thank you for sharing this query and apologies for this oversight regarding the time/space confusion. Let me help you understand the time complexity of the provided solution for Happy Number.

The isHappyNumber() function

It uses two pointers, `slowPointer` and `fastPointer`, and a while loop. In the worst case, the loop will continue until either `fastPointer` becomes 1 (happy number) or the two pointers meet. The loop involves calls to the `sumOfSquaredDigits` function.

The time complexity of the `sumOfSquaredDigits` function is O(log n) as explained after this section.

In the while loop, both pointers move towards the result (either 1 or a cycle). The movement of the pointers depends on the number of digits in the intermediate results, which is influenced by the `sumOfSquaredDigits` function.

The intermediate results are the values obtained at each step of these movements. The number of digits in these intermediate results depends on the number of digits in the original number and the specific digit squaring operations performed in the `sumOfSquaredDigits` function.

For example, letâ€™s say we have the number 19:

• `sumOfSquaredDigits(19)` might result in 82.
• `sumOfSquaredDigits(82)` might result in 68.
• `sumOfSquaredDigits(68)` might result in 100.

In this example, the intermediate results are 82 and 68. The movement of the pointers through these intermediate results continues until either `fastPointer` becomes 1 (indicating a happy number) or the two pointers meet (indicating a cycle in the sequence).

The number of digits in these intermediate results influences the time complexity of the algorithm, as the `sumOfSquaredDigits` function has a time complexity proportional to the number of digits in the input number. Therefore, the overall time complexity of the `isHappyNumber` function can be expressed as O(log n).

The sumOfSquaredDigits() function

The `sumOfSquaredDigits` function has a while loop that iterates through each digit of the input number. In the worst case, the loop will iterate through all the digits in the number, so the time complexity of this function is O(log n), where â€ślogâ€ť denotes the logarithm base 10. The worst-case scenario, where the number is not a happy number and gets stuck in a cycle, also has a time complexity of O(log n) because it is bound by the number of digits operation.

I hope this explanation helps in understanding the time complexity of this problem. Please feel free to share any more suggestions and feedbacks. Weâ€™d be happy to help.

Happy learning